Practice question · Put in order
Order the steps of the proof by contradiction that the square root of 2 is irrational.
- Assume the square root of 2 is rational and write it as a/b in lowest terms
- a and b are both even, contradicting the choice of lowest terms
- Substitute to get b squared = 2 k squared, so b is even as well
- Deduce that a is even, so write a = 2k
- Square both sides and rearrange to get a squared = 2 b squared
Hints
- The proof opens by assuming exactly the opposite of what is to be shown.
- The contradiction must strike the 'lowest terms' condition set up in the very first line.
Show the answer
- Assume the square root of 2 is rational and write it as a/b in lowest terms
- Square both sides and rearrange to get a squared = 2 b squared
- Deduce that a is even, so write a = 2k
- Substitute to get b squared = 2 k squared, so b is even as well
- a and b are both even, contradicting the choice of lowest terms
Why
The lowest-terms assumption is the trap being laid in step one: the algebra then forces both a and b to be even, so the fraction was reducible after all. Since every rational HAS a lowest-terms form, no such fraction can exist, and the square root of 2 is irrational.
Practise Rational and Irrational Numbers
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