1.1 KiB
Question 1
Prove that
\sqrt[3]{361}is irrational.
To determine whether the cube root of 361 is irrational, we need to examine whether it can be expressed as a rational number, which is a number that can be written as the quotient of two integers (where the denominator is not zero).
Let us assume that \sqrt[3]{361} is a rational number. As such, \sqrt[3]{361}=\frac{p}{q}, where p,q are co-prime integers, and q\neq0 (definition of a rational number).
Taking the cube of both sides of this equation gives
361&=\left(\frac{p}{q}\right)^{3} \\ &=\frac{p^{3}}{q^{3}} \\ p^3 &=361q^3 &(1)
Since p and q are integers, p^3 and q^3 are also integers.
Consider the prime factorisation 361=19\times19. 19 is prime, appears twice in the factorisation, and also is the only number in the factorisation. It is not in the form \frac{p^{3}}{q^{3}}. Therefore, there is no way to express 361 as the cube of a rational number.
This contradicts our original assumption that \sqrt[3]{361} is rational.
Hence, by contradiction, we can conclude that \sqrt[3]{361} is irrational.