text: Report a problem. Multiply both sides by θ, so for an angle θ radians. Paper 1 Non - Calculator - Tue 19 May 2020 Paper 2 Calculator - Thur 4 June 2020 Paper 3 Calculator - Mon 8 June 2020 All Exams are a 9am Start - If you had any additional help in your exams at school please make sure that it is known to ensure you get all the help you need to help you pass. the circumference. Give reasons for angle and length calculations involving tangents. Circle theorems - theorems Video 64 Practice Questions Textbook Exercise Circle theorems - examples Video 65 Practice Questions Textbook Exercise Circle theorems proof: right angle in a semi-circle Video 65a Practice Questions Circle theorems proof: angles at circumference and centre Video 65b Practice Questions. The Math Forum has a rich history as an online hub for the mathematics education community. The solution to 1 decimal place is w = 3. There are 20 pages in this question paper. That means that chord AB and chord GI both measure 10. Intersecting Secant Theorem Video 368b Practice Questions. binomial theorem expand integer powers of sums, e. In the above circle, OA is the perpendicular bisector of. - very nice stuff - share it -Circle Theorems Learn about Circle Theorems with this visible and informative poster. was iorn and that er and COORDINATE GEOMETRY OF THE CIRCLE EQuation of a Circle, Centre (0, 0) and Radius r A circle is a set of points (a locus) which are equidistant from a fixed point called the 'centre'. An explanation and demonstration of the theorem explaining the relationship of two intersecting chords in a circle. GEOMETRY Angles on intersecting lines, in triangles and quadrilaterals, and on parallel lines Understand acute, obtuse, reflex and right angles. Unit: A381/02. Full examples and questions for complete specification. If you need more space to complete your answer to any question, use additional answer sheets. The Rules of Circle Theorems These are completely FREE posters on The Rules of Circle Theorems. Leads to determining the square root of a number by construction. Instructions Use black ink or ball-point pen. Well, answer is quite simple, mental math is nothing but simple calculations done. Investigate the circle theorems and corollaries. The marks for each question are shown in brackets - use this as a guide as to how much time to spend on each question. What are Circle Theorems. Yr 11 Intersecting Chord Theorem Mini Test. Use geometric reasoning to answer problems giving detailed reasons. The solution to 1 decimal place is w = 3. This can be proved using congruent triangles. Intersecting Secant Theorem Video 368b Practice Questions. Online textbook and linked resources. Circle Theorems - angles on the same arc. Always show your workings. December 3, 2019. In the right-angled triangle, ON = x and NP = y. the circumference. Projector Friendly 5-a-day - March. Intersecting Chords Theorem The Intersecting Chords Theorem asserts the following very useful fact: Given a point P in the interior of a circle, pass two lines through P that intersect the circle in points A and D and, respectively, B and C. Recall that the line drawn from the centre of a circle to the midpoint of a chord is at right angles to the chord. Candidates are required to answer ALL questions. Well, answer is quite simple, mental math is nothing but simple calculations done. (We talk of the ALTERNATE segment, as being a segment on the OTHER side of a chord. Show Step-by-step Solutions. CIRCLE THEOREM WORKSHEET Exercise 1 - Introductory Questions Theorem 1: Angles Standing on the Same Arc (Chord) are Equal two circles intersecting at W md Z. Year 11 GCSE Higher Topics. 3 Reading scales 10. 1 Circle theorem rules; 1. Yr 11 Intersecting Chord Theorem Mini Test. Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - A Plus Topper Maths Solutions Problem Solving Skills Mathematics Let It Be This Or That Questions See You Math Mastering, Quantum Physics, Solution, Meant To Be, Math Equations. 4 Extra IGCSE circle theorems (intersecting chords, inside and outside circle) Similar stuff. Page Here is a grid with positive and negative numbers on the -ais and just positive numbers on the y-ais. Interestingly the wording (but not the substance) of the circle theorems content on the 1974 O level Syllabus differs notably from the 1957 Syllabus. Because AB is a chord bisected by diameter DE, two right triangles are created, as shown. We can use this idea to find a circle's center: draw a right angle from anywhere on the circle's circumference, then draw the. Yr 11 Intersecting Chord Theorem - MS. Some questions will use multiple-choice assessment. For example, angles on the same chord (are equal), angle at centre is equal to twice angle at circumferene angle on diameter is 90 o , opposite. 9 16 customer reviews. Maths4Everyone GCSE 9-1 Exam Question Practice (Trigonometry) FREE (74) Maths4Everyone Angles in Parallel Lines (Worksheets with Answers) FREE. This can be proved using congruent triangles. Some questions will use multiple-choice assessment. How to Find the Length of an Arc. The old legacy GCSE had 9-the chord rules are not in the new 1-9 spec (hence there are 7 rules now) (or at least the old Edexcel IGCSE had 9 rules; the Edexcel 1-9 has 7 rules-I can't guarantee this is the same for all Exam boards) GCSE Maths: Circle Theorems Circle Theorems show 10 more Maths question 0580/43. inverse 76. Yr 11 Intersecting Chord Theorem Mini Test. Presentation Summary : 21B: Circle theorems. Information for Candidates There are 20 pages in this question paper. A compass is an instrument used to draw circles or the parts of circles called arcs. Omission of essential working will result in loss of marks. Chapter 18 Sets. (a) Calculate the length of BE. 1 Circle theorem rules; 1. engaging contexts for questions on exponential growth/decay. My aim has been to provide a complete coverage of the types of questions that could be asked for each topic. The marks for each question are shown in brackets - use this as a guide as to how much time to spend on each question. Do exercises 14A–14L for practice. The solution to 1 decimal place is w = 3. A calculator will be allowed in this paper. 8 oe 2 M1 M1 for 7. So, A E ⋅ E C = D E ⋅ E B. Converse: The perpendicular bisector of a chord passes through the center of a circle. segment Alternate segment theorem. Chord-Chord Power Theorem: When two chords of a circle intersect, the product of the parts of one chord is equal to the product of the parts of the other chord. Mixed Pythagoras' Theorem Questions - Go Teach Maths: 1000s of free resources Number. By convention, the angle θ is measured from. INTRODUCTION 1. How are Congruent Chords related? How does the location of the vertex of an angle effect the formula for. Hours later on July 7th he passed away. to the x-axis. Paper 1 Non - Calculator - Tue 19 May 2020 Paper 2 Calculator - Thur 4 June 2020 Paper 3 Calculator - Mon 8 June 2020 All Exams are a 9am Start - If you had any additional help in your exams at school please make sure that it is known to ensure you get all the help you need to help you pass. Level 2 - Multi-step problems requiring one or more of the angle theorems. • the quadratic formula • similar figures • algebraic manipulation. A perpendicular line from a chord to the centre of the circle bisects the chord. 6 - Circle properties. 1 $\begingroup$ I'm talking about the theorems as laid out here: I'm a maths tutor and haven't found many problems my GCSE. The PDF contains both US and UK Versions of the posters. GCSE Tutorial Intersecting Chord Theorem Tangent Secant linked pair) Circle Theorems - GCSE Maths Higher - Duration: Intersecting Chords and segments intersecting outside the circle,. Qualification Accreditation Number: 500/8506/2 This is a linear specification: all assessments must be taken at the end of the course. Why not try drawing one yourself, measure the lengths and see what you get?. But they can sometimes change things up and confuse their users as well. Edexcel GCSE Mathematics (Linear) – 1MA0 CIRCLE THEOREMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. OCR GCSE in Applications of Mathematics: J925. Information for Candidates There are 20 pages in this question paper. Get an explanation of the Pythagorean Theorem, plus use these geometry worksheets to help you practice applying it. Presentation Summary : Learning goal questions. Edexcel GCSE Mathematics (Linear) – 1MA0 DIRECT & INVERSE PROPORTIONALITY Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. Intersecting Chord Theory Assessment. Pearson Edexcel - Sample Assessment Material, Specimen set 1 and Specimen set 2. Shapes; 08. Objective: On completion of the lesson the student will able to identify and use the angle sum of a triangle theorem to solve geometric problems. A n g l e B A E = 9 0 + 3 1 = 1 2 1 ° \text {Angle BAE } = 90 + 31 = 121 \degree Angle BAE = 9 0 + 3 1. Given: ABC CD bisects AB CD AB Prove: ACD BCD 1. In other words,. This book covers four of the core units that are common to BTEC National Engineering programmes. o Identify some critical steps of the process for solving practical geometry problems. Figure 6 shows a sketch of the circle C with centre N and equation (x 2)2 + (y + 1)2 = 169 4 (a) Write down the coordinates of N. The angle in a semicircle is 90 degrees. arithmetic 79. in a right angled triangle it is not too daunting to prove (or "provide a logical explanation") why the length of the altitude of a right angled triangle is the geometric mean of the two segments either side of the foot of the altitude (where it splits the hypotenuse). The Exterior Angle Theorem states that the exterior angle of a triangle: is smaller than either of the interior angle's measures. Making statements based on opinion; back them up with references or personal experience. A chord of a circle is the line segment joining two points on the circle. d a a a d a b a b d B A P Q 2. B A C D X Intersecting Chords Theorem This states that if two chords of a circle intersect as in the diagram below, AX × BX = CX × DX Proof We can prove this by forming two triangles as shown. Intersecting. 1 Introducing set notation. A calculator will be allowed in this paper. Students across 1000+ schools from Canada to Australia already using our programs Please have a look at www. Tracing paper may be used. GCSE Maths Specification and Awarding Body Information. TA and TB are tangents to the circle. For example, angles on the same chord (are equal), angle at centre is equal to twice angle at circumferene angle on diameter is 90 o , opposite. NCTM will continue to make many of the most popular parts of the Math Forum. What are they talking about? I know the intersecting chords theorem: given two chords AB and CD intersecting at X, AX * XB = CX * XD. Angle CAS = 58°. Given that AB and. the intersecting chord theorem. 