## anonymous 4 years ago How can I use this theorem "Let $$\alpha$$ be a root of the polynomial $$x^n+c_{n-1}x^{n-1}+...+c_1x+c_0$$, where the coefficients $$c_0, c_1, ..., c_n$$ are integers. Then $$\alpha$$ is either an integer or an irrational number." to show that $$\sqrt[3]{5}$$ is irrational?

1. JamesJ

The solutions of $$x^3 -5$$ must therefore be integral or irrational, and it's not hard to show they are not integers.

2. anonymous

Ah! I see. I will try that now. Thank you.

3. anonymous

Proof: The solutions of $$x^3-5$$ must be either integral or irrational. Thus, we are looking for an integer $$x$$ such that $$x^3=5$$. However, there is no such integer since $$1^3=1<5$$ and $$2^3=8>5$$. Therefore, $$x$$ must be irrational, and $$\sqrt[3]{5}$$ is irrational. $$\blacksquare$$ Is this okay?

4. JamesJ

Yes

5. anonymous

Thank you again.