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December 1st, 2011, 10:25 AM   #1
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Direct Proof?

Let a, b belong to Z. Prove that ab is even if and only if a is even or b is even.
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December 1st, 2011, 01:17 PM   #2
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Re: Direct Proof?

Quote:
Originally Posted by jrklx250s
Let a, b belong to Z. Prove that ab is even if and only if a is even or b is even.
By contrapositive direct proof : if a ,is not even and b is not even then ab is not even.

I think this easy to prove.

Then you can easily prove ; if a ,is even and b, is even then ab is even
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December 2nd, 2011, 04:04 PM   #3
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Re: Direct Proof?

Well, I doubt that a direct proof of the contrapositive would be considered a direct proof of the theorem itself.

A number is even if and only if it has a two as a prime factor. If a does not have a factor of 2, then b must. Therefore either a is even or b is even.

Notice that the fact that 2 is a prime is important. It is, for example, NOT true that "if ab is dividisible by 6 then either a is divisible by 6 or b is divisible by 6".
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December 3rd, 2011, 03:58 AM   #4
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Re: Direct Proof?

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Originally Posted by HallsofIvy
Well, I doubt that a direct proof of the contrapositive would be considered a direct proof of the theorem itself.

A number is even if and only if it has a two as a prime factor. If a does not have a factor of 2, then b must. Therefore either a is even or b is even.

Notice that the fact that 2 is a prime is important. It is, for example, NOT true that "if ab is dividisible by 6 then either a is divisible by 6 or b is divisible by 6".
An indirect proof is only a proof by contradiction

By logic we have : p=>q is equivalent to :~q =>~p ,which is equivalent to: ~p v q ([color=#FF0000]which you are using in your proof)[/color].

Don't you contradict your self ??
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