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I take for granted that $\frac{4}{2} = \frac{2}{1}$.

Today, I thought about why it must be the case. My best answers amounted to $\frac{4}{2}=2$ and $\frac{2}{1}=2$; therefore $\frac{4}{2}=\frac{2}{1}$. However, that explanation seems circular:

  • one can express $2$ as $\frac{2}{1}$.
  • As such, to say $\frac{4}{2}$ equals $\frac{2}{1}$ because both equal $2$, is nearly saying $\frac{4}{2} = \frac{2}{1}$ (the question) and $\frac{2}{1}=\frac{2}{1}$ (trivial, at best).

So why does $\frac{4}{2} = \frac{2}{1}$?

usermath
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Hal
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  • I don't think there's anything profound going on here (other than using the definition of division), but if someone can show me otherwise be my guest. – MT_ May 17 '14 at 03:45
  • Your explanation is not circular. If two numbers are equal to the same number, then those two numbers are equal (symbolically, $a = c$ and $b = c$ implies $a = b$). –  May 17 '14 at 03:45
  • @nsanger Hm. Perhaps I could also have asked my question by asking how we justify writing the number that 4/2 and 2/1 both denote in those two ways (and all the other ways one can write it). – Hal May 17 '14 at 03:50
  • Cross multiplication gives you 4 = 4 which is true. – zerosofthezeta May 17 '14 at 04:28
  • because 4 is a multiple of 2 and thus can be divided by 2. and 2 is a multiple of 2 as well and so can also be divided by 2 hence 4/2 = 2 and 2/2 = 1 so you get 2/1 and 2 divided by 1 is 2 so the answer is 2 – Kennan May 17 '14 at 04:51
  • Half of four things is all of two things. – anon May 17 '14 at 05:02
  • I think you might do well to list what properties of rational numbers you know and accept. The property you post about you clearly know, but are dubious about. But without making precise the exact nature of your confusion, I fear you may get answers which are over your head (you may have already!).

    The exercise may also eliminate the confusion itself, as often happens

     – user139388 May 17 '14 at 05:09

6 Answers6

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Actually, the reason $\boldsymbol{\frac{4}{2} = \frac{2}{1}}$ is that we define it to be so.

What do I mean? Suppose you know what the integers are, and you want to define the rational numbers from that. How do you do it? Well, you need to define

  1. What a rational number is;

  2. What it means when you write $x + y$ or $x \cdot y$ when $x, y$ are rational;

  3. What it means for two rational numbers to be equal.

For (1), you define the rational numbers as the set of ordered pairs $(p,q)$, where $(p,q)$ represents the numbers $\frac{p}{q}$, and you require that $q \ne 0$. For (2), you define addition and multiplication in the usual way. And for (3), you define $$ (p,q) = (r, s) \iff ps = qr $$ i.e., in the more familiar notation $$ \frac{p}{q} = \frac{r}{s} \iff ps = qr $$ so in particular $$ \frac{4}{2} = \frac{2}{1} \text{ since } 4 \cdot 1 = 2 \cdot 2. $$

I suppose a more interesting question is, why do we define it this way? Well, we want every rational number to have a unique additive and multiplicative inverse, we want addition and multiplication to be associative, and so on. And we want the notation $\frac{3}{4}$ to capture what we mean when we say that it is "three fourths" or "three parts out of four".

