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Why is $\sqrt{p}$ irrational when $p$ is a prime number?
ABecause every square root is irrational
BBecause prime numbers have no factors at all
CBecause assuming $\sqrt{p}$ is rational leads to a contradiction with prime factorisation
DBecause prime numbers are always odd
Answer & Solution
Correct answer: C. Because assuming $\sqrt{p}$ is rational leads to a contradiction with prime factorisation
The standard proof uses contradiction. If $\sqrt{p}$ were rational, we could write it in lowest terms and then show that $p$ divides both numerator and denominator, contradicting that the fraction is in lowest terms.
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