Stand Up & Math · Grade 9 · Activity
The People vs. √2
Proof by contradiction · Irrational numbers · Groups of 3–4 · Whiteboard
A claim has been made: "√2 can be written as a fraction." Today your group is the jury. You won't be told whether the claim is true — you're going to test it until it either survives or breaks. Before you can touch √2 itself, you need one rule of evidence.
Square every whole number from 1 to 8. For each one, note whether \(n\) is even or odd, and whether \(n^2\) is even or odd.
Complete this sentence, then check it against a couple more numbers of your own choosing: "If \(n^2\) is even, then \(n\) must be ___."
Proving "if \(n^2\) is even, then \(n\) is even" head-on is awkward. Instead, convince your group that this equivalent statement says the same thing, then prove that one instead:
"If \(n\) is odd, then \(n^2\) is odd."
Let \(n\) be odd, so \(n = 2k+1\) for some whole number \(k\). Expand \((2k+1)^2\) and show the result is odd — it should end up looking like \(2\times(\text{something}) + 1\).
Suppose — just suppose — the claim is true: \(\sqrt2\) can be written as a fraction. Then there exist whole numbers \(p\) and \(q\) with no common factor (the fraction is fully reduced) such that:
\(\sqrt2 = \dfrac{p}{q}\)
Square both sides of the equation above. Rearrange your result so it reads: \(p^2 = \) (something involving \(q\)).
Look at the equation you just found. What does it tell you about whether \(p^2\) is even or odd? Using your Phase 1 rule, what does that force to be true about \(p\) itself? Write \(p\) in the form \(2k\) once you're sure.
You showed \(p = 2k\). Substitute this into your Phase 2 equation \(p^2 = 2q^2\), and simplify as far as you can. You should be able to divide both sides by a common factor.
Look at your simplified equation. What does it tell you about \(q^2\)? Using the Phase 1 rule again, what does that force to be true about \(q\)?
Go back to how \(p\) and \(q\) were defined at the start of Phase 2. Is that condition still possible, given what you now know about both of them? Discuss with your group and write what's gone wrong on your board.
You assumed \(\sqrt2\) could be written as a fraction in lowest terms, and followed the logic honestly. Write your closing statement as if addressing the court: what did the assumption lead to, and what does that force you to conclude about \(\sqrt2\)?
Run the exact same trial on the claim "\(\sqrt4\) can be written as a fraction." Assume \(\sqrt4 = \frac{p}{q}\) in lowest terms, square, and follow the same steps you used above. Does the case collapse the same way? Find exactly where (if anywhere) the argument stops working.
Your Phase 1 rule was really about the number 2 (even vs. odd). The same style of argument works for any prime number in place of 2 — if a prime \(k\) divides \(n^2\), then \(k\) must divide \(n\).
Use this to put \(\sqrt3\) and \(\sqrt5\) on trial. Can you state, in general, which whole numbers \(N\) will always give an irrational \(\sqrt N\), and which won't?
You've worked through all four phases of The People vs. √2.
Well done — get ready for the class consolidation.