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, 02:02 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, 02:07 PMChop Suey
Try using the Rational Root Theorem :-)

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, 02:26 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, 02:44 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, 05:16 PMJaneBennet
If you are not familiar with the rational-root theorem, you can still tackle the problem directly.

Assume that is rational, so where and .

Thus

for some

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