6 cm (to 1 decimal place) In the diagram below, AB is the chord of a circle with centre O. (We talk of the ALTERNATE segment, as being a segment on the OTHER side of a chord. Information •• The total mark for this paper is 100. We can use this idea to find a circle's center: draw a right angle from anywhere on the circle's circumference, then draw the. N5 Maths Exam Questions by Topic. Circles and Angles 2. Watch the video, take notes, GCSE Tutorial Intersecting Chord Theorem Tangent Secant linked pair) This channel is managed by up and coming UK maths teachers. Anyway: Higher Tier, Section 4. Intersecting Chords Rule: (segment piece)×(segment piece) = (segment piece)×(segment piece) Theorem Proof: Theorem 2: If two secant […]. Linear programming. Triangle Formulas Sum of the interior angles of a triangle: 180° Area: …. CD CD Side 6. The converse of this theorem: Theorem 1b: If a line is drawn from the centre of a circle to the midpoint of a chord, then the line is perpendicular to the chord. December 3, 2019. Find the training resources you need for all your activities. CIRCLE THEOREM WORKSHEET Exercise 1 - Introductory Questions Theorem 1: Angles Standing on the Same Arc (Chord) are Equal two circles intersecting at W md Z. Intersecting Chord Theory Assessment. If then ]OAT =90o. Intersecting chords Tangent/secant PT2 8 compass points and 3 figure bearings Application of Pythagoras and Trig. Use these multi-sensory activities to help students learn how. Part of a curve. If inscribed angles of a circle intercept the same arc then they are congruent. We'll deal with them in groups. In the above circle, OA is the perpendicular bisector of. Once you've completed it or you need some support to complete the question, play the video as it talks you through step by step how to answer the question. We hope that the kids will also love the fun stuff and puzzles. Geometry Problem 1454. I can solve angle problems involving intersecting and parallel lines, polygons and circles including tangents. Because AB is a chord bisected by diameter DE, two right triangles are created, as shown. Viewed 164 times 1. Projector Friendly 5-a-day - March. Introducing the theorem SSM2f. Easily share your publications and get them in front of Issuu's. Question 1: Give some properties of tangents to a circle. is the measure of the smaller interior angle's measure. You must always remember the GCSE theorem that if a triangle is constructed within a circle with one side being a diameter of the circle, then it is a right angled triangle. You may use a calculator. The total mark for this paper is 100. arithmetic 79. Another way to look at it is this: if we took a segment of length r (the radius) and molded it onto the circle, the angle formed by the radii connecting the center of the circle to the endpoints of the arc would have a measure of 1 radian. The degree of angles will be the same. Understand and use the alternate segment theorem and internal and intersecting chord properties in circles. was iorn and that er and COORDINATE GEOMETRY OF THE CIRCLE EQuation of a Circle, Centre (0, 0) and Radius r A circle is a set of points (a locus) which are equidistant from a fixed point called the 'centre'. 4 Intersecting chords. N1b - Ordering decimals. The Rules of Circle Theorems These are completely FREE posters on The Rules of Circle Theorems. Key Point The equation of a circle of radius r and centre the origin is x 2+y = r. The questions cover all topics aligned to your GCSE maths revision requirement. For easily spotting this property of a circle, look out for a triangle with one of its …. 3 Versions of "angle at center" rule; 1. centre, radius, chord, diameter, Opposite angles of a cyclic circumference, tangent, arc, sector and quadrilateral add up to 180. website feedback. Similar triangles - know the properties of, use properties in questions, find missing lengths in, Solve mixed problems requiring the use of Pythagoras' theorem, trigonometry and similar shapes; Algebra 4. 130? Angles in a quadrilateral add up to 360. Firstly, recognise that since BD is a diameter, angle BAD is the angle in a semi-circle. Projector Friendly 5-a-day - June. Get the plugin now. CIRCLE THEOREM WORKSHEET Exercise 1 – Introductory Questions Theorem 1: Angles Standing on the Same Arc (Chord) are Equal Theorem 2: Angle at the Centre is Twice the Angle at the Circumference. Is that it?. For example, in the following diagram AP × PD = BP × PC Segments in circles Proof of a theorem on product of segments of chords in circles. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. AQA Modular specification reference AQA GCSE Maths Higher sets Student Book; Higher sets Teacher Guide Higher sets Teacher Guide Higher sets Practice Book A-A* Practice Book G2. These questions cover the topics in the first 11 chapters of this book, with a few Book 9 questions. The four segments we are talking about here all start at P, and some overlap each other along part of their length; PA overlaps PB, and PC overlaps PD. The engaging and attention grabbing math charts are guaranteed to improve understanding and help brighten school hallways and classrooms. The first one is as follows: A tangent line of a circle will always be perpendicular to the radius of that circle. Hence, PA/PD= PC/PB ⇒ PA × PB = PC × PD. The line SAT is the tangent at A to the circle. The sector is folded up to make a cone. This support material booklet is designed to accompany the OCR GCSE Applications of Mathematics specification for teaching from September 2010. Given a | b If m. Intersecting chords are lines that cross over each other within a circle. In other words,. Ptolemy used the theorem to create his table of chords, a trigonometric table that was applied to astronomy. Apex Learning 2014 Page of 12 –. Find the value and give reasons for missing angles on diagrams using circle theorems and triangle properties in circles. I can solve angle problems involving intersecting and parallel lines, polygons and circles including tangents. … and 6 more awesome questions! Check them out by. By convention, the angle θ is measured from. SVT is a to the circle at V, VWX and vzy are straight lines, TVY = 78 and six = 51 1) Calculate the size of. Volume and Areas Math Poster Set - Mathe Ideen 2020 Daydream Education's Math and Numeracy posters are great learning and teaching tools. In a reflex action sensory neurones carry information from a receptor to the CNS, where it is transferred over the synapse from the end of the synaptic knob (the end of the sensory neurone) to the dendrites of the relay neurone. A chord is a straight line joining 2 points on the circumference of a circle. Also, being aware that taking both U2 and U1 exams in Y10 is a time-challenge, we suggest that GCSE teaching begins in the summer term of Y9. Know what each circle theorem states Inscribed Angles - MathHelp. Information for Candidates There are 20 pages in this question paper. 4 Extra IGCSE circle theorems (intersecting chords, inside and outside circle) Similar stuff. One chord is cut into two line segments A and B. Edexcel International GCSE (9-1) Mathematics A Student Book 2: print and ebook bundle theorem 89. Year 11 GCSE probability Q - solution. This worksheet was automatically generated by the DrFrostMaths Homework Platform: students can practice this set of questions interactively by going to. Level 1 - Angles which can be found using one of the angle theorems. X is the mid-point of AB. If the rate-resource button on this page does not work, then go to your. 4 Extra IGCSE circle theorems. P40612A ©2012 Pearson Education Ltd. Online textbook and linked resources. com, find free presentations research about Intersecting Chords PPT. Mark Dawes http://www. Don't spend too long on one question. For example, in the following diagram AP × PD = BP × PC Segments in circles Proof of a theorem on product of segments of chords in circles. We hope that the kids will also love the fun stuff and puzzles. Tangents from an external point are equal. Aterro Recycling P. Answer all questions. the circumference. Exponential Functions I1. (We talk of the ALTERNATE segment, as being a segment on the OTHER side of a chord. Used to create the law of Cosines, showing students these laws don't simply appear out of thin air! catscardiganscalculus: themathkid: Excellent use of the chord theorem!. Lots of questions and examples at grades 7, 8 and 9; Edexcel Higher 1MA0/1H November 2014 Video help; Nice AS Level Maths Question. GCSE QUESTIONS. The total mark for this paper is 100. the measure of an angle formed by two chords that intersect inside a circle is equal to half the sum of the measures of the intercepted arcs theorem 9-9: angle created by two chords the measure of an angle formed by two secants, two tangents or a secant and a tangent drawn from a point outside a circle is equal to half the difference of the. The NRICH Project aims to enrich the mathematical experiences of all learners. Author: Mrs Moule, GreenMaths. A compass is an instrument used to draw circles or the parts of circles called arcs. In diagram 1, the x is half the sum of the measure of the intercepted arcs ( ABC. More formally: When two secant lines AB and CD intersect outside the circle at a point P, then. It is a continuation of our Free Poster on The Circle which can be found hereThese two posters, which come in one document, show all 8 theorems that are important for students to learn. 6 cm (to 1 decimal place) In the diagram below, AB is the chord of a circle with centre O. Projector Friendly 5-a-day - March. students should know that if f(x) = 0 when x = b a, then (ax – b) is a factor of f(x). Geometry Worksheets Geometry Worksheets for Practice and Study. Are there any relatively easy problems that make use of the Intersecting Chord Theorem? And same question for Intersecting Secants Theorem. Those pairs are named first. Introducing the theorem SSM2f. A perpendicular line drawn to a chord from the centre of the circle bisects the chord. The circle theorems; 1. Theorem 7 The angle between a tangent and a chord drawn from the point of contact is equal to any angle in the alternate segment. With regular practice, you will soon be able to handle circle theorem questions with ease and fetch excellent grades. 1 INTRODUCTION You must have measured the angles between two straight lines, let us now study the angles made by arcs and chords in a circle and a cyclic quadrilateral. chord BOD, BAD & DBO E: Angles in the same segment are equal Angle at the centre is twice the angle at the circumference. (b) Calculate the size of angle MED. Calculate the size of angle OAB. Crossing Chords Property & Proof Start. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser You may use tracing paper if needed Guidance 1. Proofs for circle theorems. Step 1: Create the problem Draw a circle, mark its centre and draw a diameter through the centre. 45 minute exam 35 % online, on demand exam. the measure of an angle formed by two chords that intersect inside a circle is equal to half the sum of the measures of the intercepted arcs theorem 9-9: angle created by two chords the measure of an angle formed by two secants, two tangents or a secant and a tangent drawn from a point outside a circle is equal to half the difference of the. advanced topics. ks2 sats questions; gcse revision sheets; gcse questions by topic; past paper solutions; 2d shapes > revision > gcse questions. Other resources by this author. GEOMETRY Angles on intersecting lines, in triangles and quadrilaterals, and on parallel lines Understand acute, obtuse, reflex and right angles. 6 - Circle properties. Chapter 18 Sets. Answer ALL the questions in the spaces provided in this question paper. Always show your workings. Gives practice in constructions and application of the intersecting chords theorem. Review Pythagoras and Trigonometry formulae and rearrange formula where the subject is a power or an inverse function. December 3, 2019. Following is a neat discovery of an alternative proof of the Pythagorean Theorem resulting from the multiple solutions to the Circle and Square problem. This can be proved using congruent triangles. ks2 sats questions; gcse revision sheets; gcse questions by topic; past paper solutions; 2d shapes > revision > gcse questions. Identify the basic parts of a circle 2. Chapter 18 Sets. Hours later on July 7th he passed away. GCSE subject criteria set out the knowledge, understanding, skills and assessment objectives common to all GCSE specifications in a given subject. The angle at the centre of the circle is twice the angle on the circumference Angles in the same segment are equal. 2 Angle properties inside a circle. Read about me, We can solve for y by the power of the point/intersecting chords theorem, which states the product of the two segments are equal, so: wx = yz 2(6) = y(3) How To Solve A Tricky GCSE Probability Problem. Huh? There is no further elaboration. To draw a circle (or arc) with a compass:. It will always form a right angle (90°) with the radius. Note that a compass is also called a pair of compasses. Circle theorem problems keyword after analyzing the system lists the list of keywords related and the list of websites with related content, in addition you can see which keywords most interested customers on the this website. 