An elementary way to prove that $\sqrt{2}+\sqrt{3}$ is irrational?

Question

Disclaimer: I know about the rational root theorem. I am trying to answer it without using it.

I did the following:

$$\sqrt{2}+\sqrt{3}=\frac{a}{b}$$

$$5+2\sqrt{6}=\frac{a^2}{b^2}$$

$$24=\frac{a^4}{b^4}-10\frac{a^2}{b^2} + 25$$

$$\frac{a^2}{b^2}\left(10-\frac{a^2}{b^2} \right)= 1$$

At this point, I am a bit confused. I thought about this: If $A=\frac{a^2}{b^2}$ then $A^{-1}=\left(10-\frac{a^2}{b^2} \right)$ and then:

$$AA^{-1}=A\left(10-A \right)=10A-A^2=1\tag{$\star$}$$

This is a polynomial with roots: $A=5\pm 2\sqrt{6}$ which means that:

$$\frac{a^2}{b^2}=5\pm 2\sqrt{6}$$

$$\frac{a^2}{2b^2}-\frac{5}{2}=\pm \sqrt{6}$$

As $\frac{a^2}{2b^2}-\frac{5}{2}$ is rational and $\sqrt{6}$ is irrational, then we obtain a contradiction. Is this correct?

Hint: by your 2nd equation it suffices to show $\sqrt 6\not\in \Bbb Q$ (which can be done the same way as $\sqrt 2,$ for which there are hundreds of prior answers here). — Bill Dubuque, Aug 23 '19 at 16:05
Duplicate of Prove that $\sqrt 2 + \sqrt 3$ is irrational and likely hundreds more. — Bill Dubuque, Aug 23 '19 at 16:09

score 1 · Answer 1 · answered Aug 23 '19 at 16:05

1

From your second line of working you should get that $$5+2\sqrt{6}=\frac{a^2}{b^2}\implies\sqrt{6}=\frac{a^2-5b^2}{2b^2}\in\mathbb{Q}$$ which is a contradiction.

answered Aug 23 '19 at 16:05

Peter Foreman

19,947

Damn. I ended up saying a lot of stuff to just say what I was already saying! facepalm. – Red Banana Aug 23 '19 at 16:07

score 1 · Answer 2 · answered Aug 23 '19 at 16:12

1

Your proof is corect but you could have stopped on the second line and get the same result.

You have $$ 5+2\sqrt{6}=\frac{a^2}{b^2}$$ which implies that $\sqrt 6$ is rational and that is a contradiction that you are looking for.

answered Aug 23 '19 at 16:12

Mohammad Riazi-Kermani

68,728

Can down-voter explain us why did you do it? – Michael Rozenberg Aug 23 '19 at 16:25
3

@MichaelRozenberg Trusted users should A) search first when a question smells like a duplicate, and B) IMHO also steer clear from stunning displays of banalities. Not saying that it is always easy to resist the temptation to answer every trivial thing, but... – Jyrki Lahtonen Aug 24 '19 at 03:56

Robert Lewis · Answer 3 · 2019-08-23T18:29:46.370

It may be addressed in vast generality, to wit:

Let

$r, s \in \Bbb N, \tag 1$

neither of which is a perfect square; if

$\sqrt r + \sqrt s = q \in \Bbb Q, \tag 2$

then

$\sqrt r = q - \sqrt s, \tag 3$

which upon squaring leads to

$r = q^2 - 2q\sqrt s + s = (q^2 + s) - 2q\sqrt s, \tag 4$

$2q\sqrt s = (q^2 + s) - r, \tag 5$

and thus

$\sqrt s = \dfrac{(q^2 + s) - r}{2q} \in \Bbb Q, \tag 6$

valid since $q \ne 0$ from (2); but this contradicts the well-known fact that an integer which is not a perfect square cannot have a rational square root, proved, for example, in the answers to this question.

Now we simply set

$r = 2, \; s = 3, \tag 7$

to obtain the specific result requested in the question title.

And so we are done.

An elementary way to prove that $\sqrt{2}+\sqrt{3}$ is irrational?

3 Answers3