(adsbygoogle = window.adsbygoogle || []).push({}); 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 = x^{2}- 2x[tex]\sqrt{5}[/tex] + 5

=> 2[tex]\sqrt{6}[/tex] = x^{2}- 2x*[tex]\sqrt{5}[/tex]

=> (2[tex]\sqrt{6}[/tex])^{2}= (x^{2}- 2x[tex]\sqrt{5}[/tex])^{2}

=> 24 = x^{4}- 4x^{3}*[tex]\sqrt{5}[/tex] + 20x^{2}

- 4x^{3}*[tex]\sqrt{5}[/tex] is irrational because 4x^{3}is rational.

x^{4}- 4x^{3}*[tex]\sqrt{5}[/tex] + 20x^{2}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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# Homework Help: Q on irrational numbers

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