Please check if this proof is ok:)?

2013-05-21 10:16 am
Prove: The product of an odd integer and an even integer is even. (Cannot understand the sense why (2k + 1) is being squared while xy is not. tenks: here's the proof,

let x be odd integer and y be even integer
Assume that xy is odd

xy = 2k + 1
xy = (2k + 1)^2
xy = (4k^2 + 4k) + 1
xy = 4(k^2 + k) + 1 let (k^2 + k) = q element of ineteger
xy = 4q + 1
but xy is even
therefore the assumption that xy is odd is false.
更新1:

How about using indirect proof? tenks

回答 (1)

2013-05-21 12:12 pm
✔ 最佳答案
There are little problems.
First, how do you infer that "xy = (2k + 1)^2" from "xy = 2k + 1"?
Second, how do you infer that "but xy is even" from "xy = 4q + 1"?

You may prove it like this instead:
Let x be odd (2k + 1) and y be even (2p), where k and p are integers.
xy = (2k + 1)(2p)
xy = 2[p(2k + 1)]
Thus, xy is even.

Edit:
You don't prove it indirectly.
If you assume xy is odd, you assume
xy = 2k + 1 where k is an integer.
But if you define xy as 2k + 1, you already agree that
2k is even whether k is even or odd, implying that
2k is even even k is odd.
There is no need to assume xy is odd while confirming unconsciously that 2k is even.


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