This deck shows how to prove a theorem three ways - directly, by contrapositive, and by contradiction - and then covers vacuous and trivial proofs, biconditionals, arguing without loss of generality, and disproof by counterexample. It targets the misconceptions that examples prove universal statements, that the converse is the same as the contrapositive, and that assuming the conclusion is a valid argument.
Subject: Foundations of Higher Mathematics · 109 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Read a theorem and name its hypothesis and conclusion, then unfold every definition.
2. Write a clean direct proof by assuming the hypothesis and chaining justified steps.
3. Recognize when a contrapositive proof is easier, and never confuse it with the converse.
4. Set up a proof by contradiction, including the classics: that the square root of two is irrational and that there are infinitely many primes.
5. Handle vacuous and trivial proofs, biconditionals, without loss of generality, and disproof by counterexample - and spot circular reasoning.
Warm-up
Discussion prompt
Before we open Proof Techniques I: Direct, Contrapositive, Contradiction: without looking back, what was the main idea of Predicate Logic & Quantifiers, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers predicates and open sentences, the universal and existential quantifiers over a stated domain, the order of nested quantifiers, and mechanical negation. It then adds the semantic layer: structures, satisfaction, validity as against satisfiability, Skolemization, and the decidability cliff. It targets the classic errors: swapping the order of quantifiers, negating "for all P" as "for all not P", forgetting the domain, and reading "there exists" as "exactly one".
Concept
Almost every theorem is a promise of the form: if the hypothesis holds, then the conclusion holds.
\[ P \Rightarrow Q \]
P is the hypothesis - what you are handed for free. Q is the conclusion - what you must earn.
hypothesis — The statement you are allowed to assume true. Every later step is permitted to lean on it.
conclusion — The statement you must reach through a chain of justified steps, using only the hypothesis and known facts.
Counterexample
Discussion prompt
Almost every theorem is a promise of the form: if the hypothesis holds, then the conclusion holds.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
P is the hypothesis - what you are handed for free. Q is the conclusion - what you must earn.
Picture it
Figure (svg): A box labeled P (given) with an arrow pointing to a box labeled Q (goal), showing an implication as a path from hypothesis to conclusion.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Picture a proof as a road. The hypothesis is where you already stand; the conclusion is the town you must reach.
Intuition
Picture a proof as a road. The hypothesis is where you already stand; the conclusion is the town you must reach.
Figure (svg): A box labeled P (given) with an arrow pointing to a box labeled Q (goal), showing an implication as a path from hypothesis to conclusion.
Every step lays one more paving stone, and each stone must be justified - by a definition, by the assumption, or by a result already proven.
Analogy
Discussion prompt
Explain A proof is a road you build by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Picture a proof as a road. The hypothesis is where you already stand; the conclusion is the town you must reach.
Concept
Before choosing any technique, unfold every definition in the statement. Replace each named word by what it literally means.
The word even is a closed door; open it. It means the number equals two times an integer.
\[ n \text{ is even} \iff n = 2k \text{ for some integer } k \]
Most stuck proofs are stuck for one reason only: a definition was left folded up.
Explain it
Discussion prompt
Explain Hypothesis mining: unfold the definitions to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Before choosing any technique, unfold every definition in the statement. Replace each named word by what it literally means.
Concept
Keep these four definitions in view for the whole lesson. Every proof here bottoms out in one of them.
even integer — An integer n with n = 2k for some integer k.
odd integer — An integer n with n = 2k + 1 for some integer k.
a divides b — There is an integer m with b = a*m. Written a | b.
rational number — A number that can be written as a ratio a/b of integers with b nonzero. A number that cannot is irrational.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of hypothesis, conclusion, even integer, odd integer, a divides b as Proof Techniques I: Direct, Contrapositive, Contradiction uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Concept
A direct proof is the straight road. You assume the hypothesis and march forward, one justified step at a time, until the conclusion appears on its own.
The skeleton never changes: assume, unfold, manipulate, arrive.
\[ \text{Assume } P \;\longrightarrow\; \text{deduce intermediate facts} \;\longrightarrow\; \text{reach } Q \]
Intuition
Forward chaining means you push from what you know toward what you want - never assume what you want and reason backward into what you know.
A good habit: write the assumption at the top of the page and the goal at the bottom, then close the gap strictly from the top down.
Ranking
Put in order
Put the moves of Worked example: the sum of two even integers is even into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. m and n are even, so m = 2a and n = 2b for some integers a and b.
Worked example
Claim: if m and n are even integers, then their sum is even.
Assume the hypothesis and unfold it
Why: m and n are even, so m = 2a and n = 2b for some integers a and b. Open the definition before doing anything else.
Add the two expressions
Why: Pure substitution, then factor out the common 2 to expose the target form.
\[ m + n = 2a + 2b = 2(a+b) \]
Match the definition of even
Why: a + b is an integer, so m + n equals two times an integer - which is exactly what even means.
Verify with a concrete case
Why: Take m = 4, n = 6: then m + n = 10 = 2*5, which is even. The general form 2(a+b) matches the example.
Picture it
Animation
Shows: Each line of the worked example "the sum of two even integers is even", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take m = 4, n = 6: then m + n = 10 = 2*5, which is even. The general form 2(a+b) matches the example.
Step zero
Discussion prompt
Worked example: the square of an odd integer is odd — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Assume the hypothesis and unfold it
Answer:
Worked example
Claim: if n is odd, then n squared is odd. We will reuse this fact later.
