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# How would a player character pay for an item that costs less than 1cp?

For reasons too complex and stupid to go into, I've recently been looking at item values and the 5e currency system. As I understand it, 1pp = 10gp = 20ep = 100sp = 1000cp, which places 1cp as the smallest representable value. However, the price per unit of Sling Bullets is 1/0.2cp (source: PHB page 150). With that in mind, how would a particularly difficult player go about purchasing a single Sling Bullet? Break a copper piece into fifths, perhaps?

Thanks for the answers, they're very helpful - for DMs. I didn't put this in the question originally because I didn't think it was important, but I can see now that was wrong. What I'm actually doing is building a relational database for D&D objects. So, when inserting a row into the item table for Sling Bullets, I wanted to follow the example of Arrows and Crossbow Bolts and put them in as a single object with a single value, although in the PHB itself only a price for Sling Bullets (20) is given.

Given the answers, I'm going to move ahead by only listing the value for an amount of an item that is at least 1cp (Sling Bullets (5) in this case). I'm going to accept an answer based on the fact that it lead me to this resolution by explaining what would be most helpful to a DM. However, if this new information leads to new suggestions, I'm open to those as well.

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As a DM, you aren't required to obey the strict prices set in the PHB. The prices listed are mostly just used as guides for creating characters on a budget; once you're in the adventuring world, goods cost whatever the local shops value them at, with the PHB listed prices as suggestions for "fair market" prices.

That being said, there's any number of ways to approach this. You could introduce a new currency that's worth 1/5 or 1/10 the value of a copper piece to give a single divisible currency for handling especially cheap goods.

Personally, I'd probably just rule that any shop that sells Sling Bullets doesn't sell them in packs of less than 5, for 1 copper piece for 5. That way, you don't have to introduce a new currency.

If the player is genuinely only interested in buying the single unit, I'd probably just sell 1 bullet for 1 copper piece. The scenarios where this would seriously inconvenience a player are pretty esoteric.

• Comments are not for extended discussion; this conversation has been moved to chat. – mxyzplk Feb 16 '19 at 11:32

