for all integers m and n where m < n if m and n are consecutive then mn is even

Question

 Prove by case method:

For all integers m and n where m < n if m and n are consecutive then mn is even

Answer 1

Since m and n are consecutive, that means that they come right after each other, like 2 and 3 or 18 and 19. This being said, this means that an odd number is being multiplied by an even number. An even number multiplied by any other number is still even, thus mn will be even.

 

The statement is true.

Answer 2

Here is another way to look at this this:

 

Prove that m(m + 1) = 2j, where j is an integer.  

 

Case I:  m is even

 

Then m = 2p, where p is an integer

 

m(m + 1) = 2p(2p + 1) = 2[p(2p + 1)] = 2j, where j = p(2p +1).

 

Case II:  m is odd

 

Then m = 2q + 1, where q is an integer

 

m(m + 1) = (2q + 1)[(2q + 1) + 1] = (2q + 1)(2q + 2) = 2(2q + 1)(q + 1) = 2j, where j = (2q + 1)(q + 1)

 

In both cases, m(m + 1) is twice an integer, and therefore is even.