Direct proof that the product of odd integers is odd

Question

Prove $P(x,y)$: If $x$ and $y$ are odd integers, then the product $xy$ must also be odd.

I need a direct proof of this.

I know that $ x $ and $y$ both have to equal to $2n+1$ in order for them to be odd. But that's all I have. Any help will be appreciated.

score 4 · Answer 1 · answered May 14 '14 at 00:24

4

Any odd number can be written as $2n-1$, for some integer $n$. Then, consider: $$(2n-1)(2m-1)=4mn-2(m+n)+1$$ $4mn-2(m+n)$ is even for all $m,n$, implying that...

answered May 14 '14 at 00:24

Why did you use $m$ instead of another $n$? – rubito May 14 '14 at 00:41
1

@rubito Because we're multiplying two different odd numbers, not the same one. – May 14 '14 at 00:58

score 0 · Answer 2 · answered May 14 '14 at 02:16

0

Hint $\, $ Multiplying by an odd $ $ preserves parity: $\ (1\!+\!2k)\,n = n + 2(kn)\,$ has same parity as $\,n.$

answered May 14 '14 at 02:16

Bill Dubuque

272,048

Said $!\bmod 2!:\ {\rm odd}\ \color{#c00}x = 1!+!2k\equiv \color{#c00}1\Rightarrow \color{#c00}xn\equiv \color{#c00}1n\equiv n,$ by the congruence product rule. By induction the same is true for $x$ any product of $k$ odds, by $,\color{#c00}1^k\equiv 1\ \ $ – Bill Dubuque Jun 26 '22 at 03:41

Direct proof that the product of odd integers is odd

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