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