Are They Rational?

Number Theory Level 2

There exist three distinct rational numbers $a$ , $b$ , and $c$ such that $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is also rational. At least how many of $\sqrt{a}$ , $\sqrt{b}$ , and $\sqrt{c}$ have to be rational?

0 1 2 3

1 solution

Michael Mendrin
May 31, 2016

Relevant wiki: Rational Numbers

Let’s suppose $\sqrt{c}$ is rational and the other two are not. Then let the sum of the other two equals a rational number $d$ , so that we have

$\sqrt{a}+\sqrt{b}=d$

or, re-arranging, we have

$2d\sqrt{b}={d}^{2}+b-a$

which implies that $\sqrt{b}$ is rational, which is a contradiction.

Let’s suppose all are irrational, and let their sum be a rational number $d$ , so that we have

$\sqrt{a}+\sqrt{b}=d-\sqrt{c}$

Squaring both sides gets us

$a+2\sqrt{ab} +b={d}^{2}-2d\sqrt{c}+d$

which implies that the following is rational

$\sqrt{ab}+\sqrt{c{d}^{2}}$

But as we’ve seen above, this leads to a contradiction.

Being that just one of them being irrational is a trivial case that won’t work, we conclude that all $3$ of the terms have to be rational.

Note: By induction, we [ought to maybe be able to] extend this argument to show that the sum of any number of irrational square roots cannot be rational.

I believe the note isn't true. We can extend it to 5 terms, but not beyond that. The reason is that when we square the terms, we introduce a lot of cross terms. With 6 terms, we can split 3 on each side. Squaring this gives us 3 cross terms on each side, and thus we haven't reduced it to a smaller case.

The general theorem does take a bit more work to prove. Of course, if you have a simple induction proof, I would love to see it.

Calvin Lin Staff - 5 years ago

Okay, fixed my comment about that, shelving this matter for another time or note.

Michael Mendrin - 5 years ago

$\text{Great use of contradiction sir}$ , $(+1)$ !

Rishabh Tiwari - 5 years ago

Are They Rational?

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