Prove that for all integers m and n, m - n is even if, and only if, both m and n are even or both m and n are odd.

Below

Step-by-step explanation:

Suppose that m and n are both even numbers.

So we can express them as the product of 2 and another number.

● n = 2×a

● m = 2×b

● m-n = 2b-2a

● m-n = 2(b-a)

m-n is an even number since it is divisible by 2.

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Suppose that both n and m are odd numbers.

● n = 2a+1

● m = 2b+1

● m-n = 2b+1-(2a+1)

● m-n = 2b+1-2a-1

● m-n = 2b-2a

● m-n = 2(b-a)

So m-n is even since it is divisible by 2.

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Suppose that m is odd and n is even ir vice versa

● n = 2a or n= 2a+1

● m = 2b+1 or m = 2b

● m-n = 2b+1-2a or m-n = 2b-2a-1

● m-n = 2(b-a) +1 or m-n = 2(b-a)-1

In both cases m-n isn't even.

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So m-n is even if and only if m and n are odd or m and are even

Case 1

both m and n are even

Therefore m/2 and n/2 are integers

Then,

m-n

=2(m/2 - n/2)

Since m/2 and n/2 are integers

Then m/2 - n/2 will be an integer

Therefore,

m-n = 2(Z)

Where Z is an integer

Since 2 is a factor of m-n

Therefore m -n is even

Case 2

Both m and n are odd

m-n

= 2(½m - ½n)

When an odd number is divided by 2 it gives an integer and a remainder of 1

Therefore

½m = Y + ½

And

½n = Z + ½

Where Y and Z are integers

Then

m-n = 2(Y+½-Z-½)

= 2(Y-Z)

Y-Z will also be an integer

m-n= 2A

Therefore m-n is even

Case 3

One is odd and the other even

m-n = 2(m/2 - n/2)

Assume m is even and n is odd

From the discussions above

m-n = 2(Y - Z - ½)

m-n = 2(A - ½)

Hence m-n is not even because when is divided by two it doesn't give an integer.

Therefore for all integers m and n, m - n is even if, and only if, both m and n are even or both m and n are odd.