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text: '[summary: We have seen various operations of arithmetic on the rational numbers; in fact the operations are not as stand-alone as they may have appeared at first, but they are part of a closely-knit structure.]\n\nWe have seen [4zx what the rational numbers are], and five things we can do with them:\n\n- [55m addition]\n- [56x subtraction]\n- [59s multiplication]\n- [5jd division]\n- [5pk comparison]\n\nThese might seem like five standalone operations, but in fact they all play nicely together in a particular way.\nYou don't need to know the fancy name, but here it is anyway: mathematicians say that the rationals **form an [ordered_field ordered field]**, meaning that the five operations above:\n\n- work in the rationals\n- slot together, behaving in certain specific ways that make it easy to calculate\n\n*You shouldn't bother learning this page particularly deeply, because none of the properties alone is very interesting; instead, try and absorb it as a whole.*\n\nWe've already seen the "instant rules" for manipulating the rationals, and where they come from.\nHere, we'll first quickly define the rational numbers themselves, and all the operations above, in "instant rule" format, just to have them all here in one place.\nThen we'll go through all the properties that are required for mathematicians to be able to say that the operations on rationals "play nicely together" in the above sense; and we'll be relying on the instant rules as our definitions, because they're totally unambiguous.\nThat way we can be much more sure that we're not making some small error.\n(Relying on an intuition about how apples work could in theory lead us astray; but the rules leave no wiggle-room or scope for interpretation.)\n\nThe letters $a, b, c, d$ should be read as standing for integers (possibly positive, negative or $0$), and $b$ and $d$ should be assumed not to be $0$ (recalling that [5jd it makes no sense] to divide by $0$).\n\n- A **rational number** is a pair of [53r integers], written as $\\frac{a}{b}$, where $b$ is not $0$, such that $\\frac{a}{b}$ is viewed as being the same as $\\frac{c}{d}$ precisely when $a \\times d = b \\times c$. %%note:This is just the instant rule for subtraction, below, together with the assumption that $\\frac{0}{x} = \\frac{0}{y}$ for any $x, y$ nonzero integers; in particular, the assumption that $\\frac{0}{b \\times d} = \\frac{0}{1}$ when $b, d$ are not $0$, which we will need later.%%\n- Addition: $$\\frac{a}{b} + \\frac{c}{d} = \\frac{a \\times d + b \\times c} {b \\times d}$$\n- Subtraction: $$\\frac{a}{b} - \\frac{c}{d} = \\frac{a}{b} + \\frac{-c}{d} = \\frac{a \\times d - b \\times c}{b \\times d}$$\n- Multiplication: $$\\frac{a}{b} \\times \\frac{c}{d} = \\frac{a \\times c}{b \\times d}$$\n- Division (where also $c$ is not $0$): $$\\frac{a}{b} \\big/ \\frac{c}{d} = \\frac{a}{b} \\times \\frac{d}{c} = \\frac{a \\times d}{b \\times c}$$\n- Comparison (where we write $\\frac{a}{b}$ and $\\frac{c}{d}$ such that both $b$ and $d$ are positive %%note:Remember, we can do that: if $b$ is negative, for instance, we may instead write $\\frac{a}{b}$ as $\\frac{-a}{-b}$, and the extra minus-sign we introduce has now flipped $b$ from being negative to being positive.%%): $\\frac{a}{b} < \\frac{c}{d}$ precisely when $\\frac{c}{d}-\\frac{a}{b}$ is positive: that is, when $$\\frac{b \\times c - a \\times d}{b \\times d} > 0$$\nwhich is in turn when $$b \\times c - a \\times d > 0$$\n\nTo be super-pedantic, it is necessary to show that the rules above are "well-defined": for instance, you don't get different answers if you apply the rules using $\\frac{2}{4}$ instead of $\\frac{1}{2}$.\nConcretely, for example, $$\\frac{2}{4} + \\frac{1}{3} = \\frac{1}{2} + \\frac{1}{3}$$ according to the instant rules.