Proving that the cube root of 5 is irrational

**AJArmstron** · August 6th 2008, 01:02 PM

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?

**Chop Suey** · August 6th 2008, 01:07 PM

Try using the Rational Root Theorem :-)

$x = \sqrt[3]{5}$

$x^3 = 5$

$x^3 - 5 = 0$
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.

**Dubulus** · August 6th 2008, 01:26 PM

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.

**AJArmstron** · August 6th 2008, 01:44 PM

Thank you for your post, but the question is prove that the cube root is irrational not the square root.

**JaneBennet** · August 6th 2008, 04:16 PM

If you are not familiar with the rational-root theorem, you can still tackle the problem directly.

Assume that $\sqrt[3]{5}$ is rational, so $\sqrt[3]{5}=\frac{a}{b}$ where $a,b\in\mathbb{Z}^+$ and $\gcd(a,b)=1$ .

Thus $\frac{a^3}{b^3}=5\ \Rightarrow\ a^3=5b^3$

$\color{white}.\hspace{20mm}.$ $\Rightarrow\ 5\mid a^3$

$\color{white}.\hspace{20mm}.$ $\Rightarrow\ 5\mid a$

$\color{white}.\hspace{20mm}.$ $\Rightarrow\ a=5c$ for some $a,b\in\mathbb{Z}$

If you then plug this back into the equation, you would find that 5 divides $b$ as well, contradicting the assumption that $a$ and $b$ are coprime.

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Proving that the cube root of 5 is irrational

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