2 GCSE Physics 3 Specimen Paper Higher Tier V1. T he second chord theorem, a generalization of the first, is stated this way in Heath's edition of Apollonius, which jumbles up Book III, Propositions 16-23: The rectangles under the segments of intersecting chords in fixed directions are as the squares of the parallel semi-diameters. Edexcel GCSE Mathematics (Linear) - 1MA0 CIRCLE THEOREMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. d a a a d a b a b d B A P Q 2. It will help you The exams are GCSE style so will also contain topics from years 7 to 9. The opposite angles are mirror images of one another. e 144: 1+4+4=9 which is a multiple of 3, and 44 is a multiple of 4, so 144 is a multiple of 12. Question 1: Give some properties of tangents to a circle. Question 9. Tangents to Circles Date_____ Period____ Determine if line AB is tangent to the circle. GCSE Tutorial Intersecting Chord Theorem Tangent Secant linked pair) Arc-Chord-Diameter Three theorems with applications / 10. If two chords intersect within a circle, the product of the measures of the segments of one will be equal to the product of the measures of the segments of the other. The Pythagorean Theorem can be used in any real life scenario that involves a right triangle having two sides with known lengths. Edexcel Specification Alignment GCSE 2015 / 2016 Exams Higher Expectation Topic Activity GM j Understand and construct geometrical proofs using circle theorems Geometry - Shape & Angle Properties Circle Theorem GM k Use 2-D representations of 3-D shapes Geometry - Shape & Angle Properties Circle Terms Rotations: Coordinate Plane. • the midpoint theorem • intersecting chord theorem. In the above circle, if the radius OB is perpendicular to the chord PQ then PA = AQ. binomial theorem expand integer powers of sums, e. In triangles OXA and OXB: OA = OB (radii) AX = XB ( is the mid-point) OX is common Therefore Triangles and are congruent (SSS) Therefore Angle OXA = angle OXB = 90°. Intersecting chords Tangent/secant PT2 8 compass points and 3 figure bearings Application of Pythagoras and Trig. Know and apply Pythagoras’ theorem to find lengths in right-angled triangles in 2D figures. Plot the shape with the following co-ordinates: (,) (5,) (,) (,) (,) (,) (,5) (,) (6,) (,5) Reflect this shape in the y-ais. The last thing you want is to worry about it too much. Projector Friendly 5-a-day - May. Alternate Exterior Angles. Angle in a Semi-circle 1. There are two questions where students will need to be provided with graphs to work on. Recall that the line drawn from the centre of a circle to the midpoint of a chord is at right angles to the chord. One Minute Gcse Maths Courses Circle Theorems 4 30 Gcse Doctor. I'm only in year ten. Investigate the circle theorems and corollaries. Pythagoras theorem Worksheet Pdf Unique Pythagorean S theorem Worksheets Looking for a Advanced Geometry Worksheets. the Factor Theorem and the Remainder Theorem. STICKY! 9-1 Exam Questions by topic … Foundation - VERSION 3. Yr 11 Intersecting Chord Theorem - MS. Edexcel GCSE Mathematics (Linear) - 1MA0 CIRCLE THEOREMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. You may use a calculator. Students across 1000+ schools from Canada to Australia already using our programs Please have a look at www. Then AL, BM and CN are concurrent if and only if 127. binomial theorem expand integer powers of sums, e. Year 11 GCSE probability Q – solution. Free essays, homework help, flashcards, research papers, book reports, term papers, history, science, politics. Use the diameter to form one side of a triangle. Intersecting chords. 20 – 30% of questions on the Higher tier and 30 – 40% of questions on the Foundation tier. 1) 16 12 8 B A Tangent 2) 6. One chord is cut into two line segments A and B. Circle Geometry Worksheets Grade 10. Level 3 - Randomised questions, some requiring finding more than one angle in a diagram. Angle in a Semi-circle 1. Proof: Apply Menelaos to this points. Intersecting Chords Theorem. You must always remember the GCSE theorem that if a triangle is constructed within a circle with one side being a diameter of the circle, then it is a right angled triangle. 3 Intersecting Chord Theorem, Tangent-secant theorem Theorem 5 (Intersecting Chord Theorem) If two chords of a circle intersect in the interior of a circle, thus determine two segments in each chord, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord. An incorrect intersecting chord theorem statement was commonly seen and thus meant that many candidates scored 0 marks. OM is the perpendicular from the centre to the chord. The marks for parts of questions are shown in round brackets: e. • Answer all questions. In the above circle, OA is the perpendicular bisector of. GCSE Math - Free download as PDF File (. pdf), Text File (. 1 $\begingroup$ I'm talking about the theorems as laid out here: I'm a maths tutor and haven't found many problems my GCSE. Because AB is a chord bisected by diameter DE, two right triangles are created, as shown. Made by expert teachers for Save My Exams. The question is asking for angle CBA, and now we know the other two angles in the triangle we can use the fact that angles in a triangle add. Following is a neat discovery of an alternative proof of the Pythagorean Theorem resulting from the multiple solutions to the Circle and Square problem. Intersecting Secant Theorem Video 368b Practice Questions. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. We'll deal with them in groups. Recall that the line drawn from the centre of a circle to the midpoint of a chord is at right angles to the chord. Year 11 GCSE Higher Topics. GCSE - AQA Link Paired Pilot Methods 2 - Higher Practice Hard Questions from Maths247 channel on YouTube, only attempt these if you have run out of revision to do on the easier stuff. We have Advanced Geometry Worksheets and the other about Benderos Printable Math it free. Apex Learning 2014 Page of 12 –. These exercises are curated for students of grade 4 through high school. I don't know anymore than you about the paper. uk Circle Theorems (H) - Version 2 January 2016 Circle Theorems (H) A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. I can solve angle problems involving intersecting and parallel lines, polygons and circles including tangents. Circle Theorem 7 link to dynamic page Previous Next > Alternate segment theorem: The angle (α) between the tangent and the chord at the point of contact (D) is equal to the angle (β) in the alternate segment*. Intersecting Chords Theorem: Given a point P in the interior of a circle, pass two lines through P that intersect the circle in points A and D and, respectively, B and C. Showing top 8 worksheets in the category - Gcse Number Summary. Real Life Applications Of Circle Theorems. one with no contradictions) can be developed if the angles are less than 180 and another type of geometry can be developed if the angles are greater than 180 and the lines are still parallel! This issue is known as the problem of Euclid's 5th postulate and it has a very interesting history. CD CD Side 6. 130? Angles in a quadrilateral add up to 360. They provide the framework within which awarding organisations create the detail of their specifications, so ensuring progression from key stage 3 national curriculum requirements and the. It will always form a right angle (90°) with the radius. Candidates are required to answer ALL questions. The intersecting chords theorem or just The chord theorem is a statement in elementary geometry that describes a relation of the four line segments created by two intersecting chords within a circle. A compass is an instrument used to draw circles or the parts of circles called arcs. Each student will work one column of 9 problems. Chapter 3 The Midpoint and Intercept theorems 4 M3. Mark Dawes http://www. The central nervous system (CNS) is made up of the spinal chord and the brain. Information •• The total mark for this paper is 100. The total mark for this paper is 100. Angle inscribed in semicircle is 90. post-1221594399232024299 2019-10. It involved setting up two intersecting right triangles; by applying Menelaus' theorem it was possible to solve one of the six sides, but only if the other five sides were known. Figure 6 shows a sketch of the circle C with centre N and equation (x 2)2 + (y + 1)2 = 169 4 (a) Write down the coordinates of N. 6 - Circle properties. Intersecting Chord Theory Assessment. segment Alternate segment theorem. History Your child will be studying the following: Understand how Hitler became Fuhrer Understand the events and outcomes of the Reichstag Fire and the Enabling Act Understand the causes of the Night of the Long Knives and the impact on. y 6 Write down the co-ordinates of the new shape (the 'image') New Shape co-ordinates: Be careful with those negatives!. When two chords intersect each other inside a circle, the products of their segments are equal. Explanation:. Stay safe and healthy. N5 Maths Exam Questions by Topic. Remember: the real M01_EDHM_RG_GCSE_8090_001-010. This specification is aimed at teachers who plan the work holistically, around the content, but not necessarily in any pre-defined order, to support the learning style of their students. Theorem 1a: If a line is drawn from the centre of a circle perpendicular to a chord, then it bisects the chord. - The fact that the gradient of a line is the negative reciprocal of the line perpendicular to it. Issuu is a digital publishing platform that makes it simple to publish magazines, catalogs, newspapers, books, and more online. and a chord. I'm sure someone has seen this before, as there are literally 100s of unique proofs of the Pythagorean Theorem, but this one was new to me. All parts were well answered. The circle theorems; 1. The last thing you want is to worry about it too much. Learn about and revise the different angle properties of circles described by different circle theorems with this BBC Bitesize GCSE Maths OCR study guide. 6 Using a calculator. Attempt every question. The intersecting chords theorem or just The chord theorem is a statement in elementary geometry that describes a relation of the four line segments created by two intersecting chords within a circle. and \overparen {DFG} ) Note: This theorem applies to the angles and arcs of chords that intersect anywhere within the circle. Circle theorems 6. Projector Friendly 5-a-day - April. Author: Created by HolyheadSchool. The old legacy GCSE had 9-the chord rules are not in the new 1-9 spec (hence there are 7 rules now) (or at least the old Edexcel IGCSE had 9 rules; the Edexcel 1-9 has 7 rules-I can't guarantee this is the same for all Exam boards) GCSE Maths: Circle Theorems Circle Theorems show 10 more Maths question 0580/43. Find the locus of the centre of a circle of radius r touching externally a circle of radius R. If you get stumped while working on a problem and can’t come up with a formula, this is the place to look. The below are all the questions from: AQA - Sample Assessment Material and Practice set 1 and 2. birdvilleschools. Projector Friendly 5-a-day - May. 221 · 20 comments Gonna be in Year 11 on September, I'm dying rn. 1 a) 16 b)1 64 c)9 256 d)27 256 2. Ask Question Asked 1 month ago. 71 × 104 = 7384; 50 × 148 = 7400; Very close! If we measured perfectly the results would be equal. Intersecting chords theorem HCF and LCM from prime factors Sine rule 3D trigonometry and Pythagoras' theorem Similar shapes Vectors Question 1 Both parts of this question were well answered. segment Alternate segment theorem. Theorem: The measure of the angle formed by 2 chords that intersect inside the circle is 1 2 the sum of the chords' intercepted arcs. - How to form the equation of a straight line. If inscribed angles of a circle intercept the same arc then they are congruent. Linear B 4365 Higher. GCSE Maths - Circle Theorems - Educational Poster - Size A2 - Tiger Moon. (they need to be prepared to. 2D SHAPES > REVISION > GCSE QUESTIONS. 