Is there a way to show that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?

  • Thread starter ~Death~
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In summary, the conversation is about solving a problem in algebra and the concept of algebraic numbers being closed under addition. The suggested method is to find a polynomial with integer coefficients and use linear algebra to prove the result.
  • #1
~Death~
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Hi,

i can't figure out this problem in my algebra book

I thought it was interesting that this thing was rational though...

anyone know how to start it?
 
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  • #2
I think i figured it out i might have been confused on the difference between algebraic and rational

idk how to show that algebraic numbers are closed under + though
 
  • #3
Erg, so there was a tiny treatment of algebraic numbers in Spivak's calculus text, but basically I have forgotten that chapter. But I think there was a trivial exercise where you show that if a > 0 is algebraic, then sqrt(a) is also algebraic. Also, intuitively, it seems true that the sum of algebraic numbers is also algebraic. On the other hand, there is probably an easier way.
 
  • #4
~Death~ said:
Hi,

i can't figure out this problem in my algebra book

I thought it was interesting that this thing was rational though...

anyone know how to start it?

In this particular case it is enough to find a polynomial (with integer coefficients) that has this as a root. It will be a polynomial of degree 8.

Start with equation [tex]x = \sqrt{2}+\sqrt{3}+\sqrt{5}[/tex] then eliminate the radicals ... isolate a radical on one side, then square. It takes three times to get rid of all three radicals, you end up with [tex]x[/tex] squared three times so you have [tex]x^8[/tex] in your polynomial.

I get
[tex]-40\,x^6+576-960\,x^2+352\,x^4+x^8 = 0[/tex]
 
  • #5
snipez90 said:
Also, intuitively, it seems true that the sum of algebraic numbers is also algebraic. On the other hand, there is probably an easier way.
The easiest way I can think is in terms of vector spaces over Q. (Meaning that you can only use rational numbers as scalars, rather than arbitrary real or complex numbers)
Theorem: A complex number a is algebraic over Q if and only if its powers 1, a, a2, a3, ... are linearly dependent.​
Can you prove this theorem? Can you see how to leverage linear algebra to prove the result you want?
 

1. What does it mean for a number to be algebraic over Q?

A number is algebraic over Q if it is a root of a polynomial with rational coefficients. In other words, it can be expressed as a finite combination of rational numbers using the operations of addition, subtraction, multiplication, and division.

2. How can we prove that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?

One way to prove this is by constructing a polynomial with rational coefficients that has sqrt(2)+sqrt(3)+sqrt(5) as a root. This can be done by using the fact that the sum and product of algebraic numbers is also algebraic.

3. Can we use the rational root theorem to show that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?

No, the rational root theorem only applies to polynomials with integer coefficients. In this case, the polynomial would have to have irrational coefficients.

4. Why is showing that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q important?

Proving that this number is algebraic over Q helps us understand the structure of rational and irrational numbers. It also has applications in fields such as number theory, algebraic geometry, and cryptography.

5. Are there other methods besides constructing a polynomial to show that sqrt(2)+sqrt(3)+sqrt(5) is algebraic over Q?

Yes, there are other methods such as using Galois theory or field extensions. Each method may have its own strengths and weaknesses, so the choice of method may depend on the specific problem at hand.

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