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!