Assume the hypothesis and unfold it
Why: n is odd, so n = 2k + 1 for some integer k.
Square and expand
Why: Substitute the odd form and expand the binomial completely.
\[ n^2 = (2k+1)^2 = 4k^2 + 4k + 1 \]
Factor out the 2 from the even part
Why: Group everything divisible by 2 so the leftover +1 stands alone.
\[ n^2 = 2(2k^2 + 2k) + 1 \]
Match the definition of odd
Why: 2k^2 + 2k is an integer, so n squared is two times an integer plus one: odd by definition.
Verify with a concrete case
Why: Take n = 5: then n squared = 25 = 2*12 + 1, which is odd. Matches the general form.
Picture it
Animation
Shows: Each line of the worked example "the square of an odd integer is odd", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Take n = 5: then n squared = 25 = 2*12 + 1, which is odd. Matches the general form.
Trap
Claim: every odd square is odd. Bad proof: 1 gives 1, 3 gives 9, 5 gives 25 - all odd, so it always holds.
Three examples, however convincing, cover three of infinitely many cases.
A universal statement asserts something about every integer. A finite list of examples can never exhaust an infinite domain, so it cannot establish a for all claim.
Prove it for an arbitrary odd n written as 2k + 1, exactly as in the worked example. One symbolic argument covers all cases at once.
\[ n = 2k+1 \;\Rightarrow\; n^2 = 2(2k^2+2k)+1 \]
Examples still earn their keep - for building intuition and for disproving false claims - but a true universal needs one general argument.
Pattern
1. State what is assumed and what is to be shown
Why: Name the hypothesis P and the conclusion Q in plain words before writing anything else.
2. Unfold every definition (hypothesis mining)
Why: Replace each named property by its literal meaning, introducing fresh variables like a, b, k.
3. Chain justified steps toward the goal
Why: Each line must follow from the assumption, a definition, algebra, or a previously proven result.
4. Recognize the target form and stop
Why: Once your last line matches the definition of Q, the proof is complete. Do not keep writing.
Elimination
Eliminate the wrong options
Which is a valid direct proof that the sum of two odd integers is even?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Writing each odd number as 2a+1 and 2b+1 and adding gives 2a+2b+2 = 2(a+b+1), two times an integer, hence even. This single argument covers every pair of odd integers.
Check
We want to prove that the sum of two odd integers is even. Read each candidate carefully.
Check your understanding
Which is a valid direct proof that the sum of two odd integers is even?
Answer: A
Why: Writing each odd number as 2a+1 and 2b+1 and adding gives 2a+2b+2 = 2(a+b+1), two times an integer, hence even. This single argument covers every pair of odd integers.
Concept
Every implication has a mirror image called its contrapositive, and the two are logically equivalent - true in exactly the same situations.
\[ (P \Rightarrow Q) \;\equiv\; (\neg Q \Rightarrow \neg P) \]
So proving the mirror statement proves the original. You are free to attack whichever direction is easier.
Picture it
Figure (svg): Diagram showing the implication P implies Q on top, and its contrapositive not-Q implies not-P below, illustrating the flip and negate operation.
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Sometimes the conclusion is slippery but its negation is concrete. The phrase n is not even becomes n is odd, and odd numbers carry a clean formula you can compute with.
Intuition
Sometimes the conclusion is slippery but its negation is concrete. The phrase n is not even becomes n is odd, and odd numbers carry a clean formula you can compute with.
To build the contrapositive: negate the conclusion, negate the hypothesis, and swap their roles.
Figure (svg): Diagram showing the implication P implies Q on top, and its contrapositive not-Q implies not-P below, illustrating the flip and negate operation.
Concept
Do not confuse the contrapositive with the converse. The converse swaps hypothesis and conclusion without negating, and it is generally NOT equivalent to the original.
\[ \text{converse: } Q \Rightarrow P \qquad \text{contrapositive: } \neg Q \Rightarrow \neg P \]
The original and its contrapositive always share the same truth value. The original and its converse need not - that gap is where many wrong proofs live.
Estimation
Predict first
Claim: for every integer n, if n squared is even then n is even. A direct attack is awkward, so use the contrapositive.
Commit before you compute: what does Worked example: if n squared is even then n is even… come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with concrete cases
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. n squared = 36 is even and n = 6 is even; n squared = 9 is odd and n = 3 is odd.
Worked example
Claim: for every integer n, if n squared is even then n is even. A direct attack is awkward, so use the contrapositive.
Form the contrapositive
Why: Negate both parts and swap. In parity terms this reads: if n is odd, then n squared is odd.
\[ (n^2 \text{ even} \Rightarrow n \text{ even}) \;\equiv\; (n \text{ odd} \Rightarrow n^2 \text{ odd}) \]
Prove the contrapositive
Why: This is exactly the earlier result: an odd n has n = 2k+1 and n squared = 2(2k^2+2k)+1, which is odd.
Conclude the original
Why: An implication and its contrapositive are logically equivalent, so proving the mirror proves the original claim.
Verify with concrete cases
Why: n squared = 36 is even and n = 6 is even; n squared = 9 is odd and n = 3 is odd. The parity link holds both ways.
Picture it
Animation
Shows: Each line of the worked example "if n squared is even then n is even (contrapositive)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: n squared = 36 is even and n = 6 is even; n squared = 9 is odd and n = 3 is odd. The parity link holds both ways.
