MHB Contrapositive Proof: Ints $m$ & $n$ - Even/Odd Combinations

Click For Summary
To prove the contrapositive statement, assume one integer, $m$, is even and the other, $n$, is odd. The sum of an even and an odd integer is always odd, which demonstrates that if $m$ and $n$ are not both even or both odd, then $m + n$ cannot be even. This confirms the contrapositive: if $m + n$ is even, then both integers must be either even or odd. Thus, the proof is complete by showing the necessary condition for the original statement.
tmt1
Messages
230
Reaction score
0
For all integers $m$ and $n$, if $m+ n$ is even then $m$ and $n$ are both even or both odd.

For a contrapositive proof, I need to show that for all ints $m$ and $n$ if $m$ and $n$ and not both even and not both odd, then $ m + n $ is not even.

How do I go about doing this?
 
Mathematics news on Phys.org
The negation of the conclusion is that exactly ONE of $m,n$ is odd.

In a proof of this type, you may assume $m$ is even, and $n$ is odd (or else we may "switch them").

Your mission, should you decide to accept it, is to prove that in this case, we have the negation of the premise:

that is, to show that $m+n$ is thus odd.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 28 ·
Replies
28
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
12K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K