# Q on irrational numbers

by Mantaray
Tags: irrational, numbers
 P: 17 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?
 PF Patron Sci Advisor Emeritus P: 16,094 Skimming it, the proof looks good.
 P: 17 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!

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