Missing information
Discussion prompt
Claim: if 3n + 2 is even, then n is even. Assuming the hypothesis directly is clumsy, so flip it.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Swap and negate: if n is odd, then 3n + 2 is odd.
Worked example
Claim: if 3n + 2 is even, then n is even. Assuming the hypothesis directly is clumsy, so flip it.
Form the contrapositive
Why: Swap and negate: if n is odd, then 3n + 2 is odd.
Assume n is odd and substitute
Why: Unfold: n = 2k + 1, then compute 3n + 2.
\[ 3n + 2 = 3(2k+1) + 2 = 6k + 5 \]
Expose the odd form
Why: Write 6k + 5 as two times an integer plus one.
\[ 6k + 5 = 2(3k + 2) + 1 \]
Conclude the original
Why: 3n + 2 is odd, proving the contrapositive, hence the original implication.
Verify with concrete cases
Why: n = 3 (odd) gives 3n + 2 = 11, odd. n = 4 (even) gives 3n + 2 = 14, even. Consistent with the claim.
Picture it
Animation
Shows: Each line of the worked example "if 3n+2 is even then n is even (contrapositive)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: n = 3 (odd) gives 3n + 2 = 11, odd. n = 4 (even) gives 3n + 2 = 14, even. Consistent with the claim.
Trap
To prove if n squared is even then n is even, a student proves if n is even then n squared is even and declares victory.
That second statement is the converse, not the contrapositive. It is true here, but it does not prove the original.
\[ \text{proved: } (n \text{ even} \Rightarrow n^2 \text{ even}) \qquad \text{needed: } (n^2 \text{ even} \Rightarrow n \text{ even}) \]
The contrapositive negates BOTH parts and swaps them: if n is odd then n squared is odd.
\[ (n^2 \text{ even} \Rightarrow n \text{ even}) \;\equiv\; (n \text{ odd} \Rightarrow n^2 \text{ odd}) \]
In other problems the converse can be false while the original is true, so proving it proves nothing. Always negate both parts; never merely swap them.
Prediction
Predict first
What is the contrapositive of if n squared is odd, then n is odd?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: If n is even, then n squared is even.
Why: The contrapositive negates both parts and swaps them: not n is odd becomes n is even, and not n squared is odd becomes n squared is even, giving if n is even then n squared is even.
Check
Consider the statement: if n squared is odd, then n is odd.
Check your understanding
What is the contrapositive of if n squared is odd, then n is odd?
Answer: A
Why: The contrapositive negates both parts and swaps them: not n is odd becomes n is even, and not n squared is odd becomes n squared is even, giving if n is even then n squared is even.
Concept
Proof by contradiction takes a bold stance: suppose the theorem is false, then show that this supposition forces an impossibility.
To prove that P implies Q, assume P together with the negation of Q, and derive something that cannot be true.
\[ \text{Assume } P \wedge \neg Q \;\longrightarrow\; \text{derive a contradiction} \]
Since the assumption led to nonsense, the assumption is wrong, so Q must hold after all.
Intuition
The mental image: you step through the door marked the claim is false, and every hallway inside collapses. If entering the room breaks mathematics, the room was never there.
Contradiction is powerful precisely because it hands you an extra assumption to exploit - the negated conclusion - which a direct proof does not give you.
Concept
Contradiction can prove that something holds without ever building the object it talks about. It rules out the alternative rather than exhibiting a witness.
non-constructive proof — A proof that establishes a statement without producing an explicit example, construction, or witness for it.
This is the deep difference from a contrapositive proof, which stays constructive. We sharpen that contrast in a moment.
Worked example
Claim: the square root of two cannot be written as a ratio of integers.
Assume the negation
Why: Suppose the square root of two IS rational. Then it equals a/b in lowest terms, with a and b integers sharing no common factor.
\[ \sqrt{2} = \frac{a}{b}, \qquad \gcd(a,b) = 1 \]
Square both sides and clear the fraction
Why: Squaring removes the root; multiplying through by b squared isolates a squared.
\[ 2 = \frac{a^2}{b^2} \;\Rightarrow\; a^2 = 2b^2 \]
Deduce that a is even
Why: a squared equals two times an integer, so a squared is even; and a square is even only when its root is even. So a = 2c for some integer c.
Substitute and deduce that b is even
Why: Replace a by 2c: then 4c squared = 2b squared, so b squared = 2c squared, making b squared even, hence b even.
\[ (2c)^2 = 2b^2 \;\Rightarrow\; b^2 = 2c^2 \]
Spring the contradiction
Why: Both a and b are even, so 2 divides both - but we assumed the fraction was in lowest terms with no common factor. Impossible.
Verify the key parity step
Why: Check the link used: if a squared is even then a is even. Its contrapositive - odd a gives odd a squared - was proven earlier, so the step is sound and the whole argument holds.
Picture it
Animation
Shows: Each line of the worked example "the square root of two is irrational (contradiction)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check the link used: if a squared is even then a is even. Its contrapositive - odd a gives odd a squared - was proven earlier, so the step is sound and the whole argument holds.
Step zero
Discussion prompt
Worked example: rational plus irrational is irrational (contradiction) — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Assume the negation
Answer:
Worked example
Claim: if r is rational and x is irrational, then their sum is irrational.