# By haggling, probably unsuccessfully.

In the real world, I can walk into a grocery store (supermarket or locally owned) that sells eggs in cartons of a dozen, pick up a carton for $1.50, calculate that that works out to 12.5¢ per egg, and try to buy 2 eggs at checkout for 25¢, but the clerk will refuse to sell me those 2 eggs because the store does not offer the purchase of individual eggs. If I want those 2 eggs, I can buy a carton of a dozen for full price. This is because a carton of only 10 eggs is essentially unsalable for the grocery store. In the game, the shopkeeper has prepared a lot (pouch, case, or whatever) of 20 sling bullets for 4cp total. (We know the bullets are sold as a lot because if they were intended to be sold individually they'd be priced at 5 bullets for 1cp, not 4 times as many for 4 times as much.) You may be able to calculate that the price per bullet is 0.2cp, but that doesn't mean the shopkeeper is willing to sell a partial lot. The shopkeeper has the right to offer whole lots of 20 bullets, take it or leave it, but selling fewer bullets than a full lot probably requires the shopkeeper to count them out, which is a waste of time they could be using to serve other customers who know how to follow commerce norms and buy what's actually on offer. Even haggling is unlikely to be successful. Suppose you offered 1cp just to buy 1 bullet. Even though that's 5 times the shopkeeper's unit price, the shopkeeper would still likely decline because the remaining 19 bullets don't constitute a full lot. The next customer will likely be uninterested in purchasing that partial lot without haggling of their own unless the shopkeeper refills the lot or discounts it, which again makes the prospect barely worth the shopkeeper's time and effort. So, let the player haggle if they want to, but the shopkeeper should behave like a real person, which probably means refusing to sell a single unit of a supply that's not being offered in single units. If you think the player can successfully haggle an agreeable total with the shopkeeper, it certainly shouldn't involve the invention of a new currency or the cutting of coins (fragments of a coin are collectively worth less than an intact coin's face value in most realistic economies), but the terms that the player and shopkeeper find agreeable depends entirely on the player's argument and shopkeeper's demeanor. • If you have an actionable suggestion for how I can improve my answer, please leave a comment. I don't see anything actionable in your comment. It looks like you just disagree with my answer. If you think my answer is unhelpful because you disagree with it, please down-vote it. If you feel passionate about helping to answer the question, please post your own answer. – Bloodcinder Feb 15 '19 at 15:57 • <regarding a number of removed comments> How to disagree with an answer: 1. Vote. 2. That is all; there is no step 2. No really, step 2 is not “comment unconstructively”. How to help improve an answer: 1. Use a comment to constructively describe the specific change(s) that the author might consider making. – SevenSidedDie Feb 15 '19 at 16:15 In ancient or medieval times, coins were sometimes cut in pieces to deal with this kind of problem. So if you do want to make this possible, simply have the shopkeeper chop up the coin. Of course, this only goes to a point. I would assume a quarter is the furthest I would take this. Which in your example would up the price a little - 5 quarters for 5 bullets - which isn't unrealistic at all, bulk prices are common. In addition, this gives you the option of filling the PCs pockets with bits and pieces of coins. Perhaps they will tire of the book keeping at some point and just agree to be a bit more hand-wavy from now on ;) An example of the practice - a half-siliqua • For some reason, I think that your answer points to how the term "farthing" entered the English language. I am not sure if it would improve your answer to include it. your call. – KorvinStarmast Feb 14 '19 at 21:12 • Did you know that rounded coins came to be favored over other shapes as it was harder to shave bits off a round coin (as opposed to anything with a corner) to fraudulently collect the metal? – JonSG Feb 15 '19 at 0:56 • I think you could take it further than a quarter. "AWK! Pieces of eight, pieces of eight!" – David Conrad Feb 15 '19 at 18:17 • This is where the slang "two bits" for a US quarter dollar comes from. One bit is an eighth of a dollar (that is, a dollar coin chopped into eighths). Thus also "pieces of eight" (and eight bits to a byte). – Marq Feb 19 '19 at 18:01 Imagine you are on a farmers market in real life. You go to a vegetable stand and say to the shopkeeper: "I would like to buy one pea, please" "You mean one pound of peas?" "No, one single pea" What would the shopkeeper do? Weight a single pea, realize it's not even a single cent and consider how to handle the situation? There are other customers waiting in line behind you. They don't have time for caring about such a ridiculous problem. So how would they deal with such a situation? A friendly shopkeeper would tell you that you can just take that single pea for free. It's such a small value they won't even notice that it's missing. A less friendly shopkeeper would tell you to stop wasting their time with such ridiculous nonsense and tell you to get lost. A pragmatic shopkeeper would tell you that you need to buy at least 50g of peas because that's the minimum their scale will measure properly. That's likely how a shopkeeper in your campaign world would react, too. • Tell the player they can have that one sling bullet as a free sample, but the shopkeep will want money if they buy more later • Tell them they need to buy them in packs of 20. Lets look at a real world example. To do that, lets replace "Sling bullet" with "piece of paper" and "cp" with "cent". Pricing is the same, so 5 pieces of paper is one cent. Now, imagine going to this store and buying one piece of paper. Are they going to let you pay .2 cents? Of course not! Instead, they'll probably do one of the following: • Charge you one cent (since anything less is impossible) • Make you buy 5 pieces of paper (since that's the standard price for one cent) • Just give it to you for free (since .2 cents is negligible) There's really no difference of options in your world, and based on the circumstances and character of the merchant, all of these seem like viable options. As a DM, you'd have to choose at your discretion which choice your merchant makes. However, from a database perspective, you can still enter the cost of each sling bullet at .2, and when calculating the price for N of any object, simply give ceiling(N * price). That at least gives you a fair asking price which should scale with the real value of the items (and maybe nice merchants will round down while stingy ones round up). • I actually can't enter the price as .2 without some significant work - I'm using an extension I've written in C to allow people to input prices/values in terms of actual currency - e.g. 2gp, 100cp etc. but the implementation only allows for integral values. – ocket8888 Feb 14 '19 at 21:42 • @ocket8888 Without knowing your implementation, it's a bit tough to work out how to get around that, but if there's a C layer between the user and your database, you could make a "fractional piece" (for example's sake, lets call it an fp) or something which is 1/100th of a cp. If someone enters .2cp, you convert that to 20fp in your database, then when they ask for the price of N of those, you convert back to cp and take the ceiling in C, but it's up to you if that's worth it. You just hide the fp coin from the user; it's simply storage. – Lord Farquaad Feb 14 '19 at 21:46 • Once I went into a store in real life, and attempted to purchase a single swedish fish from the bulk candy. They gave it to me for free. If the shop in universe sells bullets in bulk, then the same night happen to a character. – Vaelus Feb 14 '19 at 21:48 • My implementation is here btw: github.com/ocket8888/dndDB I think a simpler way to implement it would be as a struct with one uint32_t for whole number values and a uint8_t for a decimal part. Regardless, decimals are something I want to avoid if at all possible, at least for now. Maybe I'll revisit that in the future, but at the moment I'm more concerned with populating the database with values that work - even if they aren't totally perfect. – ocket8888 Feb 14 '19 at 21:55 • This questions starts to belong to normal stack exchange slowly, as it seems to be programming issue, rather than RPG one. I would suggest following Lord Farquaad advice with fixed point decimal. You have your conversion layer, so you can handle all magic there. – Artur Biesiadowski Feb 15 '19 at 10:27 If you have a player who is that penurious with his character's money, you might offer him six bullets for a copper -- don't forget to remind him that's more than 15% off! For your DB app, you should put actual prices in the table because otherwise you have to write all kinds of handling code for items with quantities (what if a player wants to buy 24?). If you want to represent the books perfectly, you add another column listing the amounts that are given in the books. So bullets are purchased in batches of 20. But you keep the information intact and can work with it. You can also drop this information and simply implement a rounding rule, so players can purchase any quantity they want, but they only pay full coins (or you can add the quarter-coins idea above). • "all kinds of handling code for items with quantities (what if a player wants to buy 24?)." any serious store application I've seen handles that by restricting what the user can enter. You can't enter 24 because that's not an option. You can buy 20 or 25. Or perhaps 4 or 5 "units" (however they are described) and each is worth 5. You can also have various sizes of units - a small pack might be 5, a big could be 50. Or round in terms of 10s, if you wish - small pack of 10 bullets is 2cp. Check out Amazon or eBay for something sold in packs of 3 or 10 or whatever, if you want an example. – VLAZ Feb 15 '19 at 11:13 • I think you misunderstand though. I'm not making any kind of 'app' - I'm just making a database that 'apps' could be built upon. So sure, maybe a store would work like that, but suppose a player buys 20 sling bullets and fires one, then later asks some 'app' built on top of my db how much his inventory is worth in total? He's right, a sensible database would be capable of providing the information to answer that, without special handling code required after data retrieval. – ocket8888 Feb 15 '19 at 15:40 • FWIW, the real world analogue to this is electronics catalogues; these resistors have a unit price of "£0.00097" digikey.co.uk/product-detail/en/stackpole-electronics-inc/… (but a minimum order quantity of a tape of 10,000) – pjc50 Feb 15 '19 at 16:05 ## When you subdivide standard lots, the price goes up Take my local pizzeria. Like many pizzerias in the US, their pricing is structured around you buying two pizzas. So • Two large pizzas are$22
• One large pizza is 18
• One slice (1/8 of a large) is 3.50

Per-slice prices are 3.50, 2.25 or 1.37... Depending on volume.

