ment 12 Help Save & Ex Required information NOTE: This is a muti part question.

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ment 12 Help Save & Ex Required information NOTE: This is a muti part question. Once an answer is submitted you wil be unable to retum to this part Prove that if n is an integer and 3m+ 2 is even, then n is even using a proof by contradiction the optons below Suppose that 3n 2 is even and that n is odd Since 3n + 2 is even, so is 3n Therefore. our supposition was wrong, hence n is even We know that if we subtract an odd number from an even number, we get an odd number As 3n is even and n is odd, 3n-nshould be odd, but 3m-n 2n is even This is a contradiction MacBook Air

Explanation / Answer

Suppose that 3n+2 is even and that n is odd
Since 3n+2 is even, so is 3n
We know that if we subtract an odd number from an even number, we get an odd number
As 3n is even and n is odd , 3n-n should be odd, but 3n - n = 2n is even, this is a contradiction
Therefore, out supposition was wrong. Hence n is even

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It follows directly that m+n = 2s for some integer s and n+ p = 2t for some integer t
Add the two equations to obtain m+p + 2n = 2s + 2t
Subtract 2n from both sides and factorize to obtain m+p = 2s + 2t - 2n = 2(s+t-n)
Since m + p is 2 times (s+t-n), which is an integer, we conclude that m+p is even

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