1 Circle theorem rules; 1. There was also a question in the 1957 paper relating to the Power of a Point Theorem, which now features in iGCSE (specifically the Intersecting Chords Theorem) but not GCSE. Search is smart and instant. Projector Friendly 5-a-day - June. Area and Perimeter [GCSE Questions] Area and Perimeter [Solutions] Angles in Polygons [GCSE Questions] Angles in Polygons [Solutions] Circle Theorems [GCSE Questions] Circle Theorems [Solutions] Intersecting Chords [GCSE Questions] Intersecting Chords [Solutions] Sectors [GCSE Questions. [1] The chord AB of C is parallel to the x-axis, lies below the x-axis and is of length 12 units as shown in Figure 6. Information •• The total mark for this paper is 100. Dear r/GCSE students, this is the type of content you can enjoy on r/6thform, we look forward to you joining us soon. A tangent meets or touches a circle only at one point, whereas the tangent line can meet a curve at more than one point, as the diagrams below illustrate. It is a continuation of our Free Poster on The Circle which can be found hereThese two posters, which come in one document, show all 8 theorems that are important for students to learn. Press J to jump to the feed. Cluster path (and word type count) Words (most frequent) ^00000 (827) to ^000010 (1832) 's ^0000110 (512) and. HIGHER CHAPTER 15 PYTHAGORAS’ THEOREM Time: 4–6 hours. INTRODUCTION 1. The diagrams show that: a) The angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. Angle in a Semi-circle 2. So, we can use the circle theorem that tells us the angle in a semi-circle is a right-angle to deduce that \text{angle BAD } = 90\degree. Now, you may have an idea of some of the basic circle theorem included in the GCSE curriculum. 3 Intersecting Chords and segments intersecting outside the. sets, logic, proofs. Pythagoras Theorem can be used to find the length of the radius 𝑟𝑟: Using Pythagoras Theorem: 𝑟𝑟2= (𝑟𝑟−2)2+ 32 (1 mark). Alternate Exterior Angles. An angle formed by an intersecting tangent and chord has its vertex "on" the circle. Objectives Get Ready Architects study the properties of circles so. It is a little easier to see this in the diagram on the right. In the circle below, the chord segments have the following lengths: D = 8, C = 3, A = 6. AQA Linear specification reference AQA GCSE Maths Higher sets Student Book; Higher sets Teacher Guide Higher sets Teacher Guide Higher sets Practice Book A-A* Practice Book G1. The marks for each question are shown in brackets - use this as a guide as to how much time to spend on each question. Remember: the real M01_EDHM_RG_GCSE_8090_001-010. Circle Theorems - angles on the same arc. We have Advanced Geometry Worksheets and the other about Benderos Printable Math it free. Common Chord to two Intersecting circle - Index : Chord. The degree of angles will be the same. Theorem: A radius or diameter that is perpendicular to a chord divides the chord into two equal parts and vice versa. Intersecting Chords Theorem The Intersecting Chords Theorem asserts the following very useful fact: Given a point P in the interior of a circle, pass two lines through P that intersect the circle in points A and D and, respectively, B and C. Students across 1000+ schools from Canada to Australia already using our programs Please have a look at www. Do Mixed Exercise 14 for consolidation. gcse exam questions on circle. birdvilleschools. You may change the number of significant figures displayed by changing the number in the box above. It will help you The exams are GCSE style so will also contain topics from years 7 to 9. Intersecting. In the right-angled triangle, ON = x and NP = y. Find the training resources you need for all your activities. It is part of a suite of GCSE qualifications offered by Edexcel. 3 Intersecting Chords and segments intersecting outside the. Intersecting Chord Theory Assessment. They provide the framework within which the awarding organisation creates the detail of the specification. Theorems Related to Chords of Circle; Cyclic Quadrilateral and Intersecting / Non-intersecting Circles; Learn more about Cyclic Quadrilateral here in detail. 95 Intersecting chords 96 Circle theorems 97 Vectors 98 Vector proof 99 Problem-solving practice 1 The questions in Now try this have been written to help you practise every topic in the book. GCSE Mathematics - Numeracy. uk Circle Theorems (H) - Version 2 January 2016 Circle Theorems (H) A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. So in this example, AD * AC = AB^2. This acclaimed series guarantees total coverage of the new Edexcel (A). Angles in a Circle and Cyclic Quadrilateral 19. I can plot and shade loci. Alternate segment theorem. The measure of an angle formed by two chords that intersect inside a circle is equal to half the sum of the measures of the intercepted arcs. Course Contents of Maths. The total mark for this paper is 100. Intersecting Chords Theorem. In the above circle, if the radius OB is perpendicular to the chord PQ then PA = AQ. Take your time, use a pencil and paper to help. Intersecting chords. 2 Converting between metric units 10. I don't know anymore than you about the paper. Intersecting chords PT Tangent/secant Include plane, axis and point Symmetry 8 compass points and 3. Opposite angles in a cyclic quadrilateral sum to 180° 5. Information •• The total mark for this paper is 100. It is part of a suite of GCSE qualifications offered by Edexcel. Explanation:. Presentation Summary : Learning goal questions. Edexcel GCSE Mathematics (Linear) – 1MA0 CIRCLE THEOREMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. They provide the framework within which awarding organisations create the detail of their specifications, so ensuring progression from key stage 3 national curriculum requirements and the. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser You may use tracing paper if needed Guidance 1. chord and segment. In diagram 1, the x is half the sum of the measure of the intercepted arcs ( ABC. Learn about and revise the different angle properties of circles described by different circle theorems with this BBC Bitesize GCSE Maths OCR study guide. Includes full solutions and score reporting. Calculate the size of angle OAB. In the right-angled triangle, ON = x and NP = y. Then ∠RST = 90 (Theorem 3, angle subtended by a diameter) Also ∠RTQ = 90 (Theorem 5, tangent is perpendicular to radius) Hence x + y = 90. Unit: A381/02. The figure referenced is below: If two chords intersect inside the circle, then they cut each other in such a way that the product of the lengths of the parts is the same for the two chords - that is,. It involves a circle theorem that crops up in IGCSE but not at GCSE, relating to the lengths of the intersections of chords. 2 Describing circle theorems;. Quadratic Formula. If you get stumped while working on a problem and can’t come up with a formula, this is the place to look. If two chords intersect within a circle, the product of the measures of the segments of one will be equal to the product of the measures of the segments of the other. CIRCLE THEOREM WORKSHEET Exercise 1 – Introductory Questions Theorem 1: Angles Standing on the Same Arc (Chord) are Equal Theorem 2: Angle at the Centre is Twice the Angle at the Circumference. Theorem: The measure of the angle formed by 2 chords that intersect inside the circle is 1 2 the sum of the chords' intercepted arcs. If a tangent segment and a secant segment. 3 Intersecting Chord Theorem, Tangent-secant theorem Theorem 5 (Intersecting Chord Theorem) If two chords of a circle intersect in the interior of a circle, thus determine two segments in each chord, the product of the lengths of the segments of one chord equals the product of the lengths of the segments of the other chord. Mar 12, 2018 - Explore sunnahh's board "Math Edexcel" on Pinterest. Learn vocabulary, terms, and more with flashcards, games, and other study tools. We have Advanced Geometry Worksheets and the other about Benderos Printable Math it free. A pair of angles opposite each other, formed by two intersecting straight lines that form an "X"-like shape, are called vertical angles or opposite angles. N1b - Ordering decimals. You can work out the length of an arc by calculating what fraction the angle is of the 360 degrees for a full circle. Dear r/GCSE students, this is the type of content you can enjoy on r/6thform, we look forward to you joining us soon. Everything About Circle Theorems - In 3 minutes! This is a graphic, simple and memorable way to remember the difference from a chord or a tangent or a segments and sectors! Circle Theorems (CXC CSEC and GCSE Math Revision) At the end of this lesson, students should be able to: 1. AB and DE are intersecting chords, so. 6 - Circle properties. CIRCLE THEOREM WORKSHEET Exercise 1 – Introductory Questions Theorem 1: Angles Standing on the Same Arc (Chord) are Equal Theorem 2: Angle at the Centre is Twice the Angle at the Circumference. Objectives Get Ready Architects study the properties of circles so. ( Gradient of straight line graphs revision) If any of this feels unfamiliar to you. Maths revision video and notes on the topic of Circle Theorems. A perpendicular line from a chord to the centre of the circle bisects the chord. It is a healthy,rewarding and a very simple exercise to map out your Aspects of Life and understand how you are dividing your life. Understand a definition of Euclid's Intersecting Chords Theorem. JustMaths was born from the passion and spirit of three full-time teachers who have a genuine belief in the power of collaboration and sharing ideas and best practice. Students learn the following theorems related to chords, secants, and tangents. Edexcel GCSE Mathematics (Linear) - 1MA0 CIRCLE THEOREMS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. There are two questions where students will need to be provided with graphs to work on. 3 Versions of "angle at center" rule; 1. Thus, using the theorem of Pythagoras, x2 +y 2= r , and this is the equation of a circle of radius r whose centre is the origin O(0,0). A debt of gratitude is owed to the dedicated staff who created and maintained the top math education content and community forums that made up the Math Forum since its inception. Intersecting chords. The total mark for this paper is 100. intersecting-chords-solutions. Given that AB and. It involved setting up two intersecting right triangles; by applying Menelaus' theorem it was possible to solve one of the six sides, but only if the other five sides were known. Solve For The Radius - Fun 1970s Math Contest Problem. Parts of a Circle - Diameter, Chord, Radius, Arc, Tangent, Intersecting Circles, Internal and External Tangents with examples, step by step solutions, definitions, Geogebra apps and worksheet - themathsteacher. Use this glossary to find definitions for common math terms. PPT – Linear GCSE Mathematics 4365 plus Level 2 Certificate Further Mathematics 8360 Route Map PowerPoint presentation | free to download - id: 6a24f6-ZDMzN. uk Simple & Compound Interest & Depreciation (F) - Version 3 January 2016 Simple & Compound Interest & Depreciation (F) A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. Intersecting chords Tangent/secant PT2 8 compass points and 3 figure bearings Application of Pythagoras and Trig. 1 Calculate the length of a missing side in a right-angled triangle using Pythagoras’ theorem. A tangent meets or touches a circle only at one point, whereas the tangent line can meet a curve at more than one point, as the diagrams below illustrate. Chapter 3 The Midpoint and Intercept theorems 4 M3. The first one is as follows: A tangent line of a circle will always be perpendicular to the radius of that circle. Draw diagrams in pencil. (a) Calculate the length of BE. Showing top 8 worksheets in the category - Intersecting Chord Theorem. Using the alternate segment theorem GCSE Subjects GCSE. If a tangent segment and a secant segment. Each chord is cut into two segments at the point of where they intersect. Understand a definition of Euclid's Intersecting Chords Theorem. Quadratic Formula. AQA Modular specification reference AQA GCSE Maths Higher sets Student Book; Higher sets Teacher Guide Higher sets Teacher Guide Higher sets Practice Book A-A* Practice Book G2. (2x y2 3yx 2)4, nd a power for which expansion full ls given conditions 10. (12 x 18) + (8 x 16. 1 Rationale. Now, the chord is split into two equal pieces, and angle AOB is bisected. IGCSE 9-1 Exam Question Practice (Intersecting Chords) 4. Circle Theorems - angles on the same arc. I've got my higher tomorrow as well. Page Here is a grid with positive and negative numbers on the -ais and just positive numbers on the y-ais. introduction to logarithms know and use the de nition of a logarithm, change logarithmic equation into an exponential one and reversly, evaluate exact values of logarithms without a calculator, e.