Consider soft drinks, assuming 16 22 32 and 44 ounce sizes. The local movie theater prices them 4.50 4.75 5.00 5.25. The convenience store prices 1.19 1.29 1.39 1.49. It's not proportional, not nearly, because the store lease is the same either way, as is the clerk's time.

So you are incorrect to subdivide prices for sling bullets. 4cp for 20 mathematically means 1cp per 5, but that certainly won't correspond to retail. More likely:

• 20 for 4cp
• 4 for 2cp
• 1 for 1cp

Needless to say, this puts the kibosh on the parsimonious fellow aiming to buy his bullets one at a time. Now, thriftiness kicks in, and he can plainly see the value is in buying by the 20.

# The exchange of a similar valued good

Many peasants may readily trade supplies that are too cheap to be concerned with their gold value. Instead, they may simply trade similar cheap items or menial jobs, so long as it doesn't seem outrageous.

Perhaps the merchant needs a bundle of sticks, for building a fire; or a collection of flower petals for their medicine.

If you're already headed in the direction maybe they'll ask you to send a message since it's not out of your way in exchange.

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Minimum purchase

Require they spend at least 1cp and offer some other cheap item in addition to the bullet. If there is nothing suitable, get them to buy something more expensive and offer to throw the bullet in for free.

Let's be real

What are these bullets made of? If they are anything more valuable than stone (e.g. lead) then they will have more intrinsic value than the copper itself. Tell them to use the copper piece as a bullet.

For purposes of the application, I agree with listing with the assumed quantity in the book. And a "per each" value, as well - at least for informational purposes.

For play purposes, use varied coin from "other realms" along with cutting coins. The RoleMaster system uses Tin pieces and Bronze pieces below Copper pieces (1cp = 10tp = 100bp)*. But you could give change in a variety of small coins, perhaps by weight and composition.

This can allow you to provide clues/hooks to adventures. (Suspicious merchant: "Where did you get this coin?") It can also allow you to burden a "particularly difficult player" with a bunch of coinage that people are willing to give him in change but none will accept (except maybe the banker who will assess it all) for payment.**

(* The cost of items differs in that system, as well, such that you actually use the Tin and Bronze pieces. IIRC, 1cp is one day's wages for a day laborer there, in a medieval setting.)

(** This is based in RL. I've worked in places where we would regularly get foreign coins in the snack bar and such. Sometimes it would be because it looked or felt enough like a US quarter or nickel or dime that it got dropped in the kitty as one by mistake. Retailers put up with it, too, and some would actually put out a box of foreign coinage and you could exchange a handful for US money.)

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Expanding on Lord Farquaad's comment, this is the solution I used for the database of creature stats, in particular, carrying capacities. Very small creatures have maximum loads that are not an integer number of pounds, so they are recorded in ounces. Every load is an integer number of ounces.

I agree with you that it's nice to avoid floating-point numbers if possible. For example, suppose that you can buy 20 of an item for 3gp. Then one is worth 20/3 gp, or about 6.66666667gp...but rendered in binary, it's 110.1010101010101010101010...

It's annoying to deal with. Unfortunately, I don't think having a separate integer for a decimal part will make it any less annoying to deal with.

The solution I used was, obviously, to record the number in a different, smaller unit --- ounces instead of pounds. That's literally what I called the field in the database, max_load_ounces. You constantly have to be converting between units anyway, especially if we're talking about coins.

Unfortunately, in the case of copper pieces rather than pounds, there isn't a natural smaller unit that conveniently turns everything into integers. (Well, the Dark Sun setting has "bits" like the farthings/fourthings KorvinStarmast mentioned, but that would probably just get more confusing and annoying to deal with, since Dark Sun has ceramic pieces in place of copper pieces, in the sense that a tenth of a silver pieces is a ceramic piece, except it sort of has ceramic pieces in place of gold pieces, it's complicated.)

Since there isn't a convenient smaller unit, I think you're fully justified in just making up one for your own convenience. If you know in advance every single one of the objects you're going to include in the database, then all you have to do is take the common denominator of all the fractions --- the least common multiple of all the denominators. But if (as I assume) you don't know everything in advance and you want a unit that will be pretty safe, you probably want one-sixtieth. Why one-sixtieth? For the same reason the ancient Mesopotamians used it. Because 60 = 2*2*3*5. 1/4 becomes 15/60. 1/3 becomes 20/60. 1/5 becomes 12/60. Sixtieths give you pretty broad coverage for the most common fractions. Sixtieths will probably work for everything you'll run into.

I was about to say you could just call the field market_price_60th_copper_pieces, but looking at your repo, you decided to get fancy and make "gp" a proper type with its own internal translation. https://github.com/ocket8888/dndDB/blob/master/gp.c already does automatic division every time a value is output --- you internally store the value in copper pieces, and translate to gold pieces on-the-fly by dividing by 100. You could just as easily store the value internally as sixtieths of copper pieces. (Or, if https://github.com/ocket8888/dndDB/blob/master/items.sql already has all the items you want and you know you won't run into any more unexpected fractions, you can take the actual common denominator instead of hedging and making the common denominator 60 to be safe.)

In any case, the good news is that if you do later run into an unplanned-for fraction --- like if for some bizarre reason the market price of something is 1/7 --- it's not too hard to translate the existing database to use a smaller unit: just multiply every price by a new prime number (in this case 7). If you have hardcoded INSERT commands as in https://github.com/ocket8888/dndDB/blob/master/items.sql , just run a SELECT for the multiples and use the output mode that spits out INSERT commands --- in SQLite this is .mode insert, I don't know how to do it in Postgres but I assume Postgres must have something analogous.

