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Q on irrational numbers

by Mantaray
Tags: irrational, numbers
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Jul25-10, 09:12 AM
P: 17
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?
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Jul25-10, 09:20 AM
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Hurkyl's Avatar
P: 16,091
Skimming it, the proof looks good.
Jul25-10, 09:28 AM
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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