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.

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.

yes the total will be same because in both of them the expression.is rearranged but their sign remains the same hence it hss just been shuffled

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step-by-step explanation: