Is \sqrt{2} + \sqrt{3} + \sqrt{5} irrational?

[Mantaray]

Homework Statement

Is \sqrt{2} + \sqrt{3} + \sqrt{5} rational?

Homework Equations

If n is an integer and not a square, then \sqrt{n} 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

The Attempt at a Solution

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

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

- 4x³*\sqrt{5} is irrational because 4x³ is rational.
x⁴ - 4x³*\sqrt{5} + 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.

\sqrt{2} + \sqrt{3} + \sqrt{5} 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!