# Thread: Using a direct method of proof.

1. ## Using a direct method of proof.

Hi guys,

Was hoping someone could help with this question - I'd appreciate it.

The question;

"Use a direct method of proof to show that if x and y are odd integers, then xy is also odd.

The given solution;

First, notice that if x is odd then x = 2m + 1 where m is an integer. Similarly,
y = 2n + 1 for some integer n. <- I understand up to this point.

Then get lost around here.
Then, xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2 (2mn + m + n) + 1 which is an odd integer."

Could someone explain what exactly is happening in that section?
Thankyou,
- Jamie.

2. Originally Posted by Skeith
Hi guys,

Was hoping someone could help with this question - I'd appreciate it.

The question;

"Use a direct method of proof to show that if x and y are odd integers, then xy is also odd.

The given solution;

First, notice that if x is odd then x = 2m + 1 where m is an integer. Similarly,
y = 2n + 1 for some integer n. <- I understand up to this point.

Then get lost around here.
Then, xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2 (2mn + m + n) + 1 which is an odd integer."

Could someone explain what exactly is happening in that section?
Thankyou,
- Jamie.
Let $u=2mn+m+n$, if $xy=2u+1$, and since $u\in\mathbb{Z}$, it follows that $xy$ is odd.

In other words, anything of the form $2u+1$ is odd if $u$ is an integer.

3. Originally Posted by Skeith
"Use a direct method of proof to show that if x and y are odd integers, then xy is also odd.

The given solution;

First, notice that if x is odd then x = 2m + 1 where m is an integer. Similarly,
y = 2n + 1 for some integer n. <- I understand up to this point.

Then get lost around here.
Then, xy = (2m + 1)(2n + 1) = 4mn + 2m + 2n + 1 = 2 (2mn + m + n) + 1 which is an odd integer.
Two times any integer is an even integer.
So ${\color{blue}2}(2mn+m+n)$ is an even integer.
Add one to that to get an odd integer.

4. Thankyou both Von and Plato - I feel I understand now. However could you check my working just to be sure?

Lets say i assign m and n values of 4 and 5 respectively. this means x = 9 and
y = 11 since it is a matter of x = 2m + 1 and 2n + 1 for y?

Then moving to the section at which I got lost,

xy = (2m + 1)(2n + 1) <- I understand now...

xy = (2m + 1)(2n + 1)
-or-
2 (2mn + m + n) + 1 <- Am I right in thinking either one or the other can be used?

Example,

4mn (m x n x 4 = 80) + 2m (8) + 2n (10) + 1 = 99
- or -
2(2mn (m x n x 2 = 40) + m + n = 49) then multiplying by 2 = 98 + 1 = 99?

Am I on the right lines or have I got lost?

Thankyou again.

5. Yes, you are on the right track.

6. Great - thanks emakarov.