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

[B-Con]

Homework Statement

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?

 

[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.

 

[HallsofIvy]

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}.

 

[B-Con]

Thanks, I got it.

By the way, saying "with integer coefficients" or at least "with rational coefficients" is necessary in this problem.

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

 

[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.

 

[HallsofIvy]

Thanks, I got it.


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

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!

 

[B-Con]

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.

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.

Sorry, I saw "monic" but didn't think about it!

Heh, no problem.