# Proving irrationality

1. Mar 15, 2005

### rsnd

how do you prove that sqrt(2) + sqrt(5) is irrational...

2. Mar 15, 2005

### matt grime

by contradiction, suppose r=sqrt(2)+sqrt(5) and r is rational. play around with that and see what you can show this implies.

3. Mar 15, 2005

### rsnd

i did get it down to 7+2root10 = m^2/n^2 where i assumed root 2 + root 5 = m/n...and before that i proved root 10 to be irrational...so i am kinda stuck on the 7+2root10 = m^2/n^2 bit...how does that mean anything...

4. Mar 15, 2005

### matt grime

why not do rearrange the equation before squaring it so that after squaring you only have, say sqrt(2) to deal with, which i presume you know is irrational.

i don't see why you're showing 7+ something is not of the from m^2/n^2, to be honest, either, and you certainly don't need to write r=m/n either.

5. Mar 15, 2005

look at this

6. Mar 15, 2005

### matt grime

I think that's aiming way too high for this question, and besides, if you're going to go and use theorems like that then why not simply invoke the fact that sqrt(2)+sqrt(3) is an algebraic integer, and hence if it were rational it would be integral, which it clearly isn't.

r-sqrt(2) = sqrt(5)
square and rearrange to get sqrt(2) rational, contradiction.

7. Mar 15, 2005

### rsnd

i had an argumetn like this

this is where i got up to...
now i am thinking that if i put everything in terms of root 10 then:
so we have irrational = "(m2/n2 - 7)/2"
so that implies that (m2/n2 - 7) is irrational and also m2/n2 is irrational and they cannot be irrationals and integers at the same time and therefore its a contradiction...makes sence??? may b not...im just insane!!!

8. Mar 15, 2005

### HallsofIvy

Staff Emeritus
Assuming you already know &radic;(2) is irrational, matt grime's suggestion, "rearrange the equation before squaring it " gives this:

Squaring both sides, r2- 2&radic;(2)r+ 2= 5 so 2&radic;(2)= r2- 3. Then &radic;(2)= (r2- 3)/2. Since the rational numbers are closed under multiplication, division, addition, and subtraction, the right hand side is rational, giving the contradiction that &radic;(2) is rational.

9. Mar 15, 2005

### Curious3141

Another way of doing it, which I think is slightly "nicer" than squaring is as follows :

Let $$\sqrt{5} + \sqrt{2} = x_1$$ and $$\sqrt{5} - \sqrt{2} = x_2$$

Assume $$x_1$$ is rational and can be expressed as $$\frac{p}{q}$$, where p and q are coprime integers.

$$\frac{1}{2}(x_1 - x_2) = \sqrt{2}$$, which is irrational, hence $$x_2$$ has to be irrational (since $$x_1$$ is rational by assumption).

But $$x_1x_2 = 3$$, giving $$x_2 = \frac{3}{x_1} = \frac{3q}{p}$$

This implies $$x_2$$ is rational, contradicting the above.

Hence $$x_1$$ is irrational (QED).

10. Mar 16, 2005

### HallsofIvy

Staff Emeritus
Yes, that is nice.

11. Mar 16, 2005

### Curious3141

Thank you.

BTW, there is a small error in your explanation of the squaring proof. It should go like so :

$$r^2 - 2\sqrt{2}r + 2 = 5$$
$$2\sqrt{2}r = r^2 - 3$$
$$\sqrt{2} = \frac{1}{2}(r - \frac{3}{r})$$

And LHS is irrational, RHS is rational, contradiction. You forgot the 'r' when you typed it out. Just a minor point, doesn't change the substance of your proof at all.

12. Mar 18, 2005

### rsnd

thanks to all! it kinda makes sence!