For the sake of completeness, it's worth noting that the worth of D&D coins have no seigniorage --- a gold coin is a fiftieth of a pound, and a pound of gold is worth 50gp. The point of minting is purely to guarantee the purity of the metal (and its standard weight) with the reputation of the minting kingdom. The fancy design is a nice bonus that makes it nigh-impossible to shave off little bits of gold from the coin without people noticing. But at the end of the day the value of a coin is its weight of precious metal, so you could use any unit of weight as a unit of value. I don't recommend that, because units of weight are bedlam. We'd have to start talking about Troy pounds of gold versus avoirdupois pounds of gold, and you probably don't want to go down that rabbit hole.

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# American Institute of Mathematical Sciences

May  2013, 7(2): 219-229. doi: 10.3934/amc.2013.7.219

## New extremal binary self-dual codes of length $68$ from $R_2$-lifts of binary self-dual codes

 1 Department of Mathematics, Fatih University, 34500, Istanbul

Received  January 2013 Published  May 2013

A lift of binary self-dual codes to the ring $R_2$ is described. By using this lift, a family of self-dual codes over $R_2$ of length $17$ are constructed. Taking the binary images of these codes, extremal binary self-dual codes of length $68$ are obtained. For the first time in the literature, extremal binary codes of length $68$ with $\gamma=4$ and $\gamma = 6$ in $W_{68,2}$ have been obtained. In addition to these, six new codes with $\gamma = 0$ and fourteen new codes with $\gamma = 2$ in $W_{68,2}$ have also been found.
Citation: Suat Karadeniz, Bahattin Yildiz. New extremal binary self-dual codes of length $68$ from $R_2$-lifts of binary self-dual codes. Advances in Mathematics of Communications, 2013, 7 (2) : 219-229. doi: 10.3934/amc.2013.7.219
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 [1] Steven T. Dougherty, Cristina Fernández-Córdoba. Codes over $\mathbb{Z}_{2^k}$, Gray map and self-dual codes. Advances in Mathematics of Communications, 2011, 5 (4) : 571-588. doi: 10.3934/amc.2011.5.571 [2] Minjia Shi, Daitao Huang, Lin Sok, Patrick Solé. Double circulant self-dual and LCD codes over Galois rings. Advances in Mathematics of Communications, 2019, 13 (1) : 171-183. doi: 10.3934/amc.2019011 [3] Stefka Bouyuklieva, Anton Malevich, Wolfgang Willems. On the performance of binary extremal self-dual codes. Advances in Mathematics of Communications, 2011, 5 (2) : 267-274. doi: 10.3934/amc.2011.5.267 [4] Masaaki Harada, Akihiro Munemasa. On the covering radii of extremal doubly even self-dual codes. Advances in Mathematics of Communications, 2007, 1 (2) : 251-256. doi: 10.3934/amc.2007.1.251 [5] Stefka Bouyuklieva, Iliya Bouyukliev. Classification of the extremal formally self-dual even codes of length 30. Advances in Mathematics of Communications, 2010, 4 (3) : 433-439. doi: 10.3934/amc.2010.4.433 [6] Masaaki Harada, Katsushi Waki. New extremal formally self-dual even codes of length 30. Advances in Mathematics of Communications, 2009, 3 (4) : 311-316. doi: 10.3934/amc.2009.3.311 [7] Gabriele Nebe, Wolfgang Willems. On self-dual MRD codes. Advances in Mathematics of Communications, 2016, 10 (3) : 633-642. doi: 10.3934/amc.2016031 [8] Masaaki Harada, Akihiro Munemasa. Classification of self-dual codes of length 36. Advances in Mathematics of Communications, 2012, 6 (2) : 229-235. doi: 10.3934/amc.2012.6.229 [9] Nikolay Yankov, Damyan Anev, Müberra Gürel. Self-dual codes with an automorphism of order 13. Advances in Mathematics of Communications, 2017, 11 (3) : 635-645. doi: 10.3934/amc.2017047 [10] Steven T. Dougherty, Cristina Fernández-Córdoba, Roger Ten-Valls, Bahattin Yildiz. Quaternary group ring codes: Ranks, kernels and self-dual codes. Advances in Mathematics of Communications, 2019, 0 (0) : 0-0. doi: 10.3934/amc.2020023 [11] Amita Sahni, Poonam Trama Sehgal. Enumeration of self-dual and self-orthogonal negacyclic codes over finite fields. Advances in Mathematics of Communications, 2015, 9 (4) : 437-447. doi: 10.3934/amc.2015.9.437 [12] Delphine Boucher. Construction and number of self-dual skew codes over $\mathbb{F}_{p^2}$. Advances in Mathematics of Communications, 2016, 10 (4) : 765-795. doi: 10.3934/amc.2016040 [13] Ekkasit Sangwisut, Somphong Jitman, Patanee Udomkavanich. Constacyclic and quasi-twisted Hermitian self-dual codes over finite fields. Advances in Mathematics of Communications, 2017, 11 (3) : 595-613. doi: 10.3934/amc.2017045 [14] Lars Eirik Danielsen. Graph-based classification of self-dual additive codes over finite fields. Advances in Mathematics of Communications, 2009, 3 (4) : 329-348. doi: 10.3934/amc.2009.3.329 [15] Ilias S. Kotsireas, Christos Koukouvinos, Dimitris E. Simos. MDS and near-MDS self-dual codes over large prime fields. Advances in Mathematics of Communications, 2009, 3 (4) : 349-361. doi: 10.3934/amc.2009.3.349 [16] Joaquim Borges, Steven T. Dougherty, Cristina Fernández-Córdoba. Characterization and constructions of self-dual codes over $\mathbb Z_2\times \mathbb Z_4$. Advances in Mathematics of Communications, 2012, 6 (3) : 287-303. doi: 10.3934/amc.2012.6.287 [17] W. Cary Huffman. Additive self-dual codes over $\mathbb F_4$ with an automorphism of odd prime order. Advances in Mathematics of Communications, 2007, 1 (3) : 357-398. doi: 10.3934/amc.2007.1.357 [18] Christos Koukouvinos, Dimitris E. Simos. Construction of new self-dual codes over $GF(5)$ using skew-Hadamard matrices. Advances in Mathematics of Communications, 2009, 3 (3) : 251-263. doi: 10.3934/amc.2009.3.251 [19] Somphong Jitman, San Ling, Ekkasit Sangwisut. On self-dual cyclic codes of length $p^a$ over $GR(p^2,s)$. Advances in Mathematics of Communications, 2016, 10 (2) : 255-273. doi: 10.3934/amc.2016004 [20] Suat Karadeniz, Bahattin Yildiz. Double-circulant and bordered-double-circulant constructions for self-dual codes over $R_2$. Advances in Mathematics of Communications, 2012, 6 (2) : 193-202. doi: 10.3934/amc.2012.6.193

