Problem 64 *

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(a)

\sqrt 2 + \sqrt 3 is rational iff \displaystyle \frac 1 {\sqrt 2 + \sqrt 3} = \sqrt 3 - \sqrt 2 is too. But then (\sqrt 3 + \sqrt 2) + (\sqrt 3 - \sqrt 2) = 3 \sqrt 2, which is rational iff \sqrt 2 is rational.

Suppose \sqrt 2 is rational then \sqrt 2 = p/q for coprime positive integers p,q. Then 2q^2 = p^2. Then the exponent of 2 in the prime factorisation of the LHS is odd, whereas that of the RHS is even, a contradiction. (since the prime factorisation of a natural number is unique by FTA) So \sqrt 2 is irrational. The sum of rationals cannot be irrational, so one of \sqrt 2 + \sqrt 3 and \sqrt 3 - \sqrt 2 is irrational. But since one is irrational iff the other is, neither are rational.

(b)

By the rational root theorem, rational roots can only be \pm 7 or \pm 1. But checking each of these 7^3 - 4 \times 7 + 7 = 49 \times 7 - 4 \times 7 + 7 = 46 \times 7 > 0, (-7)^3 + 4 \times 7 + 7 = -49 \times 7 + 4 \times 7 + 7 = -44\times 7 < 0, 1^3 - 4 + 7 = 4 and (-1)^3 + 4 + 7 = 10. So the real root is necessarily rational.

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Suppose that \log_2 3 was rational, ie. \log_2 3 = p/q for coprime non-zero integers p,q. Note that 2^1 = 2 and 2^2 = 4 implies that there 1 < \log_2 3 < 2 (more specifically, the log is positive) by the intermediate value theorem, so these p,q are either both positive or both negative. WLOG say they’re both positive.

Then 3 = 2^{p/q}, 3^{q} = 2^{p}. But this has no integer solutions p,q since the LHS is divisible by 3 and the RHS is not. (alternatively, by FTA comparing exponents we must have p = q = 0 but p, q \ne 0)

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