$\sqrt{m}+\sqrt{n}\notin\mathbb{Q}$ Irrationals, Roots Proof

Question

Let $n,m\in\mathbb{N}_0$ and $\sqrt{n}\not\in\mathbb{Q}$. Conclude that $\sqrt{n}+\sqrt{m}\not\in\mathbb{Q}$

What would you recommend me to read to be able to solve this?

Because apart from the case that if $\sqrt{m}\in\mathbb{Q}$ I am kinda stuck.

If $\sqrt{m}\in\mathbb{Q}$ then by "rational+irrational=irrational" the statement holds. (still left to prove tho..)

Presumably without using Galois theory? Not sure how you're supposed to do it... — Dustan Levenstein, Nov 27 '16 at 03:40
Similar questions have been asked here many times before: http://math.stackexchange.com/questions/457382/can-sqrtn-sqrtm-be-rational-if-neither-n-m-are-perfect-squares?noredirect=1&lq=1. — StubbornAtom, Nov 27 '16 at 07:59

score 5 · Accepted Answer · edited Jun 24 '22 at 10:02

Hint $\,\ u = \sqrt n + \sqrt m \in\Bbb Q\ \Rightarrow\ v := \sqrt n - \sqrt m = \dfrac{n-m}{\sqrt n + \sqrt m} = \dfrac{n-m}u\in \Bbb Q.$

Therefore $\ u+v = 2\sqrt n\in \Bbb Q,\,$ contradiction.

Remark $ $ The key idea behind the proof is as follows. If field $\rm\,F\,$ has two $\rm\,F$-linear independent combinations of $\rm\, \sqrt{n},\ \sqrt{m}\, $ then we can solve for $\rm\, \sqrt{n},\ \sqrt{m}\, $ in $\,\rm F.\,$ In our case we notice $\rm\ F = \mathbb Q(\sqrt{n} + \sqrt{m})\ $ contains the independent $\rm\ \sqrt{n} - \sqrt{m}\ $ since

$$\rm \sqrt{n} - \sqrt{m}\ =\ \dfrac{\ a\,-\,b}{\sqrt{n}+\sqrt{m}}\ \in\ F = \mathbb Q(\sqrt{n}+\sqrt{m}) $$

To be explicit, notice that $\rm\ u = \sqrt{n}+\sqrt{m},\ v = \sqrt{n}-\sqrt{m}\in F\ $ so solving the linear system for the roots yields $\rm\ \sqrt{n}\ =\ (u+v)/2,\ \ \sqrt{m}\ =\ (u-v)/2,\ $ both of which are clearly $\rm\,\in F,\:$ since $\rm\:u,\:v\in F\:$ and $\rm\:2\ne 0\:$ in $\rm\:F,\:$ so $\rm\:1/2\:\in F.\:$ This works over any field where $\rm\:2\ne 0,\:$ i.e. where the determinant (here $2$) of the linear system is invertible, i.e. where the linear combinations $\rm\:u,v\:$ of the square-roots are linearly independent over the base field.

This idea often proves useful , e.g. the Primitive Element Theorem works that way, obtaining two such independent combinations by Pigeonholing the infinite set $\rm\ F(\sqrt{a} + r\ \sqrt{b}),\ r \in F,\ |F| = \infty,\,$ into the finitely many fields between F and $\rm\ F(\sqrt{a}, \sqrt{b}),\,$ e.g. see PlanetMath's proof.

I can follow your argument. But i don't know how I could come up with that "out of the blue". Thanks nonetheless. — SAJW, Nov 27 '16 at 04:20

score 0 · Answer 2 · answered Nov 27 '16 at 07:37

0

In addition to Bill Dubuque's answer I made a few additions. Mainly to ensure that all denominators are not 0. Tell me if it's superfluous.

$\sqrt{n}\notin\mathbb{Q}\implies n>0\implies\sqrt{n}+\sqrt{m}>0$.

$n=m\implies \sqrt{n}+\sqrt{m}=2\sqrt{n}\notin\mathbb{Q}$.

Let $m\not=n$.

Assume $u=\sqrt{n}+\sqrt{m}$ is in $\mathbb{Q}$.

Then $\frac{n-m}{u}$ is also in $\mathbb{Q}$.

$\frac{n-m}{\sqrt{n}+\sqrt{m}}=\frac{n-m}{\sqrt{n}+\sqrt{m}}\cdot\frac{\sqrt{n}-\sqrt{m}}{\sqrt{n}-\sqrt{m}}=\sqrt{n}-\sqrt{m}=v\in\mathbb{Q}$.

$u+v=2\sqrt{n}\in\mathbb{Q}$, which is a contradiction.

Therefore $\sqrt{n}+\sqrt{m}\not\in\mathbb{Q}$.

answered Nov 27 '16 at 07:37

SAJW

1,327

It is not necessary to separate the case $, n = m.,$ since the argument I sketched works fine in that case. Note $,\sqrt n > 0,$ so $,u = \sqrt n + \sqrt m > 0,$ so all denominators are $> 0.\ \ $ – Bill Dubuque Nov 28 '16 at 16:26
@BillDubuque but if $\sqrt{n}=\sqrt{m}$ then the rationalization doesn't work. $\sqrt{n}-\sqrt{n}=0$ – SAJW Nov 28 '16 at 17:24
But there is no "rationalization" in my sketch. The equation I wrote is true even when $, n = m.\ $ It's just the special case $,v = 0,\ $ so $, u = u+v = 2\sqrt n\in \Bbb Q.\ $ – Bill Dubuque Nov 28 '16 at 17:37
@BillDubuque oh, right. Thanks for pointing that out. – SAJW Nov 28 '16 at 17:44

$\sqrt{m}+\sqrt{n}\notin\mathbb{Q}$ Irrationals, Roots Proof

2 Answers2