# Q on irrational numbers

1. Jul 25, 2010

### Mantaray

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

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

2. Relevant 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

3. 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?

2. Jul 25, 2010

### Hurkyl

Staff Emeritus
Skimming it, the proof looks good.

3. Jul 25, 2010

### 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!