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text:
Lemma 70.10.8. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Assume

1. every weakly associated point of $X$ is a point of codimension $0$, and

2. $X$ satisfies the equivalent conditions of Morphisms of Spaces, Lemma 66.49.1,

3. every codimension $0$ point of $X$ can be represented by a monomorphism $\mathop{\mathrm{Spec}}(k) \to X$.

Let $X^0 \subset |X|$ be the set of codimension $0$ points of $X$. Then we have

$\mathcal{K}_ X = \bigoplus \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{O}_{X, \eta } = \prod \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{O}_{X, \eta }$

where $j_\eta : \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta }) \to X$ is the canonical map of Schemes, Section 26.13; this makes sense because $X^0$ is contained in the schematic locus of $X$. Similarly, for every quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we obtain the formula

$\mathcal{K}_ X(\mathcal{F}) = \bigoplus \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{F}_\eta = \prod \nolimits _{\eta \in X^0} j_{\eta , *}\mathcal{F}_\eta$

for the sheaf of meromorphic sections of $\mathcal{F}$. Finally, the ring of rational functions of $X$ is the ring of meromorphic functions on $X$, in a formula: $R(X) = \Gamma (X, \mathcal{K}_ X)$.

Proof. By Decent Spaces, Lemma 67.20.3 and Section 67.6 we see that $X$ is decent1. Thus $X^0 \subset |X|$ is the set of generic points of irreducible components (Decent Spaces, Lemma 67.20.1) and $X^0$ is locally finite in $|X|$ by (b). It follows that $X^0$ is contained in every dense open subset of $|X|$. In particular, $X^0$ is contained in the schematic locus (Decent Spaces, Theorem 67.10.2). Thus the local rings $\mathcal{O}_{X, \eta }$ and the morphisms $j_\eta$ are defined.

Observe that a locally finite direct sum of sheaves of modules is equal to the product. This and the fact that $X^0$ is locally finite in $|X|$ explains the equalities between direct sums and products in the statement. Then since $\mathcal{K}_ X(\mathcal{F}) = \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{K}_ X$ we see that the second equality follows from the first.

Let $j : Y = \coprod \nolimits _{\eta \in X^0} \mathop{\mathrm{Spec}}(\mathcal{O}_{X, \eta }) \to X$ be the product of the morphisms $j_\eta$. We have to show that $\mathcal{K}_ X = j_*\mathcal{O}_ Y$. Observe that $\mathcal{K}_ Y = \mathcal{O}_ Y$ as $Y$ is a disjoint union of spectra of local rings of dimension $0$: in a local ring of dimension zero any nonzerodivisor is a unit. Next, note that pullbacks of meromorphic functions are defined for $j$ by Lemma 70.10.7. This gives a map

$\mathcal{K}_ X \longrightarrow j_*\mathcal{O}_ Y.$

Let $U \in X_{\acute{e}tale}$ be affine. By Lemma 70.10.4 the left hand side evaluates to total ring of fractions of $\mathcal{O}_ X(U)$. On the other hand, the right hand side is equal to the product of the local rings of $U$ at the codimension $0$ points, i.e., the generic points of $U$. These two rings are equal (as we already saw in the proof of Lemma 70.10.4) by Algebra, Lemmas 10.25.4 and 10.66.7. Thus our map is an isomorphism.

Finally, we have to show that $R(X) = \Gamma (X, \mathcal{K}_ X)$. This follows from the case of schemes (Divisors, Lemma 31.23.6) applied to the schematic locus $X' \subset X$. Namely, the ring of rational functions of $X$ is by definition the same as the ring of rational functions on $X'$ as it is a dense open subspace of $X$ (see above). Certainly, $R(X')$ agrees with the ring of rational functions when $X'$ is viewed as a scheme. On the other hand, by our description of $\mathcal{K}_ X$ above, and the fact, seen above, that $X^0 \subset |X'|$ is contained in any dense open, we see that $\Gamma (X, \mathcal{K}_ X) = \Gamma (X', \mathcal{K}_{X'})$. Finally, use the compatibility recorded in Lemma 70.10.5. $\square$

[1] Conversely, if $X$ is decent, then condition (c) holds automatically.

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text:
SSAT Upper Level Math : How to find the area of a square

Example Questions

← Previous 1

Example Question #21 : Area Of Polygons

The volume of a cube is 1,000 cubic centimeters. Using the conversion factor 2.5 centimeters = 1 inch, give its surface area in square inches, rounding to the nearest square inch.

100 square inches

144 square inches

75 square inches

108 square inches

96 square inches

96 square inches

Explanation:

The surface area of a cube is six times the square of its sidelength, so we find the sidelength. This is the cube root of volume 1,000, so

centimeters.

