Prove that for each real number x, $$(x+\sqrt2)$$ is irrational or $$(-x+\sqrt2)$$ is irrational.
Now I honestly just don't know where to even go with this one. I tried to do a proof by contradiction but I couldn't get anywhere and I've also reached out to some of my classmates who also seem just as clueless as I am. So really any insight would be greatly appreciated.
Thank you in advance.
Oops. I actually misread the question. I thought the question was:
If $x + \sqrt{2}$ is rational then prove $x$ is irrational. Prove the same for $-x + \sqrt{2}$.
I see the question is actually to prove that regardless as to whether $x$ is rational oro not then one, the other, or both of $x+\sqrt {2}$ or $-x + \sqrt{2}$ is irrational.
To prove that:
If $x + \sqrt{2}$ is irrational we are done.
If $x + \sqrt{2} = q$ is rational then
$x = - \sqrt{2} + q$
And $-x + \sqrt{2} = \sqrt{2} - q + \sqrt{2}= 2\sqrt{2} + q$
And as an rational(not zero)$\times$irrational + rational = irrational (see below), $2\sqrt{2} + q$ is irrational, we are done.
(the below:)
rational(not zero)$\times$ irrational + rational=rational would imply irrational = $\frac{\text{rational - rational}}{\text{rational(not zero)}}$.
Honestly, I like Jose Carlos Santos answer better, that:
$\frac {(x + \sqrt{2}) + (-x + \sqrt{2})}2 = \sqrt 2$ which is irrational. So if $(x + \sqrt{2})$ and $(-x + \sqrt{2})$ were both rational so would their average be. Which it isn't.
Which is the same idea as mine but quite elegant and eloquent in my opinion. Had I read the question correctly the first time, I wouldn't have bothered to answer.
I'm going to leave my answer as I think it goes into great detail of food for thought in general.
===== oops === my original answer--- to a slightly different question ====
Where to go?
The rationals are are closed under addition/subtraction and multiplication/division.
So if $x + \sqrt{2}= y \in \mathbb Q$ and $x \in \mathbb Q$ then $y-x \in \mathbb Q$. But $y - x = (x + \sqrt{2}) -x = \sqrt{2} \not \in \mathbb Q$.
And that's really all there is to it.
====
There is a very tiny caveat:
rational + rational = rational
and rational + irrational = irrational (as we just proved)
but irrational + irrational could be either.
I once thought "But maybe we can state, if $x,y$ are irrational then $x + y$ is irratioanal, unless $x$ and $y$ are "dependant" that is $x = r + q*y$ for some rational $r$ and $q$". I even proved that statement and thought I had adequately expressed an important property.
Unfortunatly as $x + y = q \in \mathbb Q$ means $x = q - y$ by definition, my result was actually trivial and banal. sigh.