flip it, assume the opposite, or find the one case that breaks it.
Every chapter before this one used reasoning; here we make the reasoning itself the thing we study. A conditional if p then q casts three shadows — its converse, its inverse, and its contrapositive — and exactly one of them, the contrapositive, always carries the same truth value as the original. That single fact is a quiet superpower: when a head-on proof is awkward, you may prove the equivalent contrapositive instead, or assume the conclusion false and chase the assumption into a contradiction. To knock down a sweeping "it always holds," you need just one stubborn counterexample. And when we say a sequence gets "as close as you like" to a number, that phrase is not vague hand-waving — it is a precise claim, the seed from which the limits of Stage 36 will grow. Let us reason out loud.
Start with a conditional if p then q — hypothesis p in blue, conclusion q in amber. From it we can build three relatives by swapping and negating:
Here is the headline. A statement and its contrapositive are logically equivalent: they are true together and false together, no exceptions. The same equivalence ties the converse to the inverse. But the original and its converse are not tied — one can be true while the other is false.
Only the contrapositive ¬q ⇒ ¬p is guaranteed to share the original's truth. In symbols the engine checks for us: IMP(p,q) always equals contra(p,q). Treating the converse as if it followed is the single most common slip in beginning proofs.
Take if a number is a multiple of 4, then it is even. The original is true. Its converse — "if even, then a multiple of 4" — is false (6 is even but not a multiple of 4). Its contrapositive — "if not even, then not a multiple of 4" — is true, matching the original. So original ≡ contrapositive, both true; converse ≡ inverse, both false.
If the contrapositive is just the original in disguise, then proving either one proves both. That is licence to choose the easier road. Some conclusions are hard to reach by marching forward from the hypothesis, yet their contrapositive practically proves itself.
Claim: if n² is even, then n is even. Marching forward from "n² is even" is clumsy. So prove the contrapositive instead: if n is odd, then n² is odd. Write n = 2k + 1; then n² = 4k² + 4k + 1 = 2(2k² + 2k) + 1, which is odd. The contrapositive holds, so by equivalence the original holds too. Done.
Notice the move: we never touched the original directly. We attacked its guaranteed-equivalent twin, which was friendlier, and the equivalence carried the win back. This is the everyday engine of clean proofs.
A second route uses the same equivalence in spirit. To prove a statement, assume it is false, then reason carefully until you derive a contradiction — a clash with something already known to be true. Since a true statement can never lead to a contradiction, the assumption was impossible, so the statement must be true. We first met this technique back in Stage 16 · Proof by Contradiction; here we see it as a face of the contrapositive.
The most famous example is the proof that √2 is irrational — that no fraction ab in lowest terms can square to 2 — the same number we placed on the line in Stage 6 · Irrational & Real Numbers.
A contradiction must be a real clash — two facts that cannot both hold — not just a result that "feels wrong" or that you dislike. "That's surprising" is not a contradiction. "a and b are both even, yet a/b was in lowest terms" is one.
Proving and disproving are not mirror images. To establish a universal "for every x, p(x)," you must cover every case — one example proves nothing. But to overturn it, you need just one case where p fails. That single failure is a counterexample, and it settles the matter instantly — exactly the negation rule that turns "∀x, p(x)" into "∃x, ¬p(x)."
Claim: every odd number is prime. Test the odd numbers in turn: 3 ✓, 5 ✓, 7 ✓ — tempting! But 9 = 3 · 3 is odd and not prime. One counterexample, x = 9, and the universal is false. (And 1 is not prime either, so it fails too.) The lesson: a streak of confirmations is not a proof.
One last idea opens the door to what comes next. Look at the sequence aₙ = 1 + 1/n: 2, 1.5, 1.333…, 1.25, 1.2, … . The terms never reach 1, yet they crowd toward it: pick any tiny gap you like — a thousandth, a millionth — and from some point on, every term sits inside that gap of the green line L = 1.
That phrase "as close as you like, from some point on" is itself a precise logical claim — a quantified statement of exactly the kind we have been parsing. It is the heart of the ε-language of limits in Stage 36: for every target closeness ε > 0, there exists an N beyond which |aₙ − L| < ε. Reasoning, quantifiers, and "for all / there exists" are about to pay off.
Four moves, one toolbox. Read the table once; it is the whole lesson in miniature.
| Form | Reads | Equivalent to |
|---|---|---|
| Original | p ⇒ q | Contrapositive |
| Converse | q ⇒ p | Inverse |
| Inverse | ¬p ⇒ ¬q | Converse |
| Contrapositive | ¬q ⇒ ¬p | Original |
Write the converse, inverse, and contrapositive of if a shape is a square, then it is a rectangle. Which are true?
Converse: if a rectangle, then a square — false. Inverse: if not a square, then not a rectangle — false. Contrapositive: if not a rectangle, then not a square — true. Original (true) ≡ contrapositive (true); converse (false) ≡ inverse (false).
Your friend says: "Since if it rains, the ground is wet is true, the ground being wet proves it rained." Name the error.
They assumed the converse if wet then rains, which is not equivalent to the original. A sprinkler wets the ground with no rain. Only the contrapositive — "if not wet, then it did not rain" — is guaranteed.
Prove: if n² is odd, then n is odd, by proving the contrapositive.
Contrapositive: if n is even, then n² is even. Let n = 2k; then n² = 4k² = 2(2k²), which is even. The contrapositive holds, so the original holds by equivalence.
Disprove: every prime number is odd. Give the smallest counterexample.
x = 2: it is prime but even. One counterexample makes the universal false. (Every prime past 2 is odd, but "every" must mean every.)
Outline a proof by contradiction that there is no largest natural number.
Assume there is a largest natural number N. Then N + 1 is also a natural number and N + 1 > N, so N was not the largest — a contradiction. The assumption fails, so no largest natural number exists. ∎
For aₙ = 1 + 1/n, how large must n be so that aₙ is within 0.01 of L = 1? Within 0.001?
aₙ − 1 = 1/n, so we need 1/n < 0.01, i.e. n > 100 — any n ≥ 101. For 0.001 we need n > 1000, so n ≥ 1001. For any chosen gap ε there is such an N: that is "as close as you like."
Six questions to lock it in. Tap the answer you think is right.
This lesson develops MP3 (construct viable arguments and critique the reasoning of others) and MP6 (attend to precision): students distinguish a conditional from its converse, inverse, and contrapositive, and learn that only the contrapositive is logically equivalent. Proof by contradiction and the contrapositive method support the deductive writing of HSG-CO.C.9, while the closing sequence aₙ = 1 + 1/n foreshadows the ε-language of limits (Stage 36) and connects to growth and recursive reasoning in HSF-LE and mathematical induction (Stage 26). A good home prompt: ask your learner to find a single counterexample that overturns a confident "it always works."