text:
## A strong convergence theorem for contraction semigroups in Banach spaces.(English)Zbl 1095.47016

T. Suzuki in [Proc. Am. Math. Soc. 131, No. 7, 2133–2136 (2003; Zbl 1031.47038)] established a theorem on the strong convergence to common fixed points of nonexpansive semigroups in Hilbert spaces. In the paper under review, the author generalizes Suzuki’s result for uniformly convex Banach spaces with a weakly continuous duality map.

### MSC:

 47H20 Semigroups of nonlinear operators 47H10 Fixed-point theorems 47H09 Contraction-type mappings, nonexpansive mappings, $$A$$-proper mappings, etc.

Zbl 1031.47038
Full Text:

### References:

 [1] Goebel, Uniform convexity, hyperbolic geometry, and nonexpansive mappings (1984) · Zbl 0537.46001 [2] DOI: 10.1007/BF02764907 · Zbl 0423.47024 [3] DOI: 10.1007/BF01109805 · Zbl 0149.36301 [4] Browder, Fixed point theorems for noncompact mappings in Hilbert space 53 pp 1272– (1965) · Zbl 0125.35801 [5] DOI: 10.1016/0362-546X(94)90116-3 · Zbl 0812.47058 [6] DOI: 10.1112/S0024610702003332 · Zbl 1013.47032 [7] DOI: 10.1090/S0002-9939-02-06844-2 · Zbl 1031.47038 [8] DOI: 10.1016/S0362-546X(97)00682-2 · Zbl 0935.47039 [9] DOI: 10.1016/0022-247X(80)90323-6 · Zbl 0437.47047 [10] DOI: 10.1080/01630569808816821 · Zbl 0908.47057
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.

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# session_register()问题

antidestiny 2004-03-31 07:10:33

<?php
session_register('aaa');

phpinfo();
?>

Warning: session_register(): Cannot send session cookie - headers already sent by (output started at C:\Program Files\PHP Expert Editor\php160.tmp:3) in C:\Program Files\PHP Expert Editor\php160.tmp on line 4

Warning: session_register(): Cannot send session cache limiter - headers already sent (output started at C:\Program Files\PHP Expert Editor\php160.tmp:3) in C:\Program Files\PHP Expert Editor\php160.tmp on line 4

php.ini中的session.save_path也设置了,不知道怎么还是不行????
...全文
693 点赞 收藏 8

8 条回复

cloudchen 2004-04-01

xuzuning 2004-04-01
1、session_register前一定要有session_start()
2、确保被session_register的变量是全局变量
3、session_register只适用于php4.0.x

ashchen 2004-04-01

ashchen 2004-04-01

If you want your script to work regardless of register_globals, you need to instead use the \$_SESSION array as \$_SESSION entries are automatically registered. If your script uses session_register(), it will not work in environments where the PHP directive register_globals is disabled.

register_globals: important note:
Since PHP 4.2.0, the default value for the PHP directive register_globals is off. The PHP community encourages all to not rely on this directive but instead use other means, such as the superglobals.

Caution:
This registers a global variable. If you want to register a session variable from within a function, you need to make sure to make it global using the global keyword or the \$GLOBALS[] array, or use the special session arrays as noted below.

Caution:
If you are using \$_SESSION (or \$HTTP_SESSION_VARS), do not use session_register(), session_is_registered(), and session_unregister().

ninjakoei 2004-03-31
<?php
session_start();//这个最好要用在你所有使用session的PHP网页的最顶部，
session_register('aaa');

phpinfo();
?>

feel8 2004-03-31
session_start()之前不能有输出。

If you want your script to work regardless of register_globals, you need to instead use the \$_SESSION array as \$_SESSION entries are automatically registered. If your script uses session_register(), it will not work in environments where the PHP directive register_globals is disabled.

mistjin 2004-03-31

turtlevan 2004-03-31
1.

2.

2.1w+


text:
doc-src/TutorialI/Types/numerics.tex
 author nipkow Tue Mar 03 17:05:18 2009 +0100 (2009-03-03) changeset 30224 79136ce06bdb parent 30200 0db3a35eab01 child 30242 aea5d7fa7ef5 permissions -rw-r--r--
removed and renamed redundant lemmas
     1 % $Id$

     2

     3 \section{Numbers}

     4 \label{sec:numbers}

     5

     6 \index{numbers|(}%

     7 Until now, our numerical examples have used the type of \textbf{natural

     8 numbers},

     9 \isa{nat}.  This is a recursive datatype generated by the constructors

    10 zero  and successor, so it works well with inductive proofs and primitive

    11 recursive function definitions.  HOL also provides the type

    12 \isa{int} of \textbf{integers}, which lack induction but support true

    13 subtraction.  With subtraction, arithmetic reasoning is easier, which makes

    14 the integers preferable to the natural numbers for

    15 complicated arithmetic expressions, even if they are non-negative.  The logic HOL-Complex also has the types

    16 \isa{rat}, \isa{real} and \isa{complex}: the rational, real and complex numbers.  Isabelle has no

    17 subtyping,  so the numeric

    18 types are distinct and there are functions to convert between them.

    19 Most numeric operations are overloaded: the same symbol can be

    20 used at all numeric types. Table~\ref{tab:overloading} in the appendix

    21 shows the most important operations, together with the priorities of the

    22 infix symbols. Algebraic properties are organized using type classes

    23 around algebraic concepts such as rings and fields;

    24 a property such as the commutativity of addition is a single theorem

    25 (\isa{add_commute}) that applies to all numeric types.

    26

    27 \index{linear arithmetic}%

    28 Many theorems involving numeric types can be proved automatically by

    29 Isabelle's arithmetic decision procedure, the method

    30 \methdx{arith}.  Linear arithmetic comprises addition, subtraction

    31 and multiplication by constant factors; subterms involving other operators

    32 are regarded as variables.  The procedure can be slow, especially if the

    33 subgoal to be proved involves subtraction over type \isa{nat}, which

    34 causes case splits.  On types \isa{nat} and \isa{int}, \methdx{arith}

    35 can deal with quantifiers---this is known as Presburger arithmetic---whereas on type \isa{real} it cannot.

    36

    37 The simplifier reduces arithmetic expressions in other

    38 ways, such as dividing through by common factors.  For problems that lie

    39 outside the scope of automation, HOL provides hundreds of

    40 theorems about multiplication, division, etc., that can be brought to

    41 bear.  You can locate them using Proof General's Find

    42 button.  A few lemmas are given below to show what

    43 is available.

    44

    45 \subsection{Numeric Literals}

    46 \label{sec:numerals}

    47

    48 \index{numeric literals|(}%

    49 The constants \cdx{0} and \cdx{1} are overloaded.  They denote zero and one,

    50 respectively, for all numeric types.  Other values are expressed by numeric

    51 literals, which consist of one or more decimal digits optionally preceeded by a minus sign (\isa{-}).  Examples are \isa{2}, \isa{-3} and

    52 \isa{441223334678}.  Literals are available for the types of natural

    53 numbers, integers, rationals, reals, etc.; they denote integer values of

    54 arbitrary size.

    55

    56 Literals look like constants, but they abbreviate

    57 terms representing the number in a two's complement binary notation.

    58 Isabelle performs arithmetic on literals by rewriting rather

    59 than using the hardware arithmetic. In most cases arithmetic

    60 is fast enough, even for numbers in the millions. The arithmetic operations

    61 provided for literals include addition, subtraction, multiplication,

    62 integer division and remainder.  Fractions of literals (expressed using

    63 division) are reduced to lowest terms.

    64

    65 \begin{warn}\index{overloading!and arithmetic}

    66 The arithmetic operators are

    67 overloaded, so you must be careful to ensure that each numeric

    68 expression refers to a specific type, if necessary by inserting

    69 type constraints.  Here is an example of what can go wrong:

    70 \par

    71 \begin{isabelle}

    72 \isacommand{lemma}\ "2\ *\ m\ =\ m\ +\ m"

    73 \end{isabelle}

    74 %

    75 Carefully observe how Isabelle displays the subgoal:

    76 \begin{isabelle}

    77 \ 1.\ (2::'a)\ *\ m\ =\ m\ +\ m

    78 \end{isabelle}

    79 The type \isa{'a} given for the literal \isa{2} warns us that no numeric

    80 type has been specified.  The problem is underspecified.  Given a type

    81 constraint such as \isa{nat}, \isa{int} or \isa{real}, it becomes trivial.

    82 \end{warn}

    83

    84 \begin{warn}

    85 \index{recdef@\isacommand {recdef} (command)!and numeric literals}

    86 Numeric literals are not constructors and therefore

    87 must not be used in patterns.  For example, this declaration is

    88 rejected:

    89 \begin{isabelle}

    90 \isacommand{recdef}\ h\ "\isacharbraceleft \isacharbraceright "\isanewline

    91 "h\ 3\ =\ 2"\isanewline

    92 "h\ i\ \ =\ i"

    93 \end{isabelle}

    94

    95 You should use a conditional expression instead:

    96 \begin{isabelle}

    97 "h\ i\ =\ (if\ i\ =\ 3\ then\ 2\ else\ i)"

    98 \end{isabelle}

    99 \index{numeric literals|)}

   100 \end{warn}

   101

   102

   103 \subsection{The Type of Natural Numbers, {\tt\slshape nat}}

   104

   105 \index{natural numbers|(}\index{*nat (type)|(}%

   106 This type requires no introduction: we have been using it from the

   107 beginning.  Hundreds of theorems about the natural numbers are

   108 proved in the theories \isa{Nat} and \isa{Divides}.

   109 Basic properties of addition and multiplication are available through the

   110 axiomatic type class for semirings (\S\ref{sec:numeric-axclasses}).

   111

   112 \subsubsection{Literals}

   113 \index{numeric literals!for type \protect\isa{nat}}%

   114 The notational options for the natural  numbers are confusing.  Recall that an

   115 overloaded constant can be defined independently for each type; the definition

   116 of \cdx{1} for type \isa{nat} is

   117 \begin{isabelle}

   118 1\ \isasymequiv\ Suc\ 0

   119 \rulename{One_nat_def}

   120 \end{isabelle}

   121 This is installed as a simplification rule, so the simplifier will replace

   122 every occurrence of \isa{1::nat} by \isa{Suc\ 0}.  Literals are obviously

   123 better than nested \isa{Suc}s at expressing large values.  But many theorems,

   124 including the rewrite rules for primitive recursive functions, can only be

   125 applied to terms of the form \isa{Suc\ $n$}.

   126

   127 The following default  simplification rules replace

   128 small literals by zero and successor:

   129 \begin{isabelle}

   130 2\ +\ n\ =\ Suc\ (Suc\ n)

   131 \rulename{add_2_eq_Suc}\isanewline

   132 n\ +\ 2\ =\ Suc\ (Suc\ n)

   133 \rulename{add_2_eq_Suc'}

   134 \end{isabelle}

   135 It is less easy to transform \isa{100} into \isa{Suc\ 99} (for example), and

   136 the simplifier will normally reverse this transformation.  Novices should

   137 express natural numbers using \isa{0} and \isa{Suc} only.

