- #1
Mantaray
- 17
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Homework Statement
Is [tex]\sqrt{2}[/tex] + [tex]\sqrt{3}[/tex] + [tex]\sqrt{5}[/tex] rational?
Homework 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
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?