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Proof by contradiction (1 Viewer)
- Thread starter kasser
- Start date
Kurosaki
True Fail Kid
Assume that the cube root of 2 is rational, i.e. 
where a and b are integers, and have no common factors
.
Therefore a must be even. Let a=2k, k is an integer
.
Therefore b is even. But we assumed that a and b have no common factors. Contradiction, therefore follows cube root of 2 is irrational.
Therefore a must be even. Let a=2k, k is an integer
Therefore b is even. But we assumed that a and b have no common factors. Contradiction, therefore follows cube root of 2 is irrational.