For each of the two statements and attempted proofs, (i) Is the statement true?
ID: 3007011 • Letter: F
Question
For each of the two statements and attempted proofs, (i) Is the statement true? If it is, explain why. If not, give a counter example. Each proof is wrong. Explain where the mistake(s) is/are in each proof. squareroot 2n is irrational. Proof: Contradiction Assume N, squareroot 2n = squareroot 2 middot squareroot n is rational. We know squareroot 2 is irrational (proved in class) so for squareroot 2n to be rational, squareroot n must be irrational, otherwise we would have an irrational Times a rational = a rational, which is impossible. But if we take n = 9, squareroot n = 3 which is rational. This is a contradiction. So the result is proved. Let m, n G Z. Then m + n is odd if and only if exactly one of m and n is odd (and the other one is even). Proof: Direct Assume first that m is odd and n is even. Then we can write m = 2h + 1; n = 2k for some integers h, k. So m + n = 2h+1 + 2k = 2h + 2k + l = 2(h + k) +1 which is odd. Similarly if n is odd and m is even. Conversely, assume m + n is even. Then clearly m and n are both even or they are both odd. So the result is proved.Explanation / Answer
(a)
for sqrt{2n} to be rational it correct that sqrt{n} must be irrational but that is not sufficient.
eg. sqrt{3} is irrational
But sqrt{6} is not rational.
Statement is false
sqrt{2*2}==sqrt{4}=2 which is rational
(b)
Statement is true.
If both m,n are odd means: m=2k+1,n=2r+1
m+n=2k+2r+2 is even
If both m,n are even means:m=2k,n=2r
m+n=2k+2r is even
Hence, one of m must be odd and other even.
The converse part of the proof is wrong.
Assuming m+n is wrong as we need: m+n as odd.
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