Assume the negation
Why: Suppose instead that r + x is rational. Call that sum s, a rational number.
\[ r + x = s, \qquad r, s \in \mathbb{Q} \]
Solve for x
Why: Subtract r from both sides to isolate the irrational quantity.
\[ x = s - r \]
Spring the contradiction
Why: s - r is a difference of two rationals, hence rational; so x would be rational, contradicting the assumption that x is irrational.
Verify the closure fact used
Why: Check that rationals are closed under subtraction: a/b - c/d = (ad - bc)/(bd) with bd nonzero, which is again a ratio of integers. So s - r is genuinely rational.
Picture it
Animation
Shows: Each line of the worked example "rational plus irrational is irrational (contradiction)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that rationals are closed under subtraction: a/b - c/d = (ad - bc)/(bd) with bd nonzero, which is again a ratio of integers. So s - r is genuinely rational.
Trap
Claim: if n squared is even then n is even. Bad proof: Since n is even, write n = 2k; then n squared = 4k squared is even. Done.
This begins by assuming n is even - which is the conclusion. It never uses the actual hypothesis that n squared is even.
The argument secretly proves the converse (even n gives even square) while pretending to prove the original. That circle is invalid.
Start from the hypothesis you were actually given, or use the contrapositive: assume n is odd and derive an odd square.
A quick test for circularity: cross out every line that assumes the conclusion. If nothing survives to reach the goal honestly, the proof is circular.
Check
Return to the proof that the square root of two is irrational, at the line a squared = 2 b squared.
Check your understanding
From a^2 = 2b^2 we conclude that a is even. Which fact justifies that conclusion?
Answer: A
Why: From a^2 = 2b^2, the value a^2 is two times an integer, hence even; and the contrapositive of odd gives odd square shows an even square forces an even root. So a is even.
Concept
The two techniques look similar - both bring in a negation - but they differ in what they deliver.
A contrapositive proof is constructive: it builds the chain directly from not-Q to not-P. A contradiction proof may be non-constructive: it only shows that assuming P together with not-Q breaks.
\[ \text{contrapositive: } \neg Q \Rightarrow \neg P \qquad \text{contradiction: } (P \wedge \neg Q) \Rightarrow \bot \]
If your contradiction only ever uses not-Q and never revisits P, you actually had a contrapositive proof in disguise - and the cleaner contrapositive is usually preferred.
Intuition
Think of the contrapositive as a special, disciplined case of contradiction: you assume the conclusion fails and head straight for the failure of the hypothesis, with no detours.
Contradiction gives you more room - you also get to assume the hypothesis - but that freedom is a temptation to wander. Reach for full contradiction only when the negated conclusion alone is not enough.
Estimation
Predict first
Claim: there is no largest prime; the primes go on forever.
Commit before you compute: what does Worked example: there are infinitely many primes (Euclid) come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the construction
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Check with the first primes 2, 3, 5: then N = 235 + 1 = 31, which is prime and absent from the list, exhibiting a genuinely new prime.
Worked example
Claim: there is no largest prime; the primes go on forever.
Assume the negation
Why: Suppose there are only finitely many primes and list all of them.
\[ p_1, p_2, \ldots, p_n \text{ is the complete list of all primes} \]
Construct a new number
Why: Multiply every listed prime together and add one. This N exceeds every prime on the list.
\[ N = p_1 p_2 \cdots p_n + 1 \]
N has a prime factor
Why: Every integer greater than one has at least one prime factor; call one of them q.
Show q is not on the list
Why: Dividing N by any listed prime leaves remainder one, so no p_i divides N. Hence q is a prime missing from the supposedly complete list.
\[ N \equiv 1 \pmod{p_i} \text{ for every } i \]
Spring the contradiction
Why: The list was assumed to contain every prime, yet q is a prime not on it. Impossible, so no finite list can hold them all.
Verify the construction
Why: Check with the first primes 2, 3, 5: then N = 235 + 1 = 31, which is prime and absent from the list, exhibiting a genuinely new prime.
Picture it
Animation
Shows: Each line of the worked example "there are infinitely many primes (Euclid)", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check with the first primes 2, 3, 5: then N = 235 + 1 = 31, which is prime and absent from the list, exhibiting a genuinely new prime.
Trap
Claim: if n is even then n + 1 is odd. Proof by contradiction: assume n is even and suppose n + 1 is NOT odd. Since n = 2k, we get n + 1 = 2k + 1, which is odd. Contradiction.
Look closely: the argument derived n + 1 is odd using only the hypothesis. The assumed negation was never actually used.
Wrapping a direct proof in assume the opposite adds noise, not rigor. The tell: the negated conclusion appears at the start and is never touched again.
Just write it directly: assume n is even, so n = 2k, hence n + 1 = 2k + 1 is odd. Finished.
\[ n = 2k \;\Rightarrow\; n + 1 = 2k + 1 \text{ (odd)} \]
Save contradiction for when you genuinely need the extra assumption - as in the root-two proof, where it is a fraction in lowest terms does real work.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Just write it directly: assume n is even, so n = 2k, hence n + 1 = 2k + 1 is odd. Finished.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Check
A student wants to prove: if n squared is even then n is even.
Check your understanding
Instead the student proves if n is odd then n squared is odd. Which technique is this?
Answer: A
Why: Replacing the statement by not-conclusion implies not-hypothesis and proving that is exactly proof by contrapositive; here not-even is odd and a not-even square is an odd square.
assume the whole claim is false step appears, so it is not a contradiction proof.Concept
An implication is automatically true whenever its hypothesis can never hold. There is nothing to check, because the if part never fires.
