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Q on irrational numbers 
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#1
Jul2510, 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 = 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? 


#3
Jul2510, 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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