I would like to know how to prove that $\sqrt{2} + \sqrt{5} + \sqrt{7}$ is an irrational number. I know how to do the proof for a sum of two roots. Can I just define $\sqrt{2} + \sqrt{5} :=c$ and then prove that $c$ and $c + \sqrt{7}$ are both irrational, when I treat $c$ as a single number? If this way is kind of wrong or won't work, is there another way to do the proof?
$\begingroup$
$\endgroup$
4
-
5$\begingroup$ But watch out: $\sqrt 2+\sqrt{8}-\sqrt{18}$ is rational! $\endgroup$– Hagen von EitzenApr 23, 2014 at 20:40
-
8$\begingroup$ Prove. Prove. Prove. Prove. Prove. Prove. $\endgroup$– Pedro ♦Apr 23, 2014 at 20:40
-
2$\begingroup$ Treat, rather than tread. $\endgroup$– Andrés E. CaicedoApr 23, 2014 at 20:50
-
$\begingroup$ And $c$ is a single number. What else would it be? $\endgroup$– Andrés E. CaicedoApr 23, 2014 at 20:51
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
0
One way to prove that the sum of two square roots (e.g. $\sqrt 2 + \sqrt 5$) is irrational is to suppose not, and square once. Upon rearrangement, you'll find a contradiction. Here we can do the same thing, squaring twice with some algebra: Suppose $$r = \sqrt 2 + \sqrt 5 + \sqrt7$$ were rational; then
$$r - \sqrt7 = \sqrt2 + \sqrt5$$
Squaring both sides, you'll find that
$$r^2 + 7 - 2 r \sqrt7 = 2 + 5 + 2\sqrt{10}$$
Simplify, place both radicals one one side of the equation, and square again; try to get a contradiction from here.
Of course, this doesn't always work: There are triplets of numbers such that the corresponding $r$ is rational, so carefully identify where the contradiction is.
$\begingroup$
$\endgroup$
1
(A stronger result for reference)
If $a_i$ are postive integers, then $\sum \sqrt[n]{a_i}$ is rational iff $\sqrt[n]{a_i}$ is.
answered Apr 23, 2014 at 21:08
