Proving that things are or aren't rational numbers My question comes in three parts:


*

*Suppose $x,y\in \Bbb Q$.  Prove that $2x-5y\in \Bbb Q$

*Prove that $3^{1/2}\not\in \Bbb Q$

*Suppose $x\in \Bbb Q$.   Prove that $x^2+3^{1/2}\not\in \Bbb Q$


In the third question I'm not sure how I should proceed to solve it.  I'm aware of how to prove root 3 is irrational but that question I'm not sure of.  Also the first question I don't understand how i can prove that $2x-5y$ is irrational.
 A: For (a), this follows from the more general fact that a rational number plus another rational number is again rational.  Still, it is worth going through the effort of proving this from first principles at least once.
Suppose that $x,y\in\Bbb Q$ are both rational numbers.  We wish to prove then that $2x-5y\in \Bbb Q$ is also a rational number (you wrote the word irrational there presumably by mistake.  Do not confuse the words as they mean completley opposite things)
Since $x\in \Bbb Q$, this means that there are integers $a$ and $b$ with $b\neq 0$ such that $x=\frac{a}{b}$.
Similarly, $y\in \Bbb Q$ implies that there are integers $c$ and $d$ with $d\neq 0$ such that $y=\frac{c}{d}$.  (note: I used different letters to describe $x$ than I did for $y$ since they are able to be different)
Then we have $2x-5y = 2\frac{a}{b}-5\frac{c}{d} = \frac{2ad-5bc}{bd}$.  We ask ourselves, is this a rational number?  Is it the ratio of two integers with the bottom integer nonzero?

For (c), this follows from the more general fact that a rational number plus an irrational number is irrational.  Again, it is worth approaching from first principles at least once.
We prove this via contradiction:
Suppose that $x\in \Bbb Q$ and that $x^2+\sqrt{3}\in \Bbb Q$
Then there exist integers $a,b,c,d$ with $b$ and $d$ nonzero such that $x=\frac{a}{b}$ and $x^2+\sqrt{3}=\frac{c}{d}$
But then, $\sqrt{3} = (x^2+\sqrt{3})-x^2 = \frac{c}{d}-(\frac{a}{b})^2 = \frac{b^2c - a^2d}{db^2}$.  What do we know about the rationality of $\sqrt{3}$ from the previous part?  What does this line say about the rationality of $\sqrt{3}$ though?  Is this possible as a result?  What does this imply about the original wording of the problem?
A: Hints: $1$. As $x,y \in \mathbb{Q}$ and $2,-5 \in \mathbb{Q}$, so $2x,-5y\in \mathbb{Q}$. 
$2$. If possible consider $3^{1/2}$ is rational. Then $\exists  p/q \in \mathbb{Q}$ such that $3=p^2/q^2$, then try to find a contradiction. 
$3$. If $x\in \mathbb{Q}$ then $x^2\in \mathbb{Q}$. Then what can you say about $x^2+3^{1/2}$, if $3^{1/2} \not \in \mathbb{Q}$?
