Prove that $\frac{m}{n}+\sqrt{3}\frac{p}{q}$ is irrational If $\frac{m}{n}$ and $\frac{p}{q}$ are rational. Prove that $\frac{m}{n}+\sqrt{3}\frac{p}{q}$ is irrational, given $p\neq 0$.
 A: Hint: If $\alpha + \beta \sqrt 3 = \gamma$ with $\alpha, \beta, \gamma \in \mathbb Q$, we have $\sqrt 3 = \text?$ (solve for $\sqrt 3$). Is this possible?
A: HINTS:


*

*Can you prove that $\sqrt{3}$ is irrational?

*Can you prove that if $q \neq 0$ is a non-zero rational number and $x$ is an irrational number, then $qx$ is also irrational?

*Can you prove that if $q$ is a rational number and $x$ is an irrational number, then $q+x$ is also irrational?


EDIT:
Well, actually, if I want to give you some further hints I'll give you too much information. But since you're still new to the subject I think it's not bad to see some methods first.
I'll only prove the 2nd one. I'll leave the first and third one to you to prove on your own. Suppose that $qx$ is not irrational. Therefore, $qx=p$ where $p \in \mathbb{Q}$. Since $q \neq 0$ I can divide by $q$ and get $\displaystyle x = \frac{p}{q}$. But since $p,q \in \mathbb{Q}$ and the division of two rational number is a third then $\displaystyle x = \frac{p}{q}$ which is contradiction because we assumed $x$ is irrational.
Now use the same trick for the third one. Also, I think by now you should know that $2$ or $3$ aren't any special numbers. You can prove that if $p$ is a prime number then $\sqrt{p}$ is irrational.That's a good exercise to do.
