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?

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- Aug 6th 2008, 01:02 PMAJArmstronProving that the cube root of 5 is irrational
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? - Aug 6th 2008, 01:07 PMChop Suey
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. - Aug 6th 2008, 01:26 PMDubulus
you do the proof via contradiction. assume that there exists a/b=sqrt5 where a and b are integers and are at their lowest common form (eg not 4/2 or 12/3). then using use that equation, prove that both a and b are even, thus they are divisible.

- Aug 6th 2008, 01:44 PMAJArmstronThe question says prove that the cube root of 5 is irrational
Thank you for your post, but the question is prove that the cube root is irrational not the square root.

- Aug 6th 2008, 04:16 PMJaneBennet
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.