This is a vacuous proof: the statement is true not because of the conclusion but because the hypothesis is empty.
\[ \text{if } P \text{ is always false, then } P \Rightarrow Q \text{ is true for every } Q \]
Vacuous truth feels strange the first time, but it keeps logic consistent: a promise you never have to keep is never broken.
Concept
The mirror situation: if the conclusion holds no matter what, the implication is true regardless of the hypothesis.
This is a trivial proof - trivial not because it is easy, but because the conclusion needs no help from the hypothesis.
\[ \text{if } Q \text{ is always true, then } P \Rightarrow Q \text{ is true for every } P \]
Naming these two cases stops you over-working a proof that a single observation already settles.
Ranking
Put in order
Put the moves of Worked example: a vacuously true statement into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Before touching the conclusion, ask whether any integer can satisfy the if part: it says n squared is negative.
Worked example
Claim: for every integer n, if n squared is negative, then n equals one hundred.
Inspect the hypothesis
Why: Before touching the conclusion, ask whether any integer can satisfy the if part: it says n squared is negative.
Show the hypothesis is impossible
Why: For every integer n, its square is at least zero, so it is never negative.
\[ n^2 \geq 0 \text{ for every integer } n \]
Conclude vacuously
Why: Because the hypothesis never holds, the implication is true for every n, no matter how strange the conclusion looks.
Verify the reasoning
Why: Check that the claim does not assert n = 100 outright; it asserts an implication whose if part is empty, so no counterexample can exist. The statement is vacuously true.
Picture it
Animation
Shows: Each line of the worked example "a vacuously true statement", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that the claim does not assert n = 100 outright; it asserts an implication whose if part is empty, so no counterexample can exist. The statement is vacuously true.
Intuition
Read it as if it rains inside a sealed vault, I will pay you. Since it never rains inside the vault, the promise is never violated, so it is trivially kept.
Spotting vacuous and trivial cases early can end a proof in one line, and it explains why some for all statements over an empty range are automatically true.
Concept
A statement of the form P if and only if Q packs two claims together, and BOTH must be proved.
\[ (P \iff Q) \;\equiv\; (P \Rightarrow Q) \wedge (Q \Rightarrow P) \]
Prove the forward direction and the backward direction separately. Each half may call for a different technique.
Intuition
An if and only if says the two statements travel together everywhere: whenever one is true the other is, and whenever one is false the other is. It is a round trip, not a one-way street.
Skipping a direction is the most common biconditional error, because the forward direction often feels like the whole story when it is only half.
Worked example
Claim: an integer n is even exactly when its square is even. Prove both directions.
Forward direction: even n gives even square
Why: Assume n even, so n = 2k; then n squared = 4k squared = 2(2k squared), which is even.
\[ n = 2k \;\Rightarrow\; n^2 = 2(2k^2) \]
Backward direction: even square gives even n
Why: Assume n squared even. By the contrapositive proven earlier, an odd n has an odd square, so an even square forces n to be even.
Combine the directions
Why: Both implications hold, so the biconditional holds: n is even if and only if n squared is even.
Verify with concrete cases
Why: n = 6 has square 36, both even; n = 7 has square 49, both odd. The two properties switch on and off together, as an equivalence demands.
Notation
Annotate
From Worked example: n is even if and only if n squared is even — read this one piece at a time. What is each part doing?
On: \( n = 2k \;\Rightarrow\; n^2 = 2(2k^2) \)
Trap
Claim: n is even if and only if n squared is even. Bad proof: If n is even then n = 2k and n squared = 4k squared is even. Therefore the biconditional holds.
Only the forward direction was shown. The backward direction - even square forces even n - was never addressed.
A one-directional argument proves an ordinary implication, not an equivalence. Half a round trip is not a round trip.
Prove BOTH directions and label them. Forward: even n gives even square. Backward: even square gives even n, by the contrapositive.
\[ (n \text{ even} \Rightarrow n^2 \text{ even}) \;\wedge\; (n^2 \text{ even} \Rightarrow n \text{ even}) \]
A tidy habit: write the two arrows as separate labeled paragraphs so a missing direction is impossible to overlook.
Commit first
Predict first
You proved if n is even then n squared is even. To finish the biconditional, what remains?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: Show that if n squared is even then n is even.
Why: A biconditional needs both directions. You have the forward one; the missing piece is the backward implication that an even square forces n itself to be even.
The rating matters as much as the answer: confident-and-wrong is the combination that survives revision, because nothing about it feels like it needs revisiting.
Check
You want to prove: n is even if and only if n squared is even, and you have already shown the forward direction.
Check your understanding
You proved if n is even then n squared is even. To finish the biconditional, what remains?
Answer: A
Why: A biconditional needs both directions. You have the forward one; the missing piece is the backward implication that an even square forces n itself to be even.
Concept
When a problem is symmetric in two objects, you may fix a convenient ordering and treat the other arrangement as identical. This move is called without loss of generality.
It is a labor-saving device, not a way past rigor: you must be able to justify that swapping the objects genuinely reproduces the omitted case.
Misusing it - claiming a symmetry that is not there - quietly skips real cases. Use it only when the roles are truly interchangeable.
Intuition
If a statement reads the same after you swap the names a and b, then whatever you prove for one ordering automatically holds for the other. You have not lost a case; you have only declined to write it twice.
The test: after your without loss of generality sentence, could a reader recover the skipped case just by relabeling? If yes, you are safe.
