Home › UP Board Class 10 › Mathematics › Why is $\sqrt{p}$ irrational when $p$ is a prime…
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.
Related questions
What is the derivative of $x^x$ for $x>0$?If $y=x^{\tan^{-1}x}$, then $\dfrac{dy}{dx}$ equalsThe derivative of $ in^{-1}x$ isThe derivative of $\cos^{-1}x$ isWhat is $\dfrac{d}{dx}(\tan^{-1}x)$?If $y=\dfrac{1}{x^n}$, then $\dfrac{dy}{dx}$ isWhich identity is correct?For the expression $ qrt{1-x^2}$, a useful trigonometric substitution is