   138

   139 \subsubsection{Division}

   140 \index{division!for type \protect\isa{nat}}%

   141 The infix operators \isa{div} and \isa{mod} are overloaded.

   142 Isabelle/HOL provides the basic facts about quotient and remainder

   143 on the natural numbers:

   144 \begin{isabelle}

   145 m\ mod\ n\ =\ (if\ m\ <\ n\ then\ m\ else\ (m\ -\ n)\ mod\ n)

   146 \rulename{mod_if}\isanewline

   147 m\ div\ n\ *\ n\ +\ m\ mod\ n\ =\ m%

   148 \rulenamedx{mod_div_equality}

   149 \end{isabelle}

   150

   151 Many less obvious facts about quotient and remainder are also provided.

   152 Here is a selection:

   153 \begin{isabelle}

   154 a\ *\ b\ div\ c\ =\ a\ *\ (b\ div\ c)\ +\ a\ *\ (b\ mod\ c)\ div\ c%

   155 \rulename{div_mult1_eq}\isanewline

   156 a\ *\ b\ mod\ c\ =\ a\ *\ (b\ mod\ c)\ mod\ c%

   157 \rulename{mod_mult_right_eq}\isanewline

   158 a\ div\ (b*c)\ =\ a\ div\ b\ div\ c%

   159 \rulename{div_mult2_eq}\isanewline

   160 a\ mod\ (b*c)\ =\ b * (a\ div\ b\ mod\ c)\ +\ a\ mod\ b%

   161 \rulename{mod_mult2_eq}\isanewline

   162 0\ <\ c\ \isasymLongrightarrow \ (c\ *\ a)\ div\ (c\ *\ b)\ =\ a\ div\ b%

   163 \rulename{div_mult_mult1}\isanewline

   164 (m\ mod\ n)\ *\ k\ =\ (m\ *\ k)\ mod\ (n\ *\ k)

   165 \rulenamedx{mod_mult_distrib}\isanewline

   166 m\ \isasymle \ n\ \isasymLongrightarrow \ m\ div\ k\ \isasymle \ n\ div\ k%

   167 \rulename{div_le_mono}

   168 \end{isabelle}

   169

   170 Surprisingly few of these results depend upon the

   171 divisors' being nonzero.

   172 \index{division!by zero}%

   173 That is because division by

   174 zero yields zero:

   175 \begin{isabelle}

   176 a\ div\ 0\ =\ 0

   177 \rulename{DIVISION_BY_ZERO_DIV}\isanewline

   178 a\ mod\ 0\ =\ a%

   179 \rulename{DIVISION_BY_ZERO_MOD}

   180 \end{isabelle}

   181 In \isa{div_mult_mult1} above, one of

   182 the two divisors (namely~\isa{c}) must still be nonzero.

   183

   184 The \textbf{divides} relation\index{divides relation}

   185 has the standard definition, which

   186 is overloaded over all numeric types:

   187 \begin{isabelle}

   188 m\ dvd\ n\ \isasymequiv\ {\isasymexists}k.\ n\ =\ m\ *\ k

   189 \rulenamedx{dvd_def}

   190 \end{isabelle}

   191 %

   192 Section~\ref{sec:proving-euclid} discusses proofs involving this

   193 relation.  Here are some of the facts proved about it:

   194 \begin{isabelle}

   195 \isasymlbrakk m\ dvd\ n;\ n\ dvd\ m\isasymrbrakk \ \isasymLongrightarrow \ m\ =\ n%

   196 \rulenamedx{dvd_anti_sym}\isanewline

   197 \isasymlbrakk k\ dvd\ m;\ k\ dvd\ n\isasymrbrakk \ \isasymLongrightarrow \ k\ dvd\ (m\ +\ n)

   198 \rulenamedx{dvd_add}

   199 \end{isabelle}

   200

   201 \subsubsection{Subtraction}

   202

   203 There are no negative natural numbers, so \isa{m\ -\ n} equals zero unless

   204 \isa{m} exceeds~\isa{n}. The following is one of the few facts

   205 about \isa{m\ -\ n} that is not subject to

   206 the condition \isa{n\ \isasymle \  m}.

   207 \begin{isabelle}

   208 (m\ -\ n)\ *\ k\ =\ m\ *\ k\ -\ n\ *\ k%

   209 \rulenamedx{diff_mult_distrib}

   210 \end{isabelle}

   211 Natural number subtraction has few

   212 nice properties; often you should remove it by simplifying with this split

   213 rule.

   214 \begin{isabelle}

   215 P(a-b)\ =\ ((a<b\ \isasymlongrightarrow \ P\

   216 0)\ \isasymand \ (\isasymforall d.\ a\ =\ b+d\ \isasymlongrightarrow \ P\

   217 d))

   218 \rulename{nat_diff_split}

   219 \end{isabelle}

   220 For example, splitting helps to prove the following fact.

   221 \begin{isabelle}

   222 \isacommand{lemma}\ "(n\ -\ 2)\ *\ (n\ +\ 2)\ =\ n\ *\ n\ -\ (4::nat)"\isanewline

   223 \isacommand{apply}\ (simp\ split:\ nat_diff_split,\ clarify)\isanewline

   224 \ 1.\ \isasymAnd d.\ \isasymlbrakk n\ <\ 2;\ n\ *\ n\ =\ 4\ +\ d\isasymrbrakk \ \isasymLongrightarrow \ d\ =\ 0

   225 \end{isabelle}

   226 The result lies outside the scope of linear arithmetic, but

   227  it is easily found

   228 if we explicitly split \isa{n<2} as \isa{n=0} or \isa{n=1}:

   229 \begin{isabelle}

   230 \isacommand{apply}\ (subgoal_tac\ "n=0\ |\ n=1",\ force,\ arith)\isanewline

   231 \isacommand{done}

   232 \end{isabelle}%%%%%%

   233 \index{natural numbers|)}\index{*nat (type)|)}

   234

   235

   236 \subsection{The Type of Integers, {\tt\slshape int}}

   237

   238 \index{integers|(}\index{*int (type)|(}%

   239 Reasoning methods for the integers resemble those for the natural numbers,

   240 but induction and

   241 the constant \isa{Suc} are not available.  HOL provides many lemmas for

   242 proving inequalities involving integer multiplication and division, similar

   243 to those shown above for type~\isa{nat}. The laws of addition, subtraction

   244 and multiplication are available through the axiomatic type class for rings

   245 (\S\ref{sec:numeric-axclasses}).

   246

   247 The \rmindex{absolute value} function \cdx{abs} is overloaded, and is

   248 defined for all types that involve negative numbers, including the integers.

   249 The \isa{arith} method can prove facts about \isa{abs} automatically,

   250 though as it does so by case analysis, the cost can be exponential.

   251 \begin{isabelle}

   252 \isacommand{lemma}\ "abs\ (x+y)\ \isasymle \ abs\ x\ +\ abs\ (y\ ::\ int)"\isanewline

   253 \isacommand{by}\ arith

   254 \end{isabelle}

   255

   256 For division and remainder,\index{division!by negative numbers}

   257 the treatment of negative divisors follows

   258 mathematical practice: the sign of the remainder follows that

   259 of the divisor:

   260 \begin{isabelle}

   261 0\ <\ b\ \isasymLongrightarrow \ 0\ \isasymle \ a\ mod\ b%

   262 \rulename{pos_mod_sign}\isanewline

   263 0\ <\ b\ \isasymLongrightarrow \ a\ mod\ b\ <\ b%

   264 \rulename{pos_mod_bound}\isanewline

   265 b\ <\ 0\ \isasymLongrightarrow \ a\ mod\ b\ \isasymle \ 0

   266 \rulename{neg_mod_sign}\isanewline

   267 b\ <\ 0\ \isasymLongrightarrow \ b\ <\ a\ mod\ b%

   268 \rulename{neg_mod_bound}

   269 \end{isabelle}

   270 ML treats negative divisors in the same way, but most computer hardware

   271 treats signed operands using the same rules as for multiplication.

   272 Many facts about quotients and remainders are provided:

   273 \begin{isabelle}

   274 (a\ +\ b)\ div\ c\ =\isanewline

   275 a\ div\ c\ +\ b\ div\ c\ +\ (a\ mod\ c\ +\ b\ mod\ c)\ div\ c%

   276 \rulename{zdiv_zadd1_eq}

   277 \par\smallskip

   278 (a\ +\ b)\ mod\ c\ =\ (a\ mod\ c\ +\ b\ mod\ c)\ mod\ c%

   279 \rulename{mod_add_eq}

   280 \end{isabelle}

   281

   282 \begin{isabelle}

   283 (a\ *\ b)\ div\ c\ =\ a\ *\ (b\ div\ c)\ +\ a\ *\ (b\ mod\ c)\ div\ c%

   284 \rulename{zdiv_zmult1_eq}\isanewline

   285 (a\ *\ b)\ mod\ c\ =\ a\ *\ (b\ mod\ c)\ mod\ c%

   286 \rulename{zmod_zmult1_eq}

   287 \end{isabelle}

   288

   289 \begin{isabelle}

   290 0\ <\ c\ \isasymLongrightarrow \ a\ div\ (b*c)\ =\ a\ div\ b\ div\ c%

   291 \rulename{zdiv_zmult2_eq}\isanewline

   292 0\ <\ c\ \isasymLongrightarrow \ a\ mod\ (b*c)\ =\ b*(a\ div\ b\ mod\

   293 c)\ +\ a\ mod\ b%

   294 \rulename{zmod_zmult2_eq}

   295 \end{isabelle}

   296 The last two differ from their natural number analogues by requiring

   297 \isa{c} to be positive.  Since division by zero yields zero, we could allow

   298 \isa{c} to be zero.  However, \isa{c} cannot be negative: a counterexample

   299 is

   300 $\isa{a} = 7$, $\isa{b} = 2$ and $\isa{c} = -3$, when the left-hand side of

   301 \isa{zdiv_zmult2_eq} is $-2$ while the right-hand side is~$-1$.

   302 The prefix~\isa{z} in many theorem names recalls the use of $\mathbb{Z}$ to

   303 denote the set of integers.%

   304 \index{integers|)}\index{*int (type)|)}

   305

   306 Induction is less important for integers than it is for the natural numbers, but it can be valuable if the range of integers has a lower or upper bound.  There are four rules for integer induction, corresponding to the possible relations of the bound ($\geq$, $>$, $\leq$ and $<$):

   307 \begin{isabelle}

   308 \isasymlbrakk k\ \isasymle \ i;\ P\ k;\ \isasymAnd i.\ \isasymlbrakk k\ \isasymle \ i;\ P\ i\isasymrbrakk \ \isasymLongrightarrow \ P(i+1)\isasymrbrakk \ \isasymLongrightarrow \ P\ i%

   309 \rulename{int_ge_induct}\isanewline

   310 \isasymlbrakk k\ <\ i;\ P(k+1);\ \isasymAnd i.\ \isasymlbrakk k\ <\ i;\ P\ i\isasymrbrakk \ \isasymLongrightarrow \ P(i+1)\isasymrbrakk \ \isasymLongrightarrow \ P\ i%

   311 \rulename{int_gr_induct}\isanewline

   312 \isasymlbrakk i\ \isasymle \ k;\ P\ k;\ \isasymAnd i.\ \isasymlbrakk i\ \isasymle \ k;\ P\ i\isasymrbrakk \ \isasymLongrightarrow \ P(i-1)\isasymrbrakk \ \isasymLongrightarrow \ P\ i%

   313 \rulename{int_le_induct}\isanewline

   314 \isasymlbrakk i\ <\ k;\ P(k-1);\ \isasymAnd i.\ \isasymlbrakk i\ <\ k;\ P\ i\isasymrbrakk \ \isasymLongrightarrow \ P(i-1)\isasymrbrakk \ \isasymLongrightarrow \ P\ i%

   315 \rulename{int_less_induct}

   316 \end{isabelle}

   317

   318

   319 \subsection{The Types of Rational, Real and Complex Numbers}

   320 \label{sec:real}

   321

   322 \index{rational numbers|(}\index{*rat (type)|(}%

   323 \index{real numbers|(}\index{*real (type)|(}%

   324 \index{complex numbers|(}\index{*complex (type)|(}%

   325 These types provide true division, the overloaded operator \isa{/},

   326 which differs from the operator \isa{div} of the

   327 natural numbers and integers. The rationals and reals are

   328 \textbf{dense}: between every two distinct numbers lies another.

