I seem to be having a lot of difficulty with proofs and wondered if someone can walk me through this. The question out of my textbook states:
Use a direct proof to show that if two integers have the same parity, then their sum is even.
A very similar example from my notes is as follows: Use a direct proof to show that if two integers have opposite parity, then their sum is odd. This led to:
Proposition: The sum of an even integer and an odd integer is odd.
Proof: Suppose a is an even integer and b is an odd integer. Then by our definitions of even and odd numbers, we know that integers m and n exist so that a = 2m and b = 2n+1. This means:
a+b = (2m)+(2n+1) = 2(m+n)+1 = 2c+1 where c=m+n is an integer by the closure property of addition.
Thus it is shown that a+b = 2c+1 for some integer c so a+b must be odd.
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So then for the proof of showing two integers of the same parity would have an even sum, I have thus far:
Proposition: The sum of 2 even integers is even.
Proof: Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2m???
Best Answer
"Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2m???"
Since a and b are different numbers they should be different m and n.
"Suppose a is an even integer and b is an even integer. Then by our definitions of even numbers, we know that integers m and n exist so that a=2m and b=2*n*?
And so a + b = 2m + 2n = 2(m+n) and as m+n =c for some integer c, a + b = 2c so by definition a + b is even.