Caleb Stanford
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  • You don't define $3$)--I think that is incorrect. You define the division operation as the inverse of multiplication--that's where you need to start. We don't "define" $\frac{4}{2} = 2$, we infer it from properties of multiplication and its inverse (division). Division defines the rational numbers (in terms of integers). – Jared May 17 '14 at 04:41
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    @Jared The standard approach is to define an equivalence class on the set of ordered pairs, which is the notion of equality I am referring to in (3). You cannot conclude $\frac{4}{2} = \frac{2}{1}$ if you don't know what the rational numbers are. See also the answers from Julien and user128390. – Caleb Stanford May 17 '14 at 04:45
  • You can define division in terms of this equivalence class or you can define this equivalence class in terms of division--the two are equivalent (I think). I think it's somewhat silly to say that we first think of this equivalence class when we have a well defined notion of what division is. Division is "easy" to understand (as the inverse of multiplication). To me, this leads you to say you have an equivalence class not the other way around. In a sense it's the the egg vs. the chicken, but I think we clearly know about division before the equivalence class for the rational numbers. – Jared May 17 '14 at 04:49
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    @Jared Perhaps I don't understand what you are saying, but it is genuinely impossible to define what division on the rational numbers is if you don't know what the rational numbers are in the first place. Even if it is easy to understand and you "clearly know" about it, it is not well-defined. Else, how would you define the rational numbers? – Caleb Stanford May 17 '14 at 04:51
  • in my opinion, division is what defines the rational numbers. So what is "one divided by two"? It is the value $x$ such that $1 = 2\cdot x$ (this is an inverse of multiplication). As long as the two numbers you are dividing are integers then this $x$ will be a rational number (by definition). We can do "six divided by two equals $x$": $6 = 2x \rightarrow x = 3$ (there's no way to get that value other than through guess and test) but we do know that whatever $x$ is, it will be a rational number. – Jared May 17 '14 at 05:03
  • And as another example, when we try to solve the problem $3x = 2 \rightarrow x = \frac{2}{3}$, the solution, $\frac{2}{3}$, is really just as illusory as solving the equation $x^2 = 3 \rightarrow x = \sqrt{3}$. $\sqrt{3}$ is whatever real number happens to solve the problem and likewise the rational number $\frac{2}{3}$ is whatever rational number that happens to solve the problem. Problems that give integer solutions are special just as square root problems are special, e.g. $2x = 6 \rightarrow x = 3$ and $x^2 = 25 \rightarrow x = 5$. – Jared May 17 '14 at 05:08
  • @Jared It is not a rigorous definition to define $\frac{1}{2}$ as "the value $x$ such that $1 = 2 \cdot x$". What do you mean by "value"? You mean "rational number". But the rational numbers are exactly what you are trying to define...now, it is true that for example with $\sqrt{3}$ in field theory, you define this precisely as a number $x$ such that $x^2 = 3$. But you must make this rigourous, and I have not seen a similar thing done with rational numbers, and I expect this will either (1) amount to the same equivalence class definition when it is made rigorous, or (contd) – Caleb Stanford May 17 '14 at 05:10
  • (contd) (2) be ultimately a much messier approach. – Caleb Stanford May 17 '14 at 05:11
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    @Jared The problem has to do with the notation we use for fractions. If you want to define rational numbers like you say, with $\frac{1}{2}$ denoting the value that multiplies by $2$ to get $1$, and $\frac{2}{4}$ denoting the value that multiplies by $4$ to get $2$, then you need to be able to prove these two values are equal. And in order to do so, you need to define what equal means. Thus you cannot avoid defining an equivalence class (a notion of equality), and I can think of no more straightforward way to define the equivalence class than $\frac{a}{b} = \frac{c}{d}$ if $ad = bc$. – Caleb Stanford May 17 '14 at 05:13
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    You can prove $\frac{a}{b} = \frac{c}{d} \rightarrow ad = bc$ ($b, d \neq 0$) with the definition of division as the inverse of multiplication (along with associativity, commutativity, and the identity of multiplication)--that's my point. This is not a definition, this is a property of division. And I think I understand what you are saying: that division can be defined by the above equivalence--I just don't think that's how people arrived at it (I think it's the other way). – Jared May 17 '14 at 05:27
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    @Jared Of course that's how people arrived at it but it isn't a rigorous definition in any way. "Definition of division as the inverse of multiplication" Multiplication on what? The integers? You can't define multiplication OR division OR anything on the rational numbers if you haven't defined what the rational numbers are. – Caleb Stanford May 17 '14 at 05:32
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In constructing $\mathbb Q$, you think of rational numbers as pairs $(a,b)$ of integers, where $a$ is the numerator and $b$ is the denominator -- so you think of rational numbers as elements of ${\mathbb Z}\times{\mathbb Z}$. But that's not the whole picture, a rational number actually corresponds to an equivalence class in ${\mathbb Z}\times{\mathbb Z}$, under the equivalence relation given by $(a,b)\sim(c,d)$ if and only if $ad=bc$. So in your case, $4/2=2/1$ because $4\cdot 1=2\cdot 2$. All you need in order to make sense of this is multiplication in $\mathbb Z$.

