How to convince a math teacher of this simple and obvious fact? I have in my presence a mathematics teacher, who asserts that

$$ \frac{a}{b} = \frac{c}{d} $$

Implies:

$$ a = c, \space b=d $$

She has been shown in multiple ways why this is not true:

$$ \frac{1}{2} = \frac{4}{8} $$

$$ \frac{0}{5} = \frac{0}{657} $$

For me, these seem like valid (dis)proofs by contradiction, but she isn't satisfied. She wants a 'more mathematical' proof, and I can't think of any.

I'm worried that if she isn't convinced, it may be detrimental to some students. Is there another way to systematically demonstrate the untruth of her conjecture?

EDIT: Since the answer which worked was from a comment, but each answer is also very good, I'm upvoting all of them instead of accepting a specific one. Feel free to close this question for being too open if so a moderator desires.
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 ;-)
Alternatively, $\frac{a}{b}=\frac{c}{d}$ if and only if $ad=cb$ by the definition of fraction equality. So, according to your teacher, $ad=cb\iff a=c\text{ and }d=b$, i.e., there is only one way of writing the integer $ad=cb$ as the product of two integers. However, in general we can write an integer as the product of two integers in many different ways. For example, $24$ can be written as:

$1\times 24$

$2\times 12$

$3\times 8$

$4\times 6$

$6\times 4$

$8\times 3$

$12\times 2$

$24\times 1$

I hope this helps!