# prove $\sqrt 2 + 3^{1/5}$ is irrational?

So far I've got

Suppose that $\sqrt 2+ 3^{1/5} = p/q$.

$3^{1/5} = p/q - \sqrt 2$

$3 = (p/q - \sqrt 2)^5$

$3 = (p^5/q^4 + \dots - 4 \sqrt 2)$.

since $\sqrt2$ is irrational, it can't be equal to $3$, a rational number.

Is this a legit proof? or is there any way I can more elaborate the proof?

• The idea is good. More detail would be expected. – André Nicolas Dec 17 '15 at 4:27
• You need to make sure the stuff omitted by the "...." can't "cancel out" the $4\sqrt{2}$. – JimmyK4542 Dec 17 '15 at 4:27
• You might want to expand the last step. If you can write $3$ as $r + s\sqrt{2}$, where $r$ and $s$ are rational numbers and $s$ is not zero, then your argument makes sense. – Umang Dec 17 '15 at 4:28

The general idea is good. More detail would probably be expected. We add a version that works quickly and generalizes well.

Suppose that $\sqrt{2}+3^{1/5}$ is the rational number $r$. Then $3^{1/5}=r-\sqrt{2}$. Take the fifth power of both sides. We get $$3=r^5-5r^4\sqrt{2}+20r^3-20r^2\sqrt{2}+20r-4\sqrt{2}.$$ Rearranging, we get $$B=(5r^4+20r^2+4)\sqrt{2},$$ where $B$ is rational. Since $\sqrt{2}$ is irrational, this is only possible if $5r^4+20r^2+4=0$. That cannot happen, since the expression is $\ge 4$ for all real $r$.

• If $5r^4+20r^2+4=4\sqrt{2}$ then it is both greater than 4, and its product with $\sqrt{2}$ is rational... – ziggurism Dec 17 '15 at 4:48
• The $4\sqrt{2}$ is already included in my $(5r^4+20r^2+4)\sqrt{2}$. – André Nicolas Dec 17 '15 at 4:59

It can be solved as follows: Suppose that $\sqrt 2+ 3^{1/5} = p/q$.

$$3^{1/5} = p/q-\sqrt2$$

$$3 = (p/q-\sqrt2)^5$$

$$3 = (p^5/q^5 + \dots - 4 \sqrt 2)$$

On expansion it can be written as $3=a-b\sqrt2$ where $b = 5p^4/q^4 + 20p^2/q^2 + 4$.

For $a-b\sqrt2$ to be equal to $3$, the value of b should be zero. But since b contains all the terms of even powers, b is always positive. This contradicts the assumption, so $\sqrt2 + 3^{1/5}$ is irrational.