user128390
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From wikipedia:

The rational numbers can be formally defined as the equivalence classes of the quotient set $(\mathbb{Z} × (\mathbb{Z} \setminus \{0\})) /\sim$, where the cartesian product $(\mathbb{Z} × (\mathbb{Z} \setminus \{0\}))$ is the set of all ordered pairs $(m,n)$ where $m$ and $n$ are integers, $n$ is not 0, and " $\sim$ " is the equivalence relation defined by $(m_1,n_1) \sim (m_2,n_2)$ if, and only if, $m_1n_2 − m_2n_1 = 0$.

hasnohat
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I believe every single answer so far misaddresses the OP's question.

To OP: Your reasoning is not circular. It is perfectly valid, and therefore you have answered your own question. In mathematics, when $a = c$ and $b = c$, it follows that $a = b$. This is because equality is an equivalence relation.

There is no need to bring up the definition of $\mathbb{Q}$ as a field of quotients of $\mathbb{Z}$.

user148392
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  • I agree. If a=c, and b=c, then a=b by hypothetical syllogism. As such, I might rephrase my question to ask 'How does one know whether two different arrangements of numerals express the same rational number?' – Hal May 17 '14 at 14:25
  • @Hal Do you agree that $2/1 = 2$ and $4/2 = 2$? Then you answered your own question. Do not over think. – user148392 May 17 '14 at 16:37
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The set $\mathbb{Q}$ is a group then for all $x\in \mathbb{Q}$, there is a only Inverse element such that $x+(-x)=0$. In this case note that $$\frac{4}{2}-\frac{2}{1}=0$$ then you conclude that $$\frac{4}{2}=\frac{2}{1}$$

AsdrubalBeltran
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    But how would you subtract the two fractions without assuming $\frac{2}{1} = \frac{4}{2}$? – PPP May 17 '14 at 03:43
  • By definition $\frac{a}{b}-\frac{c}{d}=\frac{ad-bc}{bd}$ for any two fractions. – AsdrubalBeltran May 17 '14 at 03:49
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    I don't see this as a definition. Maybe in other algebras, but if we want it to work perfectly with 'standard algebra', we need to make some assumptions to get this formula. – PPP May 17 '14 at 03:52
  • you accept, the definitions of usual product of rational numbers, and the usual division of rational numbers, but the addition and subtraction not? – AsdrubalBeltran May 17 '14 at 04:06
  • $\frac{a}{b} - \frac{c}{d} = \frac{ad - bc}{bd}$ is certainly not a definition--it's an inference from more elementary axioms. – Jared May 17 '14 at 04:43
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If $c \neq 0$, do you agree that $\frac{c}{c} = 1$? Do you also agree that 1 is a unit and a multiplicative identity element?

Given any rational number $\frac{a}{b}$, if we multiply both the numerator and the denominator by the same number, we're computing $\frac{a}{b} \times \frac{c}{c} = \frac{a}{b} \times 1$. So, if we have $c = 2$, and we compute $\frac{2}{1} \times \frac{c}{c}$, we have $\frac{2}{1} \times \frac{2}{2} = \frac{4}{2} = \frac{2}{1} \times 1$.

Robert Soupe
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