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Irrational Primes

Prove that the square root of a prime number is irrational.

Solution

Let \(p\) be any prime number. For \(\sqrt{p}\) to be rational, it must be expressible as the quotient of two coprime integers \((Condition 1)\).

\[\sqrt{p}=\frac{m}{n}\]

\[p=\frac{{m}^{2}}{{n}^{2}}\]

\[p{n}^{2}={m}^{2}\]

Since \(n,m,{n}^{2},{m}^{2}\) are integers, this implies that \(m\) has a factor of \(p\). Therefore, if the expression \(m=pm'\) is substituted into the third equation, then \({n}^{2}=p{(m' )}^{2}\). By a similar argument, the integer \(n\) must possess a factor of \(p\) as well. This demonstrates the fact both \(m\) and \(n\) are not coprime, which contradicts \(Condition 1\). Hence, the square root of a prime number is irrational.

Note by Steven Zheng
2 years, 9 months ago

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