SOLUTION: How can you prove that: odd + odd = even even + even = even (you initially put even + even = odd) odd + even = odd ????

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Question 9013: How can you prove that:
odd + odd = even
even + even = even (you initially put even + even = odd)
odd + even = odd
????

Answer by prince_abubu(198) About Me  (Show Source):
You can put this solution on YOUR website!
We're going to do some backwards thinking here. Hang on.

We all know that a number that ends in 0,2,4,6, or 8 is even. OK. Take ANY integer n, and you can even pick an odd number if you want. Multiply it by 2. That number n times 2 will ALWAYS turn out even. You can't help it. Now, think of it backwards: If a number A is even, it must be able to be expressed as 2*n. If A is even, then there is an integer n such that +A+=+2n+. You can think of this trick as forcing a given variable to be even.

What about forcing a variable to be odd? Let's do this forwards first. We know that if you take any integer n and multiply it by 2, the product 2*n is ALWAYS even. Add one to that product and you will ALWAYS end up with an odd result. In other words, the quantity +2n+%2B+1+ is ALWAYS odd. Thinking of it backwards, you can force the number B to be odd by declaring +B+=+2n+%2B+1+. AKA, there is such a number n such that the quantity +2n+%2B+1+ is always undoubtedly ODD.
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Now, let's try to prove that ODD + ODD = EVEN. Let's choose two odd variables C and D. If C is odd, then C must be expressed as +2n+%2B+1+, where n is any integer. If D is odd, then D must be expressed as +2m+%2B+1+. Notice that we used two different variables n and m. This shows us that C and D don't have to be the same number, just as long as they're both odd. Now, let's plug it in to our equation:

(2n + 1) + (2m + 1) ?= even? <---- START

2n + 2m + 2 ?= even? <------ START No matter what n and m are, 2n and 2m will ALWAYS be even. and the 2? 2 is even, so adding an even to an even creates an even number.

odd + odd = even <----- final answer
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If you've got two even numbers to add up, the answer is always even. Let's choose two numbers, E and F. If E and F are indeed even, then there must be integers p and q so that we can say that E+=+2p and F+=+2q

So let's put these to the test: 2p + 2q ?= even?

2(p + q) ?= even? <------------- Factor out the 2 from both terms. It seems like it doesn't matter what p and q are. Their sum will be multiplied by 2, forcing the answer to be even.

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What about odd + even = odd? Let's choose G and H as the odd and even numbers to be added. If G is odd, then it can be written as G = 2r + 1 and r is any integer. If H is even, then it can be written as H = 2s and s is ANY integer. Let's test this:

(2r + 1) + 2s ?= odd <----------------- start

2r + 2s + 1 ?= odd <---------------- rearrange

2(r + s) + 1 ?= odd <-------------- factor out the 2 from the 2r and 2s. No matter what r and s are, their sum will be multiplied by 2, forcing an even number answer. Then you've got the + 1 that forces the already even answer to be odd.