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

In summary, the homework asks you to prove that the value 2^.5 + 3^.5 is irrational, but you can't do it because all the coefficients of the polynomial have to be integers. A monic polynomial is a polynomial with only integer coefficients, so you should try to find a linear polynomial that the value is a root of.
  • #1
B-Con
26
0

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?
 
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  • #2
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
If [itex]\sqrt{2}+ \sqrt{3}[/itex] is a root of a polynomial (with integer coefficients), then the polynomial must have [itex]x-\sqrt{2}- \sqrt{3}[/itex] as a factor. Write that as [itex](x-\sqrt{2})-\sqrt{3}[/itex] 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 [itex]x-\sqrt{2}- \sqrt{3}[/itex].
 
  • #4
Thanks, I got it.

HallsofIvry said:
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.
 
  • #5
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
B-Con said:
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 [itex]x-\sqrt{2}-\sqrt{3}[/itex] was not the answer!
 
  • #7
matt grime said:
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.

HallsofIvy said:
Sorry, I saw "monic" but didn't think about it!
Heh, no problem. :tongue:
 

1. How do I find a monic polynomial with 2^.5 + 3^.5 as a root?

To find a monic polynomial with 2^.5 + 3^.5 as a root, we can use the fact that the sum of two square roots, such as 2^.5 and 3^.5, can be written as a single square root of a number. In this case, it can be written as the square root of 13. Therefore, the polynomial will have the form (x - (2^.5 + 3^.5))(x - (2^.5 - 3^.5)), which simplifies to x^2 - 13x + 1.

2. What is a monic polynomial?

A monic polynomial is a polynomial in which the coefficient of the highest degree term is equal to 1. This makes it easier to work with and solve for its roots.

3. Why is 2^.5 + 3^.5 a root of the polynomial?

2^.5 + 3^.5 is a root of the polynomial because when we substitute this value into the polynomial, we get a result of 0. This means that (2^.5 + 3^.5) is a solution to the equation x^2 - 13x + 1 = 0.

4. Can a monic polynomial have more than one root?

Yes, a monic polynomial can have more than one root. In fact, the number of roots of a polynomial is equal to its degree. In the case of our example, the polynomial is of degree 2 and has two roots: 2^.5 + 3^.5 and 2^.5 - 3^.5.

5. How can I verify that 2^.5 + 3^.5 is a root of the polynomial?

To verify that 2^.5 + 3^.5 is a root of the polynomial, we can use the method of substitution. This means that we substitute the value into the polynomial and check if the result is equal to 0. In this case, we get (2^.5 + 3^.5)^2 - 13(2^.5 + 3^.5) + 1 = 0, which confirms that it is indeed a root of the polynomial.

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