TITLE: Prove that $\sqrt[3]5 - \sqrt[4]3$ is Irrational
QUESTION [7 upvotes]: I've gone many directions and they all fail.  
The sum of two irrationals doesn't need to be irrational.
I found a proof saying: if irrational $x,y$ have a rational sum $x+y$, then $x-y$ is irrational, or vice versa. However, in this case $x+y$ and $x-y$ are irrational.  I must have misinterpreted the proof maybe.
Is the sum and difference of two irrationals always irrational?
I recognize $n\in\Bbb Q\implies \exists a,b\in\Bbb Z~(b\neq 0) : n=\dfrac ab$
Also the product of a rational and irrational number are irrational so $a = b(\sqrt[3]5 - \sqrt[4]3)$ where $a,b\in \mathbb{Z}$. I've looked on this basis as well, but so far fruitless.
Thank you,
Julian

REPLY [4 votes]: Another approach, though I like @AchilleHui’s comment a lot:
Write $\lambda^3=5$, $\mu^4=3$, and let $a=\lambda-\mu$, which we want to show is irrational. Assuming rationality of $a$, we get:
\begin{align}
3=\mu^4&=(\lambda-a)^4\\
&=5\lambda-20a+6a^2\lambda^2-4a^3\lambda+a^4\\
0&=(a^4-20a-3)\,+\,(5-4a^3)\lambda\,+\,(6a^2)\lambda^2\,.
\end{align}
Notice that the three quantities in parentheses are rational (assuming that $a$ is rational), and with the knowledge that $\{1,\lambda,\lambda^2\}$ is a $\Bbb Q$-linearly independent set, you see that $6a^2=0$, a contradiction, since $\lambda\ne\mu$.