Mathematics I / Rational and Irrational Numbers
Practice question · Put in order

Order the steps of the proof by contradiction that the square root of 2 is irrational.

Hints
  1. The proof opens by assuming exactly the opposite of what is to be shown.
  2. The contradiction must strike the 'lowest terms' condition set up in the very first line.
Show the answer
  1. Assume the square root of 2 is rational and write it as a/b in lowest terms
  2. Square both sides and rearrange to get a squared = 2 b squared
  3. Deduce that a is even, so write a = 2k
  4. Substitute to get b squared = 2 k squared, so b is even as well
  5. 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.

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