Step zero
Discussion prompt
Worked example: if a+b is odd then exactly one of them is odd — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Note the symmetry
Answer:
Worked example
Claim: for integers a and b, if a + b is odd then exactly one of a, b is odd.
Note the symmetry
Why: The statement is unchanged when we swap a and b, so we may fix an ordering of their parities without loss of generality.
Rule out both-even and both-odd
Why: If both are even, a + b = 2j + 2k is even; if both are odd, a + b = 2(j + k + 1) is even. Neither yields an odd sum.
\[ \text{both even or both odd} \;\Rightarrow\; a + b \text{ even} \]
Conclude the parities differ
Why: Since a + b is odd, a and b cannot share a parity, so one is even and one is odd - that is, exactly one is odd.
See where WLOG was used
Why: Naming which one is odd (say b) covers the mirror case (a odd) by relabeling, because the roles of a and b are interchangeable.
Verify with a concrete case
Why: a = 2, b = 3: the sum 5 is odd and exactly one of them (b) is odd. Swapping to a = 3, b = 2 gives the same conclusion.
Picture it
Animation
Shows: Each line of the worked example "if a+b is odd then exactly one of them is odd", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: a = 2, b = 3: the sum 5 is odd and exactly one of them (b) is odd. Swapping to a = 3, b = 2 gives the same conclusion.
Concept
To disprove a universal statement - a claim about every object - you need just one object where it fails. That single failing object is a counterexample.
This is the great asymmetry of proof: establishing a for all needs a general argument, but refuting it needs only one well-chosen instance.
\[ \text{to refute } \forall x\, P(x), \text{ exhibit one } x \text{ with } \neg P(x) \]
A counterexample must genuinely satisfy the hypothesis and violate the conclusion; a case that never meets the hypothesis proves nothing.
Intuition
A universal claim is a wall; a counterexample is the single loose brick that brings it down. You need not inspect every brick - one bad brick is fatal.
So when a for all statement smells wrong, do not grind on a general proof first. Hunt for a counterexample; finding one settles the matter instantly.
Missing information
Discussion prompt
Claim to test: for every whole number n starting at zero, the value below is prime.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
For n = 0, 1, 2 the values are 41, 43, 47 - all prime. The pattern is seductive, but small cases are not a proof.
Worked example
Claim to test: for every whole number n starting at zero, the value below is prime.
\[ f(n) = n^2 + n + 41 \]
Try small inputs first
Why: For n = 0, 1, 2 the values are 41, 43, 47 - all prime. The pattern is seductive, but small cases are not a proof.
Look for structure that could fail
Why: The constant term is 41. Choosing n = 40 threads a factor of 41 through every term.
Evaluate at n = 40
Why: Compute the value directly.
\[ f(40) = 40^2 + 40 + 41 = 1600 + 81 = 1681 \]
Exhibit the factorization
Why: 1681 is not prime; it is a perfect square.
\[ 1681 = 41 \times 41 = 41^2 \]
Verify the counterexample
Why: Check that n = 40 meets the hypothesis (a whole number) and breaks the conclusion (1681 is composite). One counterexample disproves the universal claim.
Picture it
Animation
Shows: Each line of the worked example "disprove a tempting always prime claim", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that n = 40 meets the hypothesis (a whole number) and breaks the conclusion (1681 is composite). One counterexample disproves the universal claim.
Trap
Claim to disprove: every prime is odd. Attempted counterexample: 9 is not prime and not odd-looking, so the claim stands.
That misses the point. To disprove every prime is odd you need a prime that is even - not a non-prime, and not some exotic giant.
Two real errors hide here: hunting for an exotic case, and offering an instance that fails the hypothesis (9 is not prime), so it tests nothing at all.
The number 2 is prime and even. It satisfies the hypothesis (prime) and violates the conclusion (odd), so it is a valid counterexample - and the smallest one, not an exotic case.
A counterexample must meet the if part and break the then part. Small, ordinary numbers are often the best place to look first.
Hypothesis
Predict first
Worked example: the product of two consecutive integers is even is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Split into cases on the parity of n
Why: Every integer is even or odd, and these two cases are exhaustive, so handling each settles the claim.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Claim: for every integer n, the product of n and n plus one is even.
\[ n(n+1) \text{ is even for every integer } n \]
Split into cases on the parity of n
Why: Every integer is even or odd, and these two cases are exhaustive, so handling each settles the claim.
Case one: n is even
Why: Then n = 2k, so n(n+1) = 2k(n+1), which is two times an integer, hence even.
Case two: n is odd
Why: Then n + 1 is even, say n + 1 = 2m, so n(n+1) = 2mn, which is even.
Combine the cases
Why: In both exhaustive cases the product is even, so the claim holds for every integer n.
Verify with concrete cases
Why: n = 3: 3 times 4 = 12, even. n = 4: 4 times 5 = 20, even. Both match the claim.
Notation
Annotate
From Worked example: the product of two consecutive integers is… — read this one piece at a time. What is each part doing?
On: \( n(n+1) \text{ is even for every integer } n \)
Pattern
Start direct
Why: Assume the hypothesis and push forward. If the conclusion falls out cleanly, you are done - this is the default first attempt.
Switch to contrapositive when the negated conclusion is concrete
Why: If not-Q gives a usable formula (like turning not even into odd), prove not-Q implies not-P instead.
