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You can never get answers with $\frac{\frac{a}{b}}{\frac{c}{0}}$ where $a, b, c$ $\ne$ $0$.

$\frac{a}{c}$$\ne$ $0$ and $\frac{b}{0}$ $=$ $undefined$.

But why is there a specific answer ($0$ definitely) if $\frac{a}{b}\cdot\frac{0}{c}$?

333-blue
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    Question completely unclear. – barak manos Dec 16 '16 at 12:53
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    @barakmanos, I think OP is asking that we can't solve $\frac{a}{b}/\frac{c}{0}$ but $\frac{a}{b}\cdot\frac{0}{c}$ is 0 (which is just rearranged form of the previous one). – Vidyanshu Mishra Dec 16 '16 at 12:55
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    Maybe, because the rule that $\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{cb}$ holds only when $\frac{c}{d}$ is defined. And $\frac{c}{0}$ is not defined. – IamThat Dec 16 '16 at 13:03
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    Denominator in your first term, $\frac{c}{0}$ is not defined, hence the whole fraction $\frac{a/b}{c/0}$ is not defined. –  Dec 16 '16 at 13:05

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The expression $\frac{a/b}{c/0}$ means:

  • Divide $a$ by $b$, call the result $x$;
  • Divide $c$ by $0$, call the result $y$;
  • Divide $x$ by $y$, this is your number.

Of course you can immediately see the problem with that: the second step is not possible to do. So this expression doesn't give a well-defined number.

However, it is true that if $b,c,d$ are all nonzero numbers, then $\frac{a/b}{c/d}$ is a well-defined number, and this number is equal to $\frac{a}{b} \cdot \frac{d}{c}$. And it is true that this last expression is well-defined when $d = 0$. However this rule can only apply when $d \neq 0$; it doesn't mean that $\frac{a/b}{c/0}$ becomes well-defined.

Here is an example of why you need to be careful. You certainly agree that $\frac{xy}{y} = x$, right? Now take $x = 2$ and $y = 0$, and you get $2 = \frac{2 \times 0}{0}$. But of course $2 \times 0 = 0$, so $2 = \frac{0}{0}$. Now I can apply the same "rule" to $3 = \frac{3 \times 0}{0} = \frac{0}{0}$, and I finally get $2 = 3$. Which is obviously nonsense.

What went wrong? Well, the "rule" $\frac{xy}{y} = y$ is only valid when $y \neq 0$, because otherwise you're dividing by zero! You can't just blindly apply "rules" without checking that the conditions for these rules are satisfied, and you can't act like undefined expressions (like $\frac{c}{0}$) mean something, otherwise you will run into trouble. The tag is full of such examples.

Najib Idrissi
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