To rewrite this as inches, divide by 2.5:

inches

The surface area of the cube in square inches is

square inches.

Example Question #1 : How To Find The Area Of A Square

The volume of a cube is 64 cubic inches. Find the side length of the cube and its surface area.

Explanation:

The volume of a cube is where is the length of one edge, so is the cube root of volume:

A cube has six faces and the surface area of a cube is . So we can write:

Surface area =

Example Question #1 : How To Find The Area Of A Square

A square has an area of 16 square inches. Give the diagonal of the square.

Explanation:

In order to determine the length of the diagonal of a square we would use the Pythagorean Theorem. First we should find the side length:

Now, "square" the length of one side and multiply by 2, then take the square root of that number to get the length of the diagonal:

Example Question #4 : How To Find The Area Of A Square

John is going to apply a fertilizer to his farm which has a dimension of 200 feet by 200 feet. Every pound of the fertilizer that he is going to use is sufficient for 40 square feet. If the fertilizer costs 2 dollars per pound, how much he should spend to fertilize his farm?

Explanation:

The area of the farm is:

square feet. So the amount of the fertilizer he needs can be calculated as:

pounds

Every pound of the fertilizer costs \$2, so he needs to spend dollars.

Example Question #5 : How To Find The Area Of A Square

The diagonal length of a square is . Find the area of the square in terms of .

Explanation:

We need to use the Pythagorean Theorem in order to solve this problem. We can write:

where is the diagonal length and is the side length. The diagonal length of the square is , so we can write:

Example Question #1 : How To Find The Area Of A Square

A square is inscribed inside a circle that has an area of  square inches. What is the area of the square?

Explanation:

Start by working backwards from the area of the circle to find its diameter.

The area of a circle is , and  (you can get this by trial and error pretty quickly, since you know it's between 3 and 4, and since 12.25 ends in a 5, you now that a 5 is involved), so the diameter of the circle is 7. This is also the diagonal of the square. Y

ou may remember that the legs of a  right triangle (of which we have two within the square, once we draw the diagonal) are equal to the hypotenuse divided by . But even if you forget this, you should recall the Pythagorean Theorem that states that . In this case,  and  are equal, so the square of each side is equal to half of , or 24.5. And the square of one side of a square is also equal to the area of the square.

Example Question #1 : How To Find The Area Of A Square

The perimter of a square is  . What is the area of the square?

Explanation:

Use the perimeter to find the side length of the square.

Now, use the side length to find the area of the square.

Example Question #791 : Geometry

The perimeter of a square is  . What is the area of the square?

Explanation:

First, use the perimeter to find the side lengths of the square.

Use this information to find the area.

Example Question #9 : How To Find The Area Of A Square

A square has the same perimeter as the above right triangle. Give the area of the square.

Explanation:

Since we know the lengths of one leg and the hypotenuse, we can calculate the length of the other leg using the Pythagorean Theorem. We can use this form:

Setting  and  equal to the lengths of the hypotenuse and the leg - 13 and 5, respectively:

The perimeter is equal to the sum of the lengths of the three sides:

This is also the perimeter of the square, so the length of each side is one fourth of this, or

The area is the square of this, or

Example Question #10 : How To Find The Area Of A Square

The perimeter of a square is equal to the circumference of a circle with area 4. What is the area of the square?

Explanation:

First, we need the radius  of the circle, which can be determined from the area of a circle formula by setting :

Simplifying the expression by splitting the radicand and rationalizing the denominator:

The circumference of the circle is  multiplied by the radius, or

This is also the perimeter of the square, so the length of each side is one fourth of this perimeter, or

The area of the square is the square of this common sidelength, or

.

← Previous 1


text:
# Divergence as transpose of gradient?

In his online lectures on Computational Science, Prof. Gilbert Strang often interprets divergence as the "transpose" of the gradient, for example here (at 32:30), however he does not explain the reason.

How is it that the divergence can be interpreted as the transpose of the gradient?

-

Here is a considerably less sophisticated point of view than some of the other answers. Recall that the dot product of vectors can be obtained by transposing the first vector. That is, $$\textbf{v}^T \textbf{w} \;=\; \begin{bmatrix}v_x & v_y & v_z\end{bmatrix}\begin{bmatrix}w_x \\ \\ w_y \\ \\w_z\end{bmatrix} \;=\; v_x w_x + v_y w_y + v_z w_z \;=\; \textbf{v}\cdot \textbf{w}.$$ (Here we are thinking of column vectors as being the "standard" vectors.)

In the same way, divergence can be thought of as involving the transpose of the $\nabla$ operator. First recall that, if $g$ is a real-valued function, then the gradient of $g$ is given by the formula $$\nabla g \;=\; \begin{bmatrix}\partial_x \\ \\ \partial_y \\ \\ \partial_z \end{bmatrix}g \;=\; \begin{bmatrix}\partial_xg \\ \\ \partial_yg \\ \\ \partial_zg \end{bmatrix}$$ Similarly, if $F=(F_x,F_y,F_z)$ is a vector field, then the divergence of $F$ is given by the formula $$\nabla^T F \;=\; \begin{bmatrix}\partial_x & \partial_y & \partial_z\end{bmatrix}\begin{bmatrix}F_x \\ \\ F_y \\ \\ F_z\end{bmatrix} \;=\; \partial_xF_x + \partial_yF_y + \partial_zF_z.$$ Thus, the divergence corresponds to the transpose $\nabla^T$ of the $\nabla$ operator.