Use contradiction when you need an extra grip
Why: If neither direction moves, assume the hypothesis and the negated conclusion together and hunt for an impossibility - the tool for irrationality and non-existence.
To disprove a universal, find one counterexample
Why: A single instance that satisfies the hypothesis but breaks the conclusion refutes the whole statement.
Real world
Discussion prompt
Outside this lesson: where does Proof Techniques I: Direct, Contrapositive, Contradiction actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Pattern: which technique should you reach for? is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck shows how to prove a theorem three ways - directly, by contrapositive, and by contradiction - and then covers vacuous and trivial proofs, biconditionals, arguing without loss of generality, and disproof by counterexample. It targets the misconceptions that examples prove universal statements, that the converse is the same as the contrapositive, and that assuming the conclusion is a valid argument.
Check
Think about the shape of the claim before picking a method.
Check your understanding
Which technique is the natural first choice to prove there is no largest prime?
Answer: A
Why: The claim asserts non-existence of a largest prime, and the clean route is Euclid's contradiction: assume a complete finite list, build a number forcing a new prime, and contradict completeness.
largest prime instance to exhibit.Concept
In a contradiction proof, everything hinges on negating the conclusion correctly. A wrong negation aims your whole argument at the wrong target.
For an existence claim the negation is a universal (there is none); for a universal claim the negation is an existence (some case fails). Getting this switch right is half the battle.
\[ \neg(\exists x\, P(x)) \equiv \forall x\, \neg P(x) \qquad \neg(\forall x\, P(x)) \equiv \exists x\, \neg P(x) \]
Write the negated conclusion explicitly before deriving anything. Guessing it wrong costs you the entire proof.
Estimation
Predict first
Claim: the base-two logarithm of three cannot be a ratio of integers. This one is a favorite in computer science.
Commit before you compute: what does Worked example: the base-two logarithm of three is… come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the parity claim
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Check both sides: two to any positive power is even, and three to any power is a product of odd numbers, hence odd.
Worked example
Claim: the base-two logarithm of three cannot be a ratio of integers. This one is a favorite in computer science.
Assume the negation
Why: Suppose it is rational, equal to a/b with a and b positive integers.
\[ \log_2 3 = \frac{a}{b}, \qquad a, b > 0 \]
Rewrite in exponential form
Why: By the definition of logarithm, two raised to a/b equals three; raise both sides to the power b to clear the fraction.
\[ 2^{a/b} = 3 \;\Rightarrow\; 2^a = 3^b \]
Compare parities
Why: The left side is even, since a is at least one so two divides it; the right side is odd, being a power of three.
\[ 2^a \text{ is even}, \qquad 3^b \text{ is odd} \]
Spring the contradiction
Why: An even number cannot equal an odd number, so the equation is impossible. Hence no such a and b exist.
Verify the parity claim
Why: Check both sides: two to any positive power is even, and three to any power is a product of odd numbers, hence odd. They can never coincide, so the argument is airtight.
Picture it
Animation
Shows: Each line of the worked example "the base-two logarithm of three is irrational", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check both sides: two to any positive power is even, and three to any power is a product of odd numbers, hence odd. They can never coincide, so the argument is airtight.
Concept
Most broken proofs fail in one of three named ways. Learn to smell each of them.
circular reasoning — Assuming the conclusion, or something equivalent to it, somewhere in the argument - so the proof leans on the very thing it claims to establish.
affirming the converse — Proving Q implies P and treating it as a proof of P implies Q. The converse is a different, generally inequivalent statement.
The third is proof by example: offering instances where a general argument is required. All three feel persuasive, which is exactly why they are dangerous.
Trap
Known fact: if a number is divisible by four then it is even. A student argues: 84 is even, therefore 84 is divisible by four.
This runs the implication backward - even implies divisible by four - which was never established and is in fact false.
\[ \text{known: } 4 \mid n \Rightarrow n \text{ even} \qquad \text{used: } n \text{ even} \Rightarrow 4 \mid n \]
Here 84 does happen to be divisible by four, but 6 is even and not divisible by four - so the reasoning itself is invalid even when the answer looks right.
Apply an implication only in its stated direction. From even you may NOT conclude divisible by four without separately proving that converse.
When you catch yourself running an if-then backward, stop: you are affirming the converse, one of the classic invalid moves.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
if n squared is even then n is even, a student proves if n is even then n squared is even and declares victory.Step zero
Discussion prompt
Worked example: a nonnegative number below every positive bound is zero — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Assume the negation
Answer:
Worked example
Claim: if x is at least zero and x is less than every positive number, then x equals zero. This underlies limit arguments in analysis.
Assume the negation
Why: Suppose x is at least zero, below every positive bound, yet x is not zero. Then x is strictly positive.
\[ x \geq 0, \quad x \neq 0 \;\Rightarrow\; x > 0 \]
Produce a specific positive bound
Why: Since x is positive, half of x is also positive, so it is one of the bounds that x must beat.
\[ \varepsilon = \tfrac{x}{2} > 0 \]
Apply the hypothesis to this bound
Why: By assumption x is less than every positive number, in particular less than x over two.
\[ x < \tfrac{x}{2} \]
Spring the contradiction
Why: Rearranging gives x over two below zero, so x is negative - contradicting that x is positive. Hence x must be zero.
\[ x < \tfrac{x}{2} \;\Rightarrow\; \tfrac{x}{2} < 0 \;\Rightarrow\; x < 0 \]
Verify the witness
Why: Check that x over two is a legitimate positive bound whenever x is positive, so the contradiction genuinely fires. Only x = 0 escapes it.
