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discrete math

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prove that if n is an integer and 3n+2 is even, then n is even using
a)a proof by contraposition
b)a proof by contradiction

I'll try part b, you'll have to refresh me on what contraposition means here.
Here is the claim we start with
If n is an integer and 3n+2 is even, then n is even.
Reduction as absurdum or proof by contradiction begins by assuming the conclusion is false and then showing this contradicts one of the premises, thereby showing the conclusion is true.
Suppose n is odd, then 3n is odd since the product of odd integers is an odd int. Every odd int. + and even int. is odd. Show this by adding 2k+1 + 2m = 2(k+m)+1 = an odd number. Therefore 3n+2 is an odd number, but this contradicts the assumption that 3n+2 is even. Therefore if 3n+2 is even then n is even.
I think contraposition would be: If n is even then 3n+2 is even. You should be able to do this I think.

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