This transpose notation is often advantageous. For example, the formula $$\nabla^T (gF) \;=\; (\nabla^T g)F \,+\, g(\nabla^T F)$$ (where $\nabla^T g$ is the transpose of the gradient of $g$) seems much more obvious than $$\text{div} (gF) \;=\; (\text{grad } g)\cdot F \,+\, g\;\text{div } F.$$ Indeed, this is the formula that leads to the integration by parts used in the video: $$\int\!\!\int g (\nabla^T F)\,dx\,dy \;=\; -\int\!\!\int (\nabla g)^T F \,dx\,dy.$$

-
I've been looking at those formulas for years, and that is probably the most intuitive form I've seen. To be clear, since $g$ is a "scalar", then $\nabla^T g = (\nabla g)^T$, or have I misinterpreted it? –  rcollyer Jun 12 '11 at 20:20
@rcollyer: Yes, since $g$ is a "scalar", $g^T$ is the same as $g$. Also, note that transpose usually reverses the order of multiplication, but it isn't necessary in this case since "scalars" and "vectors" commute. –  Jim Belk Jun 12 '11 at 20:25

A "dual pair" in functional analysis consists of a topological vector space E and its dual space $E'$ of continuous linear functionals, or some subspace of this.

That is for real vector spaces, for every element $e \in E$ and $e' \in E'$, we can write $$\langle e, e' \rangle \in \mathbb{R}$$ Example: Let $E$ be a Hilbert space, then $E = E'$ and the dual pairing is given by the scalar product.

In the case at hand we have two function spaces and the dual pairing is defined to be $$\int_{\Omega} u(x, y) v(x, y) d x d y$$ When you have some operator $$T: E \to E$$ it is often possible to define the "transposed operator" T' to be the operator $$T': E' \to E'$$ by the requirement that $$\langle T e, e' \rangle = \langle e, T' e' \rangle$$ for all e, e'. In the context of Hilbert spaces, it is more common to talk about "adjoint operators". The name "transpose" is motivated by the fact that for linear operators on finite dimensional vector spaces, the transpose is given by the transposed (conjugate, for complex ground field) matrix of the matrix that represents $T$ with respect to a fixed basis.

In the case at hand, when we write down $$\int_{\Omega} (- div \; grad u(x, y)) v(x, y) d x d y$$ you'll see that this is the same as $$\int_{\Omega} (grad \; u(x, y)) \cdot (grad \; v(x, y)) d x d y$$ by integration by parts, if the boundary terms are zero. The $\cdot$ denotes the canonical scalar product of vectors in $\mathbb{R}^n$. So, if the boundary terms are zero, we have

$$\langle - div \; e, e' \rangle = \langle e, grad \; e' \rangle$$ where - strictly speaking - the dual pairing on each side is different, because the first is a dual pairing of functions with values in $\mathbb{R}$, while the second is for functions with values in $\mathbb{R}^2$. But neglecting this technical detail, the operator grad is in this sense the transposed operator of the operator div.

-

Here is a discrete analogue of the situation in which one really can literally take the transpose of a matrix that is analogous to gradient and get a matrix that is analogous to divergence.

Let $G$ be a finite graph with vertex set $V$ and edge set $E$, and let $\mathbb{R}^V, \mathbb{R}^E$ be the vector spaces of functions $V \to \mathbb{R}$ resp. $E \to \mathbb{R}$. (Actually $\mathbb{R}^E$ is slightly more complicated than this: we want to be able to refer to an edge from $u$ to $v$ as both $uv$ and as $vu$, subject to the condition that $f(uv) = -f(vu)$.) Given a function $f \in \mathbb{R}^V$, which we think of as a discrete analogue of a scalar function on $G$, we can define

$$\text{grad}(f)(uv) = f(v) - f(u)$$

and this gives a function $\text{grad}(f) \in \mathbb{R}^E$, which we think of as a discrete analogue of the gradient, but which is more typically known as the (oriented) incidence matrix of $G$. Note here the "fundamental theorem of discrete line integrals": if $v_1 \to v_2 \to ... \to v_n$ is a path, then

$$\sum_{i=1}^{n-1} \text{grad}(f)(v_{i+1} v_i) = f(v_n) - f(v_1)$$

as expected. Now, both spaces $\mathbb{R}^V, \mathbb{R}^E$ come equipped with inner products given by

$$\langle a, b \rangle_V = \sum_{v \in V} a(v) b(v)$$

and

$$\langle a, b \rangle_E = \sum_{e \in E} a(e) b(e)$$

respectively. Generally speaking, if $A, B$ are a pair of inner product spaces and $T : A \to B$ is a linear operator, then under nice conditions there exists a unique linear operator $T^{\dagger} : B \to A$ such that

$$\langle Ta, b \rangle_B = \langle a, T^{\dagger} b \rangle_A.$$

$T^{\dagger}$ is the adjoint of $T$, and this is the abstract definition of the transpose of a matrix. You can verify that if you pick orthonormal bases of $A, B$ and write $T$ in terms of those bases, then $T^{\dagger}$ is precisely the transpose of $T$ in the usual sense. Thus the operator $\text{grad} : \mathbb{R}^V \to \mathbb{R}^E$ has an adjoint. If we simply take the transpose of the matrix representing $\text{grad}$ (with respect to the orthonormal bases given by the functions which are equal to $1$ on a particular vertex or edge and $0$ otherwise) we get that for $g \in \mathbb{R}^E$,

$$\text{div}(g)(u) = \sum_{uv \in E} g(uv).$$

If we think of $g$ as a flow on the graph $G$, then this is precisely a measure of how much is flowing in or out of a particular vertex, so is an appropriate discrete analogue of the divergence, and in fact a discrete analogue of the divergence theorem holds.

In multivariable calculus, something similar is happening as above, except that the spaces and inner products are more complicated.

