I am trying to prove or disprove that the cube root of 5 is an irrational number.
I have tried using a proof by contradiction, although am not convinced that I have a good answer.
Can anyone help please?
Try using the Rational Root Theorem :-)
$\displaystyle x = \sqrt[3]{5}$
$\displaystyle x^3 = 5$
$\displaystyle x^3 - 5 = 0$
Cube root of 5 is a root. Find all the possible rational roots. If they don't satisfy the equation, then there are no rational roots => cube root of 5 is irrational.
If you are not familiar with the rational-root theorem, you can still tackle the problem directly.
Assume that $\displaystyle \sqrt[3]{5}$ is rational, so $\displaystyle \sqrt[3]{5}=\frac{a}{b}$ where $\displaystyle a,b\in\mathbb{Z}^+$ and $\displaystyle \gcd(a,b)=1$.
Thus $\displaystyle \frac{a^3}{b^3}=5\ \Rightarrow\ a^3=5b^3$
$\displaystyle \color{white}.\hspace{20mm}.$ $\displaystyle \Rightarrow\ 5\mid a^3$
$\displaystyle \color{white}.\hspace{20mm}.$ $\displaystyle \Rightarrow\ 5\mid a$
$\displaystyle \color{white}.\hspace{20mm}.$ $\displaystyle \Rightarrow\ a=5c$ for some $\displaystyle a,b\in\mathbb{Z}$
If you then plug this back into the equation, you would find that 5 divides $\displaystyle b$ as well, contradicting the assumption that $\displaystyle a$ and $\displaystyle b$ are coprime.