\nIt's actually true that they *are* well-defined in this sense, but we won't show it here; consider them to be exercises.\n\nAdditionally, each integer can also be viewed as a rational number: namely, the integer $n$ can be viewed as $\\frac{n}{1}$, being "$n$ copies of the $\\frac{1}{1}$-chunk".\n\nThe properties required for an "ordered field"—that is, a nicely-behaved system of arithmetic in the rationals—fall naturally into groups, which will correspond to main-level headings below.\n\n[toc:]\n\n# How addition behaves\n\n## Addition always spits out a rational number %%note:Confucius say that man who run in front of bus get tired; man who run behind bus get exhausted. On the other hand, mathematicians say that the rationals are *closed* under addition.%%\n\nYou might remember a certain existential dread mentioned in the [4zx intro to rational numbers]: whether it made sense to add rational numbers at all.\nThen we saw in [55m] that in fact it does make sense.\n\nHowever, in this page we are working from the instant rules above, so if we want to prove that "addition spits out a rational number", we actually have to prove the fact that $\\frac{a \\times d + b \\times c} {b \\times d}$ is a rational number; no mention of apples at all.\n\nRemember, a rational number here (as defined by instant rule!) is a pair of integers, the bottom one being non-zero.\nSo we just need to check that $a \\times d + b \\times c$ is an integer and that $b \\times d$ is a non-zero integer, where $a, b, c, d$ are integers and $b, d$ are also not zero.\n\nBut that's easy: the product of integers is integer, and the product of integers which aren't zero is not zero.\n\n## There must be a rational $0$ such that adding $0$ to anything doesn't change the thing %%note:Mathematicians say that there is an [-54p] for addition.%%\n\nThis tells us that if we want to stay where we are, but we absolutely must add something, then we can always add $0$; this won't change our answer.\nMore concretely, $\\frac{5}{6} + 0 = \\frac{5}{6}$.\n\nStrictly speaking, I'm pulling a bit of a fast one here; if you noticed how, then *very* well done.\nA rational number is a *pair* of integers, such that the bottom one is not $0$; and I've just tried to say that $0$ is our rational such that adding $0$ doesn't change rationals.\nBut $0$ is not a pair of integers!\n\nThe trick is to use $\\frac{0}{1}$ instead.\nThen we can verify that $$\\frac{0}{1} + \\frac{a}{b} = \\frac{0 \\times b + a \\times 1}{1 \\times b} = \\frac{0 + a}{b} = \\frac{a}{b}$$ as we require.\n\n## Every rational must have an anti-rational under addition %%note:Mathematicians say that addition is *invertible*.%%\n\nConcretely, every rational $\\frac{a}{b}$ must have some rational $\\frac{c}{d}$ such that $\\frac{a}{b} + \\frac{c}{d} = \\frac{0}{1}$.\nWe've [56x already seen] that we can define $-\\frac{a}{b}$ to be such an "anti-rational", but remember that in order to fit in with our instant rules above, $-\\frac{a}{b}$ is also not a pair of integers; we should instead use $\\frac{-a}{b}$.\n\nThen $$\\frac{a}{b} + \\frac{-a}{b} = \\frac{a \\times b + (-a) \\times b}{b \\times b} = \\frac{0}{b \\times b}$$\n\nFinally, we want $\\frac{0}{b \\times b} = \\frac{0}{1}$; but this is immediate by the instant-rule definition of "rational number", since $0 \\times 1 = 0 \\times (b \\times b)$ (both being equal to $0$).\n\n## Addition doesn't care which way round we do it %%note:Mathematicians say that addition is [3jb commutative].%%\n\nConcretely, this is the fact that $$\\frac{a}{b} + \\frac{c}{d} = \\frac{c}{d} + \\frac{a}{b}$$\nThis is intuitively plausible because "addition" is just "placing apples next to each other", and if I put five apples down and then seven apples, I get the same number of apples as if I put down seven and then five.\n\nBut we have to use the instant rules now, so that we can be sure our definition is completely watertight.