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text:
# Choice of predictable (or jointly measurable) eigenvalues and eigenvectors of nuclear-operator-valued stochastic process

Let $q^{ij}$, $i,j\in\mathbb{N}$, be predictable real-valued stochastic processes. Let $(e^i)$, $i\in\mathbb{N}$ be an ONB of a separable Hilbert space $H$. Assume that $Q=\sum_{i,j=1}^\infty q^{ij}e_i\otimes e_j$ is a predictable, positive definite, self-adjoint, nuclear operator. If you do not 'like' predictability, forget about it, and replace the concept with jointly measurable. How would I go about proving that there exists a choice of both eigenvalues and eigenvectors which are predictable? A more general result is welcomed. Thank you in advance.

Let's consider, say, the whole set of pairs (eigenvalue, projection on the eigenspace) as a point process in a Polish space (with respect to strong topology on projections), prove its measurability, then extract its points one-by-one in a measurable way (which can be done directly for any locally finite point process in a Polish space), obtain bases for the eigenspaces by simple orthogonalization, and convert them into a single sequence.

Perhaps the most difficult part here is to prove that the point process is measurable. To achieve this I would use a well-known theorem of Lusin and Souslin (see Kehris "Classical descriptive set theory" (15.1) and (15.2)):

An injective Borel mapping of standard Borel spaces has Borel image and is an isomorphism onto its image.

Note that we have a double characterization of nuclear operators - as elements of the Polish space $S^1(H)$ of nuclear operators, and as suitable integer-valued measures on $\mathbb{R} \times P(H)$, where $P(H)$ is the Polish space of projections on $H$. That the conditions on the point processes define a Borel subset can be verified directly (the measurable structure on the space of measures is generated by evaluations, and it is standard Borel). Another easy thing to verify directly is that the map "point process $\mapsto$ operator" is Borel (here it is convenient to generate the measurable structure on $S^1$ by evaluations $A \mapsto \left\langle Ax,y\right\rangle,x,y\in H$). Hence by the Lusin-Souslin theorem, the inverse mapping "operator $\mapsto$ point process" is Borel.

I stumbled upon a proof in a book by Zabczyk and Peszat, entitled "Stochastic Partial Differential Equations driven by Levy Noise." I will provide the proof for the convenience of others.

This can be proved using the Kuratwoski-Ryll-Nardzesk section theorem:

Assume that $K$ is a compact metric space. Let $(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{P}})$ be a probability space, and let $\psi: K\times \tilde{\Omega}\rightarrow \mathbb{R}$ be a function such that, for every $x\in K$, $\psi(x,\cdot)$ is measurable and, for every $\tilde{\omega}\in \tilde{\Omega}$, $\psi(\cdot,\tilde{\omega})$ is continuous. Then there exists a $K$-valued random variable $\xi: \tilde{\Omega}\rightarrow K$ such that $$\psi(\xi(\tilde{\omega}),\tilde{\omega})=\sup_{k\in K} \psi(x,\tilde{\omega}),$$ for all $\tilde{\omega}\in \tilde{\Omega}$.

Let $Q_t(\omega)$ be a predictable process taking values in the space of positive nuclear operators. Also, let $Q$ be self-adjoint. Being nuclear, the random operator is also compact. Using the above theorem, given $T<\infty$, we can apply the above theorem to $\tilde{\Omega}=\Omega\times[0,T],$ $\tilde{\mathcal{F}}=\mathcal{P}_{[0,T]}$(the predictable sigma-algebra), and $\tilde{P}=T^{-1}\mathbb{P}\otimes dt$, to the set $K:=\{x\in U: |x|_H\le 1\}$ endowed with the weak topology (so compact by Alagou) and to the function $$\psi(x,\omega,t)=\psi(x,\tilde{\omega})=(Q_t(\omega)x,x)_H.$$ Then there exists a $H$-valued process $g^1_t(\omega)$, $t\in [0,T]$, $\omega\in \Omega$ such that $$(Q_t(\omega)g^1_t(\omega),g^1_t(\omega))_H=\sup (Q_t(\omega)y,y)=\gamma^1_t(\omega),$$ where the supremum is taken over all $y\in H$ with $|y|_H\le 1$. Using the random variables $\gamma^1, \gamma^2,\ldots, \gamma^n$ and $g^1,g^2,\ldots,g^n$, we can repeat this procedure for the operator-valued process $$Q_t(\omega)-\sum_{j=1}^n \gamma^j(\omega,t)g^j(\omega,t)\otimes g^j(\omega,t).$$ Note that the first eigenvalue is either $||Q_t(\omega)||_{op}$ or $-||Q_t(\omega)||_{op}$ and that the process ends because $\lim{n\rightarrow\infty} \lambda_n(t,\omega)=0$, where, for each $(t,\omega)$, $\lambda_n(t,\omega)$ are the eigenvalues of $Q_t(\omega)$.


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# Minimum of a linear congruence sub-sequence

I have the following little problem : let $a,b,u,v$ be four given integers with $\gcd(a,b)=1$. Now I would like to find the minimum of the linear congruence subsequence $\{ax \pmod b : u \le x \le v\}$. Of course, we have $0<u<v<b$ otherwise it is trivial.

I have some sort of algorithm but wonder if there have already been some work on such a problem ? some references ?

Thanks in advance, Eric

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 Welcome to MSE! Please note the LaTeX formatting I did to your question, LaTeX being the language of choice to write mathematics here. – Andreas Caranti Mar 2 at 13:37 I wonder whether you can make it better than $O(v - u)$. – Andreas Caranti Mar 2 at 14:07 I have an algorithm which is log(b) worst case. It is a funny little thing but was wondering if some mathematical work out there that has different approaches or may be even better. And thanks for the LaTeX editing. Just discovered the site. Wasn't aware of the LaTeX possibility. – Eric Andres Mar 2 at 14:49 That'a interesting. I'm not aware of any work in this direction, but then I'm no expert at all. As to the editing, you're welcome, it's a community here. – Andreas Caranti Mar 2 at 15:25