Question

Prove the following statement by contrapositive or by contradiction:

For all integers m and n, if m + n is even then m, n are both even or m, n are both odd.

Answer 1

Proof by Contrapositive

The contraposition of the statement is

If 'm is odd and n is even' or 'm is even and n is odd' then 'm+n is odd'.

a)At first we will take one of the statement as 'm is odd and n is even'

As we know the ways to write a even and odd number

m is odd mean

m=2a+1,

n=2b,

where a and b are any integers

Now substitute the values of m and n in (m+n)

m+n = (2a+1)+2b = 2a+2b+1 = 2(a+b)+1 Suppose c =a+b

m+n = 2c+1, which is odd

Since the contrapostive statement is true than the original statement is also true.

Similarly we can take m is even and n is odd, for that

m=2a and n=2b+1, substitute the values of m and n in m+n, we get

m+n = 2a+(2b+1) = 2a+2b+1 = 2(a+b)+1 Suppose c =a+b

m+n = 2c+1, which is odd

Again, Since the contrapostive statement is true than the original statement is also true.