\n\nSo here we go: $$\\frac{a}{b} + \\frac{c}{d} = \\frac{a \\times d + b \\times c}{b \\times d} = \\frac{c \\times b + d \\times a}{d \\times b} = \\frac{c}{d} + \\frac{a}{b}$$\nwhere we have used the fact that multiplication of *integers* doesn't care about the order in which we do it, and similarly addition of *integers*.\n\n## Addition doesn't care about the grouping of terms %%note:Mathematicians say that addition is [3h4 associative].%%\n\nHere, we mean that $$\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\frac{e}{f} = \\frac{a}{b} + \\left( \\frac{c}{d} + \\frac{e}{f} \\right)$$\nThis is an intuitively plausible fact, because "addition" is just "placing apples next to each other", and if I put down three apples, then two apples to the right, then one apple to the left, I get the same number ($6$) of apples as if I had put down one apple on the left, then three apples in the middle, then two on the right.\n\nBut we have to use the instant rules now, so that we can be sure our definition is completely watertight.\nSo here goes, starting from the left-hand side:\n$$\\left(\\frac{a}{b} + \\frac{c}{d}\\right) + \\frac{e}{f} = \\frac{a \\times d + b \\times c}{b \\times d} + \\frac{e}{f} = \\frac{(a \\times d + b \\times c) \\times f + (b \\times d) \\times e}{(b \\times d) \\times f}$$\n\nSince addition and multiplication of *integers* don't care about grouping of terms, this is just $$\\frac{a \\times d \\times f + b \\times c \\times f + b \\times d \\times e}{b \\times d \\times f}$$\n\nNow going from the right-hand side:\n$$\\frac{a}{b} + \\left( \\frac{c}{d} + \\frac{e}{f} \\right) = \\frac{a}{b} + \\frac{c \\times f + d \\times e}{d \\times f} = \\frac{a \\times (d \\times f) + b \\times (c \\times f + d \\times e))}{b \\times (d \\times f)}$$\nand similarly we can write this $$\\frac{a \\times d \\times f + b \\times c \\times f + b \\times d \\times e}{b \\times d \\times f}$$\nwhich is the same as we got by starting on the left-hand side.\n\nSince we showed that both the left-hand and the right-hand side are equal to the same thing, they are in fact equal to each other.\n\n# How multiplication behaves\n\n## Multiplication always spits out a rational number %%note:Mathematicians say that the rationals are *closed* under multiplication.%%\nThis one is very easy to show: $\\frac{a}{b} \\times \\frac{c}{d} = \\frac{a \\times c}{b \\times d}$, but $b \\times d$ is not zero when $b$ and $d$ are not zero, so this is a valid rational number.\n\n## There must be a rational $1$ such that multiplying a thing by $1$ doesn't change the thing %%note:Mathematicians say that multiplication has an [-54p].%%\n\nFrom our motivation of what multiplication was ("do unto $\\frac{a}{b}$ what you would otherwise have done unto $1$"), it should be clear that the number $1$ ought to work here.\nHowever, $1$ is not strictly speaking a rational number according to the letter of the "instant rules" above; so instead we use $\\frac{1}{1}$.\n\nThen $$\\frac{1}{1} \\times \\frac{a}{b} = \\frac{1 \\times a}{1 \\times b} = \\frac{a}{b}$$ since $1 \\times n = n$ for any integer $n$.\n\n## Every nonzero rational must have a corresponding rational such that when we multiply them, we get $1$ %%note:Mathematicians say that every nonzero rational has an *inverse*. (Multiplication is not quite *invertible*, because $0$ has no inverse.)%%\n[todo: recall the intuition from multiplication]\n\nUsing the instant rule, though, we can take our nonzero rational $\\frac{a}{b}$ (where neither $a$ nor $b$ is zero; this is forced by the requirement that $\\frac{a}{b}$ be nonzero), and use $\\frac{b}{a}$ as our corresponding rational.\nThen $$\\frac{a}{b} \\times \\frac{b}{a} = \\frac{a\\times b}{b \\times a} = \\frac{a \\times b}{a \\times b} = \\frac{1}{1}$$\n\n## Multiplication doesn't care which way round we do it %%note:Mathematicians say that multiplication is [3jb commutative].