# Finding a monic polynomial with 2^.5 + 3^.5 as a root

1. Oct 18, 2007

### B-Con

1. The problem statement, all variables and given/known data
Number Theory homework. I'm supposed to prove that the value
2^.5 + 3^.5
is irrational by finding a monic polynomial that the aforementioned number is a root of. This would be trivial if I were supposed to prove that just 2^.5 was irrational because its monic polynomial equation would be x^2 -2 = 0.

But I can't just raise 2^.5 + 3^.5 to a power and then subtract it to get my desired polynomial because all coefficients of a monic polynomial have to be integers, and 2^.5 + 3^.5 is obviously not an integer.

What method should I use to go about finding such a monic polynomial?

Last edited: Oct 18, 2007
2. Oct 18, 2007

### matt grime

You should just do it, in the words of Tim Gowers. Of course raising it to some power and then subtracting it won't work. Why should it? But what's stopping you doing other things. A polynomial is just a linear dependence over Q of the various powers of that number. Try to find one.

3. Oct 18, 2007

### HallsofIvy

Staff Emeritus
If $\sqrt{2}+ \sqrt{3}$ is a root of a polynomial (with integer coefficients), then the polynomial must have $x-\sqrt{2}- \sqrt{3}$ as a factor. Write that as $(x-\sqrt{2})-\sqrt{3}$ and think "(x-a)(x+a)= x2- a2".

By the way, saying "with integer coefficients" or at least "with rational coefficients" is necessary in this problem. Without that, an obvious answer is the linear polynomial $x-\sqrt{2}- \sqrt{3}$.

4. Oct 19, 2007

### B-Con

Thanks, I got it.

Isn't that what a monic polynomial is by definition? Sorry, didn't mean to be ambiguous.

5. Oct 19, 2007

### matt grime

Now you've got it, I should point out that the reference to Gowers wasn't completely out of the blue. He has a web page with this very problem on it done purely in terms of linear algebra.

6. Oct 19, 2007

### HallsofIvy

Staff Emeritus
Sorry, I saw "monic" but didn't think about it! Anyway, you weren't all that "ambiguous". I was pretty sure $x-\sqrt{2}-\sqrt{3}$ was not the answer!

7. Oct 19, 2007

### B-Con

Thanks for the tip, I'll check it out. I figured Gowers was simply someone who I hadn't heard of before.

By the way, I like the quote in your sig.

Heh, no problem. :tongue: