Chapter 4 - Elementary Number Theory and Methods of Proof - Exercise Set 4.7 - Page 213: 22

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**Proof:** Suppose, for the sake of contradiction, that \(n\) is an integer that is not a perfect square but that \(\sqrt{n}\) is rational. Then there exist positive integers \(m\) and \(k\) in lowest terms (i.e. \(\gcd(m,k)=1\)) such that \[ \sqrt{n}=\frac{m}{k}. \] Squaring both sides gives \[ n=\frac{m^2}{k^2} \quad \Longrightarrow \quad n k^2 = m^2. \] Now write the prime factorizations: - Let \[ n = p_1^{a_1} \, p_2^{a_2} \, \cdots \, p_r^{a_r}, \] where the exponents \(a_i\) are nonnegative integers. Since \(n\) is not a perfect square, at least one exponent \(a_j\) is odd. - Since \(m^2\) is a perfect square, every prime in its factorization appears with an even exponent. - Similarly, \(k^2\) is a perfect square, so every prime in its factorization appears with an even exponent. Thus, in the product \(n k^2\), each prime \(p_i\) appears with an exponent \(a_i + 2b_i\) (where \(2b_i\) comes from \(k^2\)). For the particular prime \(p_j\) for which \(a_j\) is odd, the exponent \(a_j + 2b_j\) is odd because the sum of an odd number and an even number is odd. But \(n k^2 = m^2\) is a perfect square, so every prime in its factorization must have an even exponent. This contradiction shows that our assumption that \(\sqrt{n}\) is rational must be false. Thus, if \(n\) is any integer that is not a perfect square, then \(\sqrt{n}\) is irrational.