   329 This property follows from the division laws, since if $x\not=y$ then $(x+y)/2$ lies between them:

   330 \begin{isabelle}

   331 a\ <\ b\ \isasymLongrightarrow \ \isasymexists r.\ a\ <\ r\ \isasymand \ r\ <\ b%

   332 \rulename{dense}

   333 \end{isabelle}

   334

   335 The real numbers are, moreover, \textbf{complete}: every set of reals that

   336 is bounded above has a least upper bound.  Completeness distinguishes the

   337 reals from the rationals, for which the set $\{x\mid x^2<2\}$ has no least

   338 upper bound.  (It could only be $\surd2$, which is irrational.) The

   339 formalization of completeness, which is complicated,

   340 can be found in theory \texttt{RComplete} of directory

   341 \texttt{Real}.

   342

   343 Numeric literals\index{numeric literals!for type \protect\isa{real}}

   344 for type \isa{real} have the same syntax as those for type

   345 \isa{int} and only express integral values.  Fractions expressed

   346 using the division operator are automatically simplified to lowest terms:

   347 \begin{isabelle}

   348 \ 1.\ P\ ((3\ /\ 4)\ *\ (8\ /\ 15))\isanewline

   349 \isacommand{apply} simp\isanewline

   350 \ 1.\ P\ (2\ /\ 5)

   351 \end{isabelle}

   352 Exponentiation can express floating-point values such as

   353 \isa{2 * 10\isacharcircum6}, but at present no special simplification

   354 is performed.

   355

   356 \begin{warn}

   357 Type \isa{real} is only available in the logic HOL-Complex, which is

   358 HOL extended with a definitional development of the real and complex

   359 numbers.  Base your theory upon theory \thydx{Complex_Main}, not the

   360 usual \isa{Main}, and set the Proof General menu item \pgmenu{Isabelle} $>$

   361 \pgmenu{Logics} $>$ \pgmenu{HOL-Complex}.%

   362 \index{real numbers|)}\index{*real (type)|)}

   363 \end{warn}

   364

   365 Also available in HOL-Complex is the

   366 theory \isa{Hyperreal}, which define the type \tydx{hypreal} of

   367 \rmindex{non-standard reals}.  These

   368 \textbf{hyperreals} include infinitesimals, which represent infinitely

   369 small and infinitely large quantities; they facilitate proofs

   370 about limits, differentiation and integration~\cite{fleuriot-jcm}.  The

   371 development defines an infinitely large number, \isa{omega} and an

   372 infinitely small positive number, \isa{epsilon}.  The

   373 relation $x\approx y$ means $x$ is infinitely close to~$y$.''

   374 Theory \isa{Hyperreal} also defines transcendental functions such as sine,

   375 cosine, exponential and logarithm --- even the versions for type

   376 \isa{real}, because they are defined using nonstandard limits.%

   377 \index{rational numbers|)}\index{*rat (type)|)}%

   378 \index{real numbers|)}\index{*real (type)|)}%

   379 \index{complex numbers|)}\index{*complex (type)|)}

   380

   381

   382 \subsection{The Numeric Type Classes}\label{sec:numeric-axclasses}

   383

   384 Isabelle/HOL organises its numeric theories using axiomatic type classes.

   385 Hundreds of basic properties are proved in the theory \isa{Ring_and_Field}.

   386 These lemmas are available (as simprules if they were declared as such)

   387 for all numeric types satisfying the necessary axioms. The theory defines

   388 dozens of type classes, such as the following:

   389 \begin{itemize}

   390 \item

   391 \tcdx{semiring} and \tcdx{ordered_semiring}: a \emph{semiring}

   392 provides the associative operators \isa{+} and~\isa{*}, with \isa{0} and~\isa{1}

   393 as their respective identities. The operators enjoy the usual distributive law,

   394 and addition (\isa{+}) is also commutative.

   395 An \emph{ordered semiring} is also linearly

   396 ordered, with addition and multiplication respecting the ordering. Type \isa{nat} is an ordered semiring.

   397 \item

   398 \tcdx{ring} and \tcdx{ordered_ring}: a \emph{ring} extends a semiring

   399 with unary minus (the additive inverse) and subtraction (both

   400 denoted~\isa{-}). An \emph{ordered ring} includes the absolute value

   401 function, \cdx{abs}. Type \isa{int} is an ordered ring.

   402 \item

   403 \tcdx{field} and \tcdx{ordered_field}: a field extends a ring with the

   404 multiplicative inverse (called simply \cdx{inverse} and division~(\isa{/})).

   405 An ordered field is based on an ordered ring. Type \isa{complex} is a field, while type \isa{real} is an ordered field.

   406 \item

   407 \tcdx{division_by_zero} includes all types where \isa{inverse 0 = 0}

   408 and \isa{a / 0 = 0}. These include all of Isabelle's standard numeric types.

   409 However, the basic properties of fields are derived without assuming

   410 division by zero.

   411 \end{itemize}

   412

   413 Hundreds of basic lemmas are proved, each of which

   414 holds for all types in the corresponding type class. In most

   415 cases, it is obvious whether a property is valid for a particular type. No

   416 abstract properties involving subtraction hold for type \isa{nat};

   417 instead, theorems such as

   418 \isa{diff_mult_distrib} are proved specifically about subtraction on

   419 type~\isa{nat}. All abstract properties involving division require a field.

   420 Obviously, all properties involving orderings required an ordered

   421 structure.

   422

   423 The class \tcdx{ring_no_zero_divisors} of rings without zero divisors satisfies a number of natural cancellation laws, the first of which characterizes this class:

   424 \begin{isabelle}

   425 (a\ *\ b\ =\ (0::'a))\ =\ (a\ =\ (0::'a)\ \isasymor \ b\ =\ (0::'a))

   426 \rulename{mult_eq_0_iff}\isanewline

   427 (a\ *\ c\ =\ b\ *\ c)\ =\ (c\ =\ (0::'a)\ \isasymor \ a\ =\ b)

   428 \rulename{mult_cancel_right}

   429 \end{isabelle}

   430 \begin{pgnote}

   431 Setting the flag \pgmenu{Isabelle} $>$ \pgmenu{Settings} $>$

   432 \pgmenu{Show Sorts} will display the type classes of all type variables.

   433 \end{pgnote}

   434 \noindent

   435 Here is how the theorem \isa{mult_cancel_left} appears with the flag set.

   436 \begin{isabelle}

   437 ((c::'a::ring_no_zero_divisors)\ *\ (a::'a::ring_no_zero_divisors) =\isanewline

   438 \ c\ *\ (b::'a::ring_no_zero_divisors))\ =\isanewline

   439 (c\ =\ (0::'a::ring_no_zero_divisors)\ \isasymor\ a\ =\ b)

   440 \end{isabelle}

   441

   442

   443 \subsubsection{Simplifying with the AC-Laws}

   444 Suppose that two expressions are equal, differing only in

   445 associativity and commutativity of addition.  Simplifying with the

   446 following equations sorts the terms and groups them to the right, making

   447 the two expressions identical.

   448 \begin{isabelle}

   449 a\ +\ b\ +\ c\ =\ a\ +\ (b\ +\ c)

   450 \rulenamedx{add_assoc}\isanewline

   451 a\ +\ b\ =\ b\ +\ a%

   452 \rulenamedx{add_commute}\isanewline

   453 a\ +\ (b\ +\ c)\ =\ b\ +\ (a\ +\ c)

   454 \rulename{add_left_commute}

   455 \end{isabelle}

   456 The name \isa{add_ac}\index{*add_ac (theorems)}

   457 refers to the list of all three theorems; similarly

   458 there is \isa{mult_ac}.\index{*mult_ac (theorems)}

   459 They are all proved for semirings and therefore hold for all numeric types.

   460

   461 Here is an example of the sorting effect.  Start

   462 with this goal, which involves type \isa{nat}.

   463 \begin{isabelle}

   464 \ 1.\ Suc\ (i\ +\ j\ *\ l\ *\ k\ +\ m\ *\ n)\ =\

   465 f\ (n\ *\ m\ +\ i\ +\ k\ *\ j\ *\ l)

   466 \end{isabelle}

   467 %

   468 Simplify using  \isa{add_ac} and \isa{mult_ac}.

   469 \begin{isabelle}

   470 \isacommand{apply}\ (simp\ add:\ add_ac\ mult_ac)

   471 \end{isabelle}

   472 %

   473 Here is the resulting subgoal.

   474 \begin{isabelle}

   475 \ 1.\ Suc\ (i\ +\ (m\ *\ n\ +\ j\ *\ (k\ *\ l)))\

   476 =\ f\ (i\ +\ (m\ *\ n\ +\ j\ *\ (k\ *\ l)))%

   477 \end{isabelle}

   478

   479

   480 \subsubsection{Division Laws for Fields}

   481

   482 Here is a selection of rules about the division operator.  The following

   483 are installed as default simplification rules in order to express

   484 combinations of products and quotients as rational expressions:

   485 \begin{isabelle}

   486 a\ *\ (b\ /\ c)\ =\ a\ *\ b\ /\ c

   487 \rulename{times_divide_eq_right}\isanewline

   488 b\ /\ c\ *\ a\ =\ b\ *\ a\ /\ c

   489 \rulename{times_divide_eq_left}\isanewline

   490 a\ /\ (b\ /\ c)\ =\ a\ *\ c\ /\ b

   491 \rulename{divide_divide_eq_right}\isanewline

   492 a\ /\ b\ /\ c\ =\ a\ /\ (b\ *\ c)

   493 \rulename{divide_divide_eq_left}

   494 \end{isabelle}

   495

   496 Signs are extracted from quotients in the hope that complementary terms can

   497 then be cancelled:

   498 \begin{isabelle}

   499 -\ (a\ /\ b)\ =\ -\ a\ /\ b

   500 \rulename{minus_divide_left}\isanewline

   501 -\ (a\ /\ b)\ =\ a\ /\ -\ b

   502 \rulename{minus_divide_right}

   503 \end{isabelle}

   504

   505 The following distributive law is available, but it is not installed as a

   506 simplification rule.