Picture it
Animation
Shows: Each line of the worked example "a nonnegative number below every positive bound is zero", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Check that x over two is a legitimate positive bound whenever x is positive, so the contradiction genuinely fires. Only x = 0 escapes it.
Prediction
Predict first
A student proves the sum of two odd numbers is even by writing Assume the sum is even. Then it equals 2m, which is even. Done. What is the flaw?
Answer it in your own words, now, with nothing to choose from. The options are on the next slide — and picking the right one off a list is an easier skill than producing it.
Correct: It assumes the conclusion (that the sum is even) instead of deriving it.
Why: The argument opens by assuming the sum is even, which is exactly what must be proved, so it leans on its own conclusion. That circularity means it establishes nothing.
Check
Read the following proof carefully before answering.
Check your understanding
A student proves the sum of two odd numbers is even by writing Assume the sum is even. Then it equals 2m, which is even. Done. What is the flaw?
Answer: A
Why: The argument opens by assuming the sum is even, which is exactly what must be proved, so it leans on its own conclusion. That circularity means it establishes nothing.
Concept
A proof is a piece of communication. Announce your technique up front so the reader knows which road you are on.
Standard openers: Assume or Suppose for the hypothesis; Suppose, for contradiction, before a contradiction; By contrapositive, it suffices to show before a flip.
Close with a clear end marker - the letters Q E D or a small filled square - so the reader knows the argument is complete.
Explain it
Discussion prompt
Explain Signposting: writing a proof the reader can follow to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
A proof is a piece of communication. Announce your technique up front so the reader knows which road you are on.
Intuition
An unlabeled proof forces the reader to reverse-engineer your intent. One opening sentence naming the technique removes all that friction.
This is more than etiquette: writing suppose for contradiction commits you to using the negated conclusion, and readers can hold you to it - which catches the secretly-direct trap automatically.
Analogy
Discussion prompt
Explain Tell the reader which technique you chose by analogy to something with no Foundations of Higher Mathematics in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
An unlabeled proof forces the reader to reverse-engineer your intent. One opening sentence naming the technique removes all that friction.
Ranking
Put in order
Put the moves of Worked example: divisibility is transitive into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. a divides b means b = a times m; b divides c means c = b times n, for some integers m and n.
Worked example
Claim: if a divides b and b divides c, then a divides c.
Assume the hypothesis and unfold both parts
Why: a divides b means b = a times m; b divides c means c = b times n, for some integers m and n.
\[ b = a m, \qquad c = b n \]
Substitute one into the other
Why: Replace b in the second equation using the first, then regroup the factors.
\[ c = b n = (a m) n = a(mn) \]
Match the definition of divides
Why: mn is an integer, so c equals a times an integer, which means a divides c.
Verify with a concrete case
Why: a = 2, b = 6, c = 18: two divides six and six divides eighteen, and indeed two divides eighteen since 18 = 2 times 9. Matches the claim.
Picture it
Animation
Shows: Each line of the worked example "divisibility is transitive", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: a = 2, b = 6, c = 18: two divides six and six divides eighteen, and indeed two divides eighteen since 18 = 2 times 9. Matches the claim.
Elimination
Eliminate the wrong options
Which statement is best refuted by a single counterexample rather than proved in general?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Only a false universal yields to a counterexample. Every odd number is prime fails at 9, which is odd but composite; the other three statements are true and require general proofs.
Check
Only a false universal can be refuted by a single instance. Pick the one that fits.
Check your understanding
Which statement is best refuted by a single counterexample rather than proved in general?
Answer: A
Why: Only a false universal yields to a counterexample. Every odd number is prime fails at 9, which is odd but composite; the other three statements are true and require general proofs.
Concept
Direct, contrapositive, and contradiction are not rival schools; they are three roads across the same terrain, from hypothesis to conclusion.
Direct walks straight. Contrapositive takes the mirror road when the negated conclusion is firmer ground. Contradiction enters the forbidden room and watches it collapse.
Around them sit the supporting moves: vacuous and trivial shortcuts, both directions for a biconditional, symmetry via without loss of generality, and a single counterexample to fell a false universal.
Counterexample
Discussion prompt
Direct, contrapositive, and contradiction are not rival schools; they are three roads across the same terrain, from hypothesis to conclusion.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
Direct walks straight. Contrapositive takes the mirror road when the negated conclusion is firmer ground. Contradiction enters the forbidden room and watches it collapse.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — Pattern: the anatomy of any proof · Pattern: which technique should you reach for? · A theorem is an implication · A proof is a road you build · Hypothesis mining: unfold the definitions. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can read a theorem as an implication, name its hypothesis and conclusion, and unfold every definition before choosing a technique.
You can prove directly, flip to the contrapositive when it is cleaner, and set up a contradiction when you need the extra assumption - and you know which classics use which.
\[ P \Rightarrow Q, \qquad \neg Q \Rightarrow \neg P, \qquad (P \wedge \neg Q) \Rightarrow \bot \]
And you can catch the three classic failures - circular reasoning, affirming the converse, and proof by example - and disprove a false universal with one counterexample.
| Situation | Reach for |
|---|---|
| Conclusion falls out from the hypothesis | Direct proof |
| Negation of the conclusion is concrete | Contrapositive |
| You need an extra assumption, or prove non-existence | Contradiction |
The claim is a false for all | One counterexample |
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