%%\nThis had its own section on the [59s multiplication page], and its justification there was by rotating a certain picture.\nHere, we'll use the instant rule:\n$$\\frac{a}{b} \\times \\frac{c}{d} = \\frac{a \\times c}{b \\times d} = \\frac{c \\times a}{d \\times b} = \\frac{c}{d} \\times \\frac{a}{b}$$\nwhere we have used that multiplication of *integers* doesn't care about order.\n\n## Multiplication doesn't care about the grouping of terms %%note:Mathematicians say that multiplication is [3h4 associative].%%\n\nThis is much harder to see from our intuition, because while addition is a very natural operation ("put something next to something else"), multiplication is much less natural (it boils down to "make something bigger").\nIn a sense, we're trying to show that "make something bigger and then bigger again" is the same as "make something bigger by a bigger amount".\nHowever, it turns out to be the case that multiplication *also* doesn't care about the grouping of terms.\n\nUsing the instant rule for multiplication, it's quite easy to show that $$\\frac{a}{b} \\times \\left(\\frac{c}{d} \\times \\frac{e}{f} \\right) = \\left(\\frac{a}{b} \\times \\frac{c}{d} \\right) \\times \\frac{e}{f}$$\n\nIndeed, working from the left-hand side:\n$$\\frac{a}{b} \\times \\left(\\frac{c}{d} \\times \\frac{e}{f} \\right) = \\frac{a}{b} \\times \\frac{c \\times e}{d \\times f} = \\frac{a \\times (c \\times e)}{b \\times (d \\times f)}$$\nwhich is $\\frac{a \\times c \\times e}{b \\times d \\times f}$ because multiplication of *integers* doesn't care about the grouping of terms.\n\nOn the other hand, $$\\left(\\frac{a}{b} \\times \\frac{c}{d} \\right) \\times \\frac{e}{f} = \\frac{a \\times c}{b \\times d} \\times \\frac{e}{f} = \\frac{(a \\times c) \\times e}{(b \\times d) \\times f} = \\frac{a \\times c \\times e}{b \\times d \\times f}$$\n\nThese two are equal, so we have shown that the left-hand and right-hand sides are both equal to the same thing, and hence they are the equal to each other.\n\n# How addition and multiplication interact\n\n## Multiplication "filters through" addition %%note:Mathematicians say that multiplication *distributes over* addition.%%\n\nWhat we will show is that $$\\left(\\frac{c}{d} + \\frac{e}{f}\\right) \\times \\frac{a}{b} = \\left(\\frac{a}{b} \\times \\frac{c}{d}\\right) + \\left(\\frac{a}{b} \\times \\frac{e}{f}\\right)$$\n\nThis is intuitively true: the left-hand side is "make $\\frac{a}{b}$, but instead of starting out with $1$, start with $\\left(\\frac{c}{d} + \\frac{e}{f}\\right)$"; while the right-hand side is "make $\\frac{a}{b}$, but instead of starting out with $1$, start with $\\frac{c}{d}$; then do the same but start with $\\frac{e}{f}$; and then put the two together".\nIf we draw out diagrams for the right-hand side, and put them next to each other, we get the diagram for the left-hand side.\n(As an exercise, you should do this for some specific values of $a,b,c,d,e,f$.)\n\nWe'll prove it now using the instant rules.\n\nThe left-hand side is $$\\left(\\frac{c}{d} + \\frac{e}{f}\\right) \\times \\frac{a}{b} = \\frac{c \\times f + d \\times e}{d \\times f} \\times \\frac{a}{b} = \\frac{(c \\times f + d \\times e) \\times a}{(d \\times f) \\times b}$$\nwhich is $\\frac{c \\times f \\times a + d \\times e \\times a}{d \\times f \\times b}$.\n\nThe right-hand side is $$\\left(\\frac{a}{b} \\times \\frac{c}{d}\\right) + \\left(\\frac{a}{b} \\times \\frac{e}{f}\\right) = \\frac{a \\times c}{b \\times d} + \\frac{a \\times e}{b \\times f} = \\frac{(a \\times c) \\times (b \\times f) + (b \\times d) \\times (a \\times e)}{(b \\times d) \\times (b \\times f)} = \\frac{(a \\times c \\times b \\times f) + (b \\times d \\times a \\times e)}{(b \\times d \\times b \\times f)}$$\nNotice that everything on the top has a $b$ in it somewhere, so we can use this same "filtering-through" property that we already know the *integers* have:\n$$\\frac{b \\times [(a \\times c \\times f) + (d \\times a \\times e)]}{b \\times (d \\times b \\times f)}$$\nAnd finally we know that multiplying a fraction's numerator and denominator both by $b$ doesn't change the fraction:\n$$\\frac{(a \\times c \\times f) + (d \\times a \\times e)}{d \\times b \\times f}$$\n\nThis is just the same as the left-hand side, after we rearrange the product terms.