Why does $\frac{4}{2} = \frac{2}{1}$? 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 $\frac{4}{2} = \frac{2}{1}$?
 A: 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


*

*What a rational number is;

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

*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".
A: 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 $\mathbb Z$.
A: 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, $m_1n_2 − m_2n_1 = 0$.
A: 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 $\mathbb{Z}$.
A: 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 $$\frac{4}{2}=\frac{2}{1}$$
A: 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 $\frac{2}{1} \times \frac{2}{2} = \frac{4}{2} = \frac{2}{1} \times 1$.
