How to convince a math teacher of this simple and obvious fact?

Solution 1:

You can prove that all the numbers are equal ;-)

Let's assume that for all $a,b,c,d \in \mathbb{R}$, $b \neq 0$, $d \neq 0$ we have

$$ \frac{a}{b} = \frac{c}{d}\quad \text{ implies }\quad a = c\ \text{ and }\ b = d. \tag{$\spadesuit$}$$

Now take any two numbers, say $p$ and $q$, and write

$$\frac{p}{p} = \frac{q}{q}.$$

Using claim $(\spadesuit)$ we have $p = q$. For the special case, where one of them equals zero (e.g. $q$), use $$\frac{2p}{2p} = \frac{p+q}{p+q}.$$

I hope this helps ;-)

Solution 2:

Given $a, c \in \mathbb{Z}$ and $b, d \in \mathbb{N}$, suppose that \begin{align} \frac{a}{b} = \frac{c}{d} \Longrightarrow a = c, \, b = d. \end{align} Thus, \begin{align} \frac{a}{b} = 1 \cdot \frac{a}{b} = \frac{2}{2} \cdot \frac{a}{b} = \frac{2a}{2b} \Longrightarrow b = 2 b, \end{align} and $1 = 2$ (as $b$ is non-zero), which is absurd.

Solution 3:

Say $$\frac { a }{ b } =\frac { c }{ d } =k,$$ then $$a=bk,\\ c=dk.$$ Sum up $$\left( a+c \right) =\left( b+d \right) k.$$ You find $$\\ \frac { a+c }{ b+d } =k=\frac { a }{ b } =\frac { c }{ d }. $$ Which implies that you can find another number which is equal to $\frac { a }{ b } .$

Solution 4:

If $\frac{a}{b}$ is an integer $n$, then:

$\frac{a}{b}=\frac{n}{1}$.

In other words, if $\frac{a}{b}$ is an integer, we also know $b=1$ if your teacher were correct. However, $\frac{a}{b}$ is an integer if and only if $b$ divides $a$ and we have fractions such as $\frac{4}{2}=2$ for which the denominator is not equal to $1$.