Was hoping someone could help with this question - I'd appreciate it.
"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 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)
2 (2mn + m + n) + 1 <- Am I right in thinking either one or the other can be used?
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?