\n\n# How comparison interacts with the operations\n\n## How comparison interacts with addition\nThe founding principle for how we came up with the definition of "comparison" was that adding the same thing to two sides of a balance scale didn't change the balance.\n\nIn terms of the instant rules, that becomes: if $\\frac{a}{b} < \\frac{c}{d}$, then $\\frac{a}{b} + \\frac{e}{f} < \\frac{c}{d} + \\frac{e}{f}$.\n\nBut by our instant rule, this is true precisely when $$0 < \\frac{c}{d} + \\frac{e}{f} - (\\frac{a}{b} + \\frac{e}{f})$$ and since multiplication by $-1$ filters through addition, that is precisely when $$0 < \\frac{c}{d} + \\frac{e}{f} - \\frac{a}{b} - \\frac{e}{f}$$\nwhich (since addition doesn't care about which way round we do the additions, and subtraction is just a kind of addition) is precisely when $$0 < \\frac{c}{d} - \\frac{a}{b} + \\frac{e}{f} - \\frac{e}{f}$$\ni.e. when $0 < \\frac{c}{d} - \\frac{a}{b}$; i.e. when $\\frac{a}{b} < \\frac{c}{d}$.\n\n## How comparison interacts with multiplication\n\nThe aim here is basically to show that we can't multiply two things and get an anti-thing.\nWritten out in the notation, we wish to show that if $0 < \\frac{a}{b}$ and if $0 < \\frac{c}{d}$ then $0 < \\frac{a}{b} \\times \\frac{c}{d}$, since the test for anti-ness is "am I less than $0$?".\n\nSince $0 < \\frac{a}{b}$, there are two options: either both $a$ and $b$ are positive, or they are both negative.\n(If one is negative and one is positive, then the fraction will be negative.)\n\nLikewise either both $c$ and $d$ are positive, or both are negative.\n\nSo we have four options in total:\n\n- $a, b, c, d$ are positive\n- $a, b, c, d$ are negative\n- $a, b$ are positive; $c, d$ are negative\n- $a, b$ are negative; $c, d$ are positive.\n\nIn the first case, we have $\\frac{a \\times c}{b \\times d}$ positive, because all of $a, b, c, d$ are.\n\nIn the second case, we have $a \\times c$ positive and $b \\times d$ also positive (because both are two negative integers multiplied together); so again the fraction $\\frac{a \\times c}{b \\times d}$ is positive.\n\nIn the third case, we have $a \\times c$ negative and $b \\times d$ also negative (because both are a positive number times a negative number); so the fraction $\\frac{a \\times c}{b \\times d}$ is a negative divided by a negative.\nTherefore it is positive, because we can multiply the numerator and the denominator by $-1$ to turn it into a positive divided by a positive.\n\n%%hidden(Example):\nWe'll consider $\\frac{1}{3}$ and $\\frac{-2}{-5}$.\nThen the product is $\\frac{1}{3} \\times \\frac{-2}{-5} = \\frac{-2}{-15}$; but that is the same as $\\frac{2}{15}$.\n%%\n\nIn the fourth case, we can do the same as the above, or we can be a bit sneaky: using the fact from earlier that multiplication doesn't care about order, we can note that $\\frac{a}{b} \\times \\frac{c}{d}$ is the same as $\\frac{c}{d} \\times \\frac{a}{b}$.\nBut now $c$ and $d$ are positive, and $a$ and $b$ are negative; we've already shown (in the previous case) that if the first fraction is "positive divided by positive", and the second fraction is "negative divided by negative".\nBy swapping $a$ for $c$, and $b$ for $d$, we can use a result we've already proved to obtain this final case.\n\nHence we've shown all four possible cases, and so the final result follows.',
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