Prove that \(\sqrt{23}\) is irrational. (3 marks)
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\(\text{Proof (See Worked Solutions)} \)
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\(\text{Proof by contradiction:} \)
\(\text{Assume}\ \sqrt{23}\ \text{is rational} \)
\( \sqrt{23} = \dfrac{p}{q}\ \ \text{where}\ p, q \in \mathbb{Z}\ \ \text{with no common factor except 1} \)
\(23\) | \(= \dfrac{p^2}{q^2} \) | |
\(23q^2\) | \(=p^2\) |
\(\Rightarrow \text{23 is a factor of}\ p^2 \)
\(\Rightarrow \text{23 is a factor of}\ p \)
\( \exists k \in \mathbb{Z}\ \ \text{such that}\ \ p=23k \)
\(23q^2\) | \(=(23k)^2 \) | |
\(q^2\) | \(=23k^2 \) |
\(\Rightarrow \text{23 is a factor of}\ q^2 \)
\(\Rightarrow \text{23 is a factor of}\ q \)
\(\therefore \text{HCF}\ \geq 23 \)
\(\therefore \text{By contradiction,}\ \sqrt{23}\ \text{is rational} \)