# Irrational demonstration

1. Sep 25, 2012

### inverse

Demonstrate that $\sqrt{2}+\sqrt{3}$ is irrational.

Thanks

2. Sep 25, 2012

### Eval

This feels like homework. However, I will give a proof just to make sure I still can:

For purposes of contradiction, assume $\sqrt{2}+\sqrt{3}$ is rational. That is, there are some integers n and m (m≠0) such that $\sqrt{2}+\sqrt{3}=\frac{n}{m}$ and gcd(n,m)=1 (in other words, n and m are relatively prime). We have, then, that:
$\sqrt{2}+\sqrt{3}=\frac{n}{m}$
$m(\sqrt{2}+\sqrt{3})=n$
$m(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})=n(\sqrt{2}-\sqrt{3})$
$m(4+9)=n(\sqrt{2}-\sqrt{3})$
$13m=n(\sqrt{2}-\sqrt{3})$

Since 13 is prime and m does not divide n, 13|n and $m|(\sqrt{2}-\sqrt{3})$. This implies, then, that $\sqrt{2}-\sqrt{3}$ is in fact an integer. Disregarding the fact that this is not the case, we still have, then, that $l=\sqrt{2}-\sqrt{3}$ for some l, but $\sqrt{2}+\sqrt{3}=l+2\sqrt{3}$. Since $\sqrt{12}$ is irrational and l is rational, their sum is irrational, further contradicting the original assumption.

To use the latter contradiction, you would need to further show that $\sqrt{12}$ is irrational, so I would stick with the former contradiction (that $\sqrt{2}-\sqrt{3}$ is an integer)

3. Sep 25, 2012

### inverse

I fail to understand how I show from the expression $\sqrt{2}-\sqrt{3}$that must comply something for m and n is irrational.

4. Sep 25, 2012

### Norwegian

If Eval had done the first part correctly, he/she would have arrived at the otherwise immediate
√3 - √2 = m/n. Adding this equation to √3 + √2 = n/m, you obtain that √3 is rational, which is a contradiction.

Another way is just squaring both sides of √3 + √2 = n/m, giving √6 rational, and again a contradiction.

5. Sep 25, 2012

### micromass

Staff Emeritus
Moved to homework. Please do not give out full solutions.

6. Sep 25, 2012

### inverse

It is not to homwork, is to pass an exam.

Thanks

7. Sep 25, 2012

### micromass

Staff Emeritus
It is a homework-type question. It still belongs in homework and the homework rules apply.

8. Sep 25, 2012

### inverse

I have proposed a model for solving methodology see, because if I have a reference (Worked) can not practice more similar exercises.

9. Sep 26, 2012

### Eval

This post contains incorrect info (after the quote, of course)
To avoid misinforming the person asking the question, I would like to note that a conjugate is not an inverse. A conjugate of (a+b) is (a-b) whereas the inverse of (a+b) is (a-b)/(a2-b2). Applying this:

If √3 + √2 = n/m, then 1/(√3 + √2) = m/n. Then, multiplying the top and bottom of the lefthand side by its conjugate, we get (√3 - √2)/(92+22) = (√3 - √2)/13 = m/n, not √3 - √2 = m/n. Then you can add this to (√3 + √2)/13 = n/(13m) to get that (√3)/13 = m/n+n/13m = (13m2+n2)/(13nm), so √3 = (13m2+n2)/(nm) implying that √3 is the ratio of two integers, which is a contradiction. This is no more immediate than the proof I gave :P

Also, squaring both sides of √3 + √2 = n/m would imply that √6 is rational after several more steps and is probably the easiest method. However, in all the methods given by Norwegian, I would like to point out that you would further have to show that √3 and √6 are also irrational, respectively, so you will have some more work to do.

The easiest way to prove this would be to assume that they are rational (so the ration of two integers).

Last edited: Sep 26, 2012
10. Sep 26, 2012

### Norwegian

Hi Eval,
I invite you to reconsider the above. In particular, I want you to observe the elementary identity (√3 + √2)(√3 - √2)=1 (which by the way shows that conjugation equals inverse in this case).

11. Sep 26, 2012

### Eval

Ah, right, sorry. It is not exactly an identity, but yes, you are right. It is because I was still thinking (a+b) as opposed to (sqrt(a)+sqrt(b)). The identitiy there is a-b, which in thios case happens to be 1 (which is extremely useful).

Thanks for the catch, I'll edit my previous post so that I don't mislead people. Blame it on the lack of sleep I have had, my brain is failing me :/