Stand Up & Math · Teacher Answers
The People vs. √2
Not for distribution · Proof by contradiction · Full worked solutions
This is a genuine proof, not a computation exercise — the payoff is logical, not numerical. Expect this to feel unfamiliar even to your strongest students; the unfamiliarity is the point. Resist the urge to explain the contradiction for them at any gate — the entire activity is built so the contradiction is something a group discovers together and has to argue about before you confirm it.
"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."
Students work entirely on their whiteboard. Display phase content on student devices only after each gate is unlocked.
Even/odd behaviour of squares
Every even \(n\) gives an even \(n^2\); every odd \(n\) gives an odd \(n^2\).
Groups should notice the pattern within the first 4–5 values. Push them to state it as a general rule (Task 1b) rather than just "yeah it alternates."
"If \(n^2\) is even, then \(n\) must be even."
Let \(n = 2k+1\) (odd), for some integer \(k\).
This is \(2\times(\text{integer})+1\), which is odd. So odd \(n\) ⟹ odd \(n^2\).
"If n odd, then n² odd" and "if n² even, then n even" are contrapositives of each other — logically identical statements. This is very likely the first time most students have used a contrapositive on purpose. It's fine (expected, even) if groups need you to walk through why these two statements are the same before they attempt the algebra — do this once as a mini-conversation, not a lecture, e.g. "if it's not true that n is even, could n² possibly be even?"
Require: correct expansion of \((2k+1)^2\), correctly identified as odd, AND some acknowledgment (even informal) that this proves the even case too.
Setting the trap
\(p^2 = 2q^2\) is 2 × (an integer), so \(p^2\) is even. By the Phase 1 rule, \(p\) must be even.
Require: \(p^2=2q^2\) correctly derived, and \(p\) correctly identified as even with the Phase 1 rule cited as the reason — not just asserted.
Following the substitution
\(q^2 = 2k^2\) is even by the same reasoning, so (Phase 1 rule again) \(q\) must also be even.
\(p\) and \(q\) were assumed to share no common factor (lowest terms). But both are now shown to be even — they share a factor of 2. That's a direct contradiction of the starting assumption.
Require the group to explicitly name what contradicts what — "both p and q are even, but we assumed they had no common factor." A vague "it doesn't work" isn't enough; push them to say the actual contradiction.
Closing argument
The assumption "√2 can be written as a fraction in lowest terms" leads to a logical impossibility. Since the assumption cannot be true, its opposite must be: √2 cannot be written as a fraction — it is irrational.
Accept any phrasing that correctly identifies: (1) we assumed rationality, (2) that led to a contradiction, (3) therefore the assumption was false, (4) therefore √2 is irrational. This is their first experience of a full proof by contradiction — the structure matters as much as the content.
Assume \(\sqrt4 = \frac{p}{q}\) in lowest terms.
This does not force \(q\) to be even — it just gives \(k=q\), i.e. \(p = 2q\), i.e. \(\frac{p}{q}=2\). No contradiction appears; the case simply confirms \(\sqrt4=2\), which is true and unremarkable.
This is the heart of the enrichment: students see that the SAME machinery, applied honestly, doesn't always produce a contradiction — it only does so when the number under the root isn't a perfect square. That distinction is usually left completely implicit in how this proof is taught; making students find it themselves is what makes this a genuinely advanced task rather than a memorized script.
The Phase 1 rule generalizes: for any prime \(k\), if \(k \mid n^2\) then \(k \mid n\). This genuinely requires \(k\) to be prime — it fails for composite numbers (e.g. \(4 \mid 36 = 6^2\), but \(4 \nmid 6\)). Running the identical 4-phase argument with 3 or 5 in place of 2 produces the same style of contradiction, since 3 and 5 are prime and not perfect squares.
Strong groups may only get as far as "it works for 3 and 5 too" without stating the general theorem — that's a completely reasonable stopping point for a first encounter. If a group reaches the full generalization unprompted, that's genuinely university-level territory; feel free to tell them so.
1. Cold-call a group to state the Phase 1 rule and why the contrapositive trick was necessary.
2. Have a different group walk the whole class through the contradiction in Phase 3 — the "both even" moment is the crux; make sure everyone actually sees it.
3. Ask: "Where exactly did the √4 case stop working?" — this is the question that separates surface-level pattern-matching from real understanding.
4. If time allows, name-drop that this exact proof is attributed to the ancient Greeks and that the discovery reportedly unsettled the Pythagoreans, who believed all numbers were ratios of whole numbers.