Q on irrational numbers

Mantaray

1. The problem statement, all variables and given/known data

Is [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] rational?

2. Relevant equations

If n is an integer and not a square, then [tex]\sqrt{n}[/tex] is irrational

For a rational number a and an irrational number b,

a + b is irrational
a * b is irrational if a is not equal to 0

3. The attempt at a solution

Assume that [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] = x, with x being a rational number.

[tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] = x - [tex]\sqrt{5}[/tex]
=> ([tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex])² = (x - [tex]\sqrt{5}[/tex])²
=> 2 + 2[tex]\sqrt{6}[/tex] + 3 = x² - 2x[tex]\sqrt{5}[/tex] + 5
=> 2[tex]\sqrt{6}[/tex] = x² - 2x*[tex]\sqrt{5}[/tex]
=> (2[tex]\sqrt{6}[/tex])² = (x² - 2x[tex]\sqrt{5}[/tex])²
=> 24 = x⁴ - 4x³*[tex]\sqrt{5}[/tex] + 20x²

- 4x³*[tex]\sqrt{5}[/tex] is irrational because 4x³ is rational.
x⁴ - 4x³*[tex]\sqrt{5}[/tex] + 20x² is thus irrational.

The left hand side of the equation is rational, as 24 is a rational number.

This is a contradiction, thus our assumption was false, x cannot be a rational number.

[tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] is thus irrational

Is this a valid proof, or should the equation be worked out further?

Hurkyl

Skimming it, the proof looks good.

Mantaray

I looked at some threads on the internet, but worked the equation out to an eighth degree equation, so I wasn't sure this way was actually correct. Thanks a lot!