   507 \begin{isabelle}

   508 (a\ +\ b)\ /\ c\ =\ a\ /\ c\ +\ b\ /\ c%

   509 \rulename{add_divide_distrib}

   510 \end{isabelle}

   511

   512

   513 \subsubsection{Absolute Value}

   514

   515 The \rmindex{absolute value} function \cdx{abs} is available for all

   516 ordered rings, including types \isa{int}, \isa{rat} and \isa{real}.

   517 It satisfies many properties,

   518 such as the following:

   519 \begin{isabelle}

   520 \isasymbar x\ *\ y\isasymbar \ =\ \isasymbar x\isasymbar \ *\ \isasymbar y\isasymbar

   521 \rulename{abs_mult}\isanewline

   522 (\isasymbar a\isasymbar \ \isasymle \ b)\ =\ (a\ \isasymle \ b\ \isasymand \ -\ a\ \isasymle \ b)

   523 \rulename{abs_le_iff}\isanewline

   524 \isasymbar a\ +\ b\isasymbar \ \isasymle \ \isasymbar a\isasymbar \ +\ \isasymbar b\isasymbar

   525 \rulename{abs_triangle_ineq}

   526 \end{isabelle}

   527

   528 \begin{warn}

   529 The absolute value bars shown above cannot be typed on a keyboard.  They

   530 can be entered using the X-symbol package.  In \textsc{ascii}, type \isa{abs x} to

   531 get \isa{\isasymbar x\isasymbar}.

   532 \end{warn}

   533

   534

   535 \subsubsection{Raising to a Power}

   536

   537 Another type class, \tcdx{ringppower}, specifies rings that also have

   538 exponentation to a natural number power, defined using the obvious primitive

   539 recursion. Theory \thydx{Power} proves various theorems, such as the

   540 following.

   541 \begin{isabelle}

   542 a\ \isacharcircum \ (m\ +\ n)\ =\ a\ \isacharcircum \ m\ *\ a\ \isacharcircum \ n%

   543 \rulename{power_add}\isanewline

   544 a\ \isacharcircum \ (m\ *\ n)\ =\ (a\ \isacharcircum \ m)\ \isacharcircum \ n%

   545 \rulename{power_mult}\isanewline

   546 \isasymbar a\ \isacharcircum \ n\isasymbar \ =\ \isasymbar a\isasymbar \ \isacharcircum \ n%

   547 \rulename{power_abs}

   548 \end{isabelle}%%%%%%%%%%%%%%%%%%%%%%%%%

   549 \index{numbers|)}


text:
# What’s up with McNemar’s test?

A quick note from EACL: some papers related to LSDSem workshop (Bugert et al. 2017; Zhou et al. 2015) use McNemar’s test to establish statistical significance and I find it very odd.

McNemar’s test examine “marginal (probability) homogeneity” which in our case is whether two systems yield (statistically) the same performance. According to the source code I found on Github, the way it works is:

1. Obtain predictions of System 1 and System 2
2. Compare them to gold labels to fill this table:
Sys1 correct Sys1 wrong a b c d
3. Compute the test statistics: $\chi^2 = {(b-c)^2 \over b+c}$ and p-value
4. If p-value is less than a certain level (e.g. the magical 0.05), we reject the null hypothesis which is p(Sys1 correct) == p(Sys2 correct)

As it happens in the papers, the difference is statistically significant and therefore results are meaningful. Happy?

Not so fast. Continue reading

# A paper is the tip of an iceberg

I was reading Clark and Manning (2016) and studying their code. The contrast is just amazing.

This is what the paper has to say:

This is what I found after 1 hour of reading a JSON file and writing down all layers of the neural net (the file is data/models/all_pairs/architecture.json, created when you run the experiment):

Without the source code, this would be a replication nightmare for sure.

References

Clark, K., & Manning, C. D. (2016). Improving Coreference Resolution by Learning Entity-Level Distributed Representations. Proceedings of the 54th Annual Meeting of the Association for Computational Linguistics, 643–653. http://doi.org/10.18653/v1/P16-1061

# Long paper accepted for EACL 2017

Title: Tackling Error Propagation through Reinforcement Learning: A Case of Greedy Dependency Parsing

Conference: EACL 2017 (European Chapter of the Association for Computational Linguistics), at Valencia, 3-7 April 2017.

Abstract:
Error propagation is a common problem in NLP. Reinforcement learning explores erroneous states during training and can therefore be more robust when mistakes are made early in a process. In this paper, we apply reinforcement learning to greedy dependency parsing which is known to suffer from error propagation. Reinforcement learning improves accuracy of both labeled and unlabeled dependencies of the Stanford Neural Dependency Parser, a high performance greedy parser, while maintaining its efficiency. We investigate the portion of errors which are the result of error propagation and confirm that reinforcement learning reduces the occurrence of error propagation.

Full article: arXiv:1702.06794

Slides: view online

# Presentation at CLIN 27

Last week, I had a good time at CLIN 27. The city was pretty, with a cute morning market and tasty croissants. The snow was kind to me (sometimes I does like snow if it is gentle). The poem presentation from Tim van de Cruys was funny and I met some old friends. I brought to CLIN my own side project where I explore, explain and (slightly) improve word2vec:

# Hyperparameter tuning in SURFsara HPC Cloud

Hyperparameter tuning is difficult, not because it’s terribly complicated but obtaining enough resource is often not easy. I’m lucky enough to work at Vrije Universiteit and therefore can access the SURFsara HPC Cloud with not too much effort. Compared to Amazon EC2 (the only other cloud solution I have tried before), the functionality is rather basic but I think suits the needs of many researchers. Using the web interface or OpenNebula API, you can easily customize an image, attach hard drive, launch 10 instances and access any of them using a public key. What else do you need to run your experiments? Continue reading

# Reproducing Chen & Manning (2014)

Neural dependency parsing is attractive for several reasons: first, distributed representation generalizes better, second, fast parsing unlocks new applications, and third, fast training means parsers can be co-trained with other NLP modules and integrated into a bigger system.

Chen & Manning (2014) from Stanford were the first to show that neural dependency parsing works and Google folks were quick to adopt this paradigm to improve the state-of-the-art (e.g. Weiss et al., 2015).

Though Stanford open-sourced their parser as part of CoreNLP, they didn’t release the code of their experiments. As anybody in academia probably knows, reproducing experiments is non-trivial, even extremely difficult at times. Since I have painstakingly gone through the process, I think it’s a good idea to share with you.

# Skip-gram negative sampling as (unshifted) PMI matrix factorization

In previous post, we arrived at two formulas showing the equivalence between SGNS and shifted PMI:

$p(D|w,c) = \sigma(w \cdot c) = \frac{1}{1 + e^{-w \cdot c}}$    (1)

$p(D|w,c) = \frac{1}{1 + ke^{-\mathrm{PMI}(w,c)}}$    (2)

Apparently, the reason for the “shift” is that in (1) there’s no while in (2) there is. The “shift” is not just an ugly patch in the formula but it might also have a negative effect on the quality of learned embeddings. Continue reading

# A new proof of the equivalence of word2vec’s SGNS and Shifted PPMI

[removed section]

At the heart of the argument was Levy and Goldberg’s proof that minimizing the loss of Skip-gram negative sampling (SGNS) is effectively approximating a shifted PPMI matrix. Starting with the log-likelihood, they worked their way to local objective for each word-context pair and compare its derivative to zero to arrive at a function of PPMI. One might rightly question if the loss function is essential in this proof or there is a deeper link between the two formalizations?

My answer: Yes, there is. Continue reading

# Similarity, co-occurrence, functional relation, part-whole relation, subcategorization, what else?

In word sense disambiguation and named-entity disambiguation, an important assumption is that a document consists of related concepts and entities.

There are millions of concepts and entities, what makes some related but not others? This question is difficult and I don’t have the definitive answer. But it is a good start to list some classes of relatedness. Continue reading

# Knowledge base completion 101

Knowledge base completion (KBC) is not a standard task in natural language processing nor in machine learning. A search on Google scholar results in only over 100 article containing this phrase. Although it is similar to link prediction, “a long-standing challenge in modern information science” (Lü & Zhou, 2011), it has received much less attention.

However KBC is potentially an important step towards natural language understanding and recent advances in representation learning have enabled researchers to learn larger datasets with improved precision. Actually, a half of KBC articles were published in or after 2010. Continue reading


text:
# mxnet.ndarray.Correlation¶

mxnet.ndarray.Correlation(data1=None, data2=None, kernel_size=_Null, max_displacement=_Null, stride1=_Null, stride2=_Null, pad_size=_Null, is_multiply=_Null, out=None, name=None, **kwargs)

Applies correlation to inputs.

The correlation layer performs multiplicative patch comparisons between two feature maps.

Given two multi-channel feature maps $$f_{1}, f_{2}$$, with $$w$$, $$h$$, and $$c$$ being their width, height, and number of channels, the correlation layer lets the network compare each patch from $$f_{1}$$ with each patch from $$f_{2}$$.

For now we consider only a single comparison of two patches. The ‘correlation’ of two patches centered at $$x_{1}$$ in the first map and $$x_{2}$$ in the second map is then defined as:

$c(x_{1}, x_{2}) = \sum_{o \in [-k,k] \times [-k,k]} <f_{1}(x_{1} + o), f_{2}(x_{2} + o)>$

for a square patch of size $$K:=2k+1$$.

Note that the equation above is identical to one step of a convolution in neural networks, but instead of convolving data with a filter, it convolves data with other data. For this reason, it has no training weights.

Computing $$c(x_{1}, x_{2})$$ involves $$c * K^{2}$$ multiplications. Comparing all patch combinations involves $$w^{2}*h^{2}$$ such computations.

Given a maximum displacement $$d$$, for each location $$x_{1}$$ it computes correlations $$c(x_{1}, x_{2})$$ only in a neighborhood of size $$D:=2d+1$$, by limiting the range of $$x_{2}$$. We use strides $$s_{1}, s_{2}$$, to quantize $$x_{1}$$ globally and to quantize $$x_{2}$$ within the neighborhood centered around $$x_{1}$$.

The final output is defined by the following expression:

$out[n, q, i, j] = c(x_{i, j}, x_{q})$

where $$i$$ and $$j$$ enumerate spatial locations in $$f_{1}$$, and $$q$$ denotes the $$q^{th}$$ neighborhood of $$x_{i,j}$$.

Defined in src/operator/correlation.cc:L198

Parameters
• data1 (NDArray) – Input data1 to the correlation.

• data2 (NDArray) – Input data2 to the correlation.

• kernel_size (int (non-negative), optional, default=1) – kernel size for Correlation must be an odd number

• max_displacement (int (non-negative), optional, default=1) – Max displacement of Correlation

• stride1 (int (non-negative), optional, default=1) – stride1 quantize data1 globally

• stride2 (int (non-negative), optional, default=1) – stride2 quantize data2 within the neighborhood centered around data1


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# Forms

Occasional Visitor

# Forms

I made a 12 questions form up and it only shows the first page, questions 1-3 in the form.  It shows the first page, then submit, then it ends.