Hey guys. I forget where I found this problem but it goes as follows: Prove that [tex]\sqrt[3]{2}[/tex] cannot be represented in the form p+q[tex]\sqrt{r}[/tex] where p,q, and r are rational numbers.(adsbygoogle = window.adsbygoogle || []).push({});

It is easy to show that [tex]\sqrt[3]{2}[/tex] is irrational, so it cannot be put in the form m/n, where m and n are integers. However, I do not know where to go from here. I figured that since [tex]\sqrt[3]{2}[/tex] is irrational it cannot be put in any form using rational numbers, but I am not sure.

Any help is appreciated, thanks in advance.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Proof with the cube root of 2

Loading...

Similar Threads - Proof cube root | Date |
---|---|

I Doubling a Cube | Friday at 11:41 PM |

B Did President Garfield really come up with an alternate proof? | Mar 7, 2018 |

B Fermat's Last Theorem; unacceptable proof, why? | Feb 15, 2018 |

I Proof without words for Heron's formula | Jan 19, 2018 |

I Proof of an Inequality | Dec 19, 2017 |

**Physics Forums - The Fusion of Science and Community**