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In the equation

$$\frac{n-1}{n(n-1)(n-2)!} = \frac{1}{n(n-2)!} $$

can I cancel out the factors $(n-1)$'s in the numberator and denominator, so the equation is equal?

I've learned that you can't cross anything if there is a + or - sign in the fraction?

amWhy
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Amir Šaran
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    You can do that, provided $n\neq 1$ (which I suppose it is, otherwise $(n-2)!$ typically does not make sense) – projectilemotion Nov 15 '18 at 19:53
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    The more precise statement to the nonrule you've learned is that you can't cross things out if addition or subtraction is in the way. In this case, they aren't in the way. The thing you're simplifying is very similar to $$\frac{y}{xyz}=\frac{1}{xz}$$ –  Nov 15 '18 at 19:59
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    Instead of trying to apply a (false) rule, think about what's really going on here. You have a fraction, and some number ($n-1$) is a factor of both the numerator and denominator. So of course you can cancel it. – rogerl Nov 15 '18 at 20:03
  • So I can just put the n-1 in parentheses and pretend that it's just one number? – Amir Šaran Nov 15 '18 at 20:05
  • Yes. If it helps, what if you pretended $n$ had a specific value, like $\pi$. Then you'd have $\frac{\pi-1}{\pi(\pi-1)(\pi-2)}$. Is it clear now that you can cancel them? – rogerl Nov 15 '18 at 20:16

1 Answers1

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Yes the cancellation is allowed

$$\frac{n-1}{n(n-1)(n-2)!} = \frac{1}{n(n-2)!}$$

provided that we exclude case $n=1$ from the solutions.

For example the the following equation

$$\frac{(x-1)^2}{(x-1)}=0 \stackrel{x-1\neq0}\implies x-1=0$$

has not solution because we need to exlude the case $x=1$.

user
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