If $\frac{a}{b}$ is fully reduced, then $a$ is odd or $b$ is odd.

Asked Jul 02 '21 at 01:01

Active Jul 02 '21 at 01:04

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Is the following contrapositive proof correct?

Proposition: If $\frac{a}{b}$ is fully reduced, then $a$ is odd or $b$ is odd.

Proof: Suppose that $a$ is even and $b$ is even, then $\exists k,\ell \in \mathbb{Z},a = 2k,b=2\ell$, and therefore $$ \frac{a}{b} = \frac{2k}{2\ell} $$ is not fully reduced.

edited Jul 02 '21 at 01:04

asked Jul 02 '21 at 01:01

mhdadk

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5

... because one could cancel the common factor $2$ to get $\frac{k}{l}$. – Geoffrey Trang Jul 02 '21 at 01:03
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@AryaMcCarthy good catch. – mhdadk Jul 02 '21 at 01:05
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Outside of those minor flaws, this is exactly correct. Very good – Stephen Donovan Jul 02 '21 at 01:06
Yep, that's it. You can put it in other words but that's it. – fleablood Jul 02 '21 at 03:08

If $\frac{a}{b}$ is fully reduced, then $a$ is odd or $b$ is odd.

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