Verified answer

1) Suppose to the contrary that √3 were rational.

Then, √3 = m/n for some positive integers m, n with gcd(m, n) = 1.

Square both sides:

3 = (m/n)^2 = m^2 / n^2

==> m^2 = 3n^2.

Since the right side of the equation is a multiple of 3, we see that 3 | m^2

(shorthand for m^2 is a multiple of 3).

==> 3 | m, because 3 is a prime.

Writing m = 3k for some positive integer k, we have (3k)^2 = 3n^2

==> 9k^2 = 3n^2

==> 3k^2 = n^2.

So as before, we have 3 | n^2 and thus 3 | n.

This is a contradiction, because now 3 | gcd(m, n) = 1, which is absurd.

Hence, √3 is irrational.

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2) Let m be an even integer and n be an odd integer.

So, m = 2r and n = 2s+1 for some integers r and s.

Then, m + n = 2r + (2s+1) = 2(r+s) + 1.

Since r+s is an integer (call it k), we have m+n = 2k+1; hence m+n is odd.

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I hope this helps!