What a Proof Is

Study Sheet

What a Proof Is

Statements, definitions, and the difference between checking and proving

Statements and Definitions

Concept
A proof is a chain of implications

A proof is a sequence of statements, each of which is an assumption, a definition, a previously proved fact, or a logical consequence of earlier lines, ending at the claim. Nothing else is allowed — not examples, not diagrams that "look right", not "it is obvious". The reader must be able to check every step without trusting you.

The strange thing about proofs, for someone meeting them for the first time, is how little they resemble the way we are convinced of things in daily life. Checking twenty cases convinces; it does not prove, because the twenty-first might fail. What a proof provides is a reason that covers every case at once. The reason usually comes from a definition: to prove something about even numbers, you will write n=2kn = 2k, because that is what "even" means, and everything follows from that sentence.

Tip
Know the definitions cold

Even: n=2kn = 2k. Odd: n=2k+1n = 2k + 1. a∣ba \mid b: b=akb = ak. Rational: p/qp/q with integers p,qp, q, q≠0q \ne 0. Prime: p>1p > 1 with no divisors other than 11 and pp. Every proof about these objects begins by unpacking the definition.

Example
Checking versus proving

Claim: the sum of two odd numbers is even. Checking: 3+5=83 + 5 = 8, 7+11=187 + 11 = 18 — convincing, but not a proof. Proof: let the numbers be 2a+12a + 1 and 2b+12b + 1, where aa and bb are integers (the definition of odd, with a different letter for each number). Their sum is 2a+2b+2=2(a+b+1)2a + 2b + 2 = 2(a + b + 1). Since a+b+1a + b + 1 is an integer, the sum is 22 times an integer, which is the definition of even. ■\blacksquare

Example
Odd squares and the number 8

Claim: the square of every odd integer leaves remainder 11 when divided by 88. Proof. Let n=2k+1n = 2k + 1 with kk an integer. Then n2=4k2+4k+1=4k(k+1)+1n^2 = 4k^2 + 4k + 1 = 4k(k + 1) + 1. Of the two consecutive integers kk and k+1k + 1 one is even, so k(k+1)=2mk(k + 1) = 2m for some integer mm. Therefore n2=8m+1n^2 = 8m + 1. ■\blacksquare

Each line has a reason. The first is the definition of odd. The second is algebra, and the factoring 4k(k+1)4k(k + 1) is chosen on purpose: it exposes a product of consecutive integers, which we know something about. The third line is itself a small proof by cases (if kk is even, kk is the even one; if kk is odd, k+1k + 1 is). Checking 32=9=8+13^2 = 9 = 8 + 1, 52=25=24+15^2 = 25 = 24 + 1 and 72=49=48+17^2 = 49 = 48 + 1 is how you would guess the claim; the factoring is what proves it.

Side note
Where it goes wrong

Using the same letter for two different unknowns. "Let the odd numbers be 2k+12k + 1 and 2k+12k + 1" proves only that an odd number added to ITSELF is even. Two numbers that are not given as equal need two letters.

Try it
Try it: write it out

Prove that the product of two odd numbers is odd.

Show the answer

Let the numbers be 2a+12a + 1 and 2b+12b + 1 with a,ba, b integers. Their product is 4ab+2a+2b+1=2(2ab+a+b)+14ab + 2a + 2b + 1 = 2(2ab + a + b) + 1. Since 2ab+a+b2ab + a + b is an integer, the product is 22 times an integer plus 11, which is odd. ■\blacksquare

Statements and Implications

Concept
What "if PP then QQ" promises
PTTFFQTFTFP ⇒ QTFTT
The only row where the promise is broken: PP true, QQ false.

A statement is a sentence that is either true or false. The implication "if PP then QQ" (written P⇒QP \Rightarrow Q) is false in exactly one situation: PP true and QQ false. When PP is false the implication makes no promise, and it counts as true.

Think of an implication as a promise: "if you finish the problem set, you may leave early". The promise is broken only if you finish and are kept in; if you never finish, nothing was promised. Mathematics needs this convention for a sharper reason. The theorem "for every integer nn, if 6∣n6 \mid n then nn is even" should be true — yet for n=9n = 9 the hypothesis is false. If a false hypothesis made the implication false, every theorem with a hypothesis would be "refuted" by the numbers it is not about. So a counterexample to P⇒QP \Rightarrow Q must make PP true and QQ false; nothing else counts.

Tip
Converse, inverse, contrapositive

From P⇒QP \Rightarrow Q come three relatives: the converse Q⇒PQ \Rightarrow P, the inverse not P⇒not Q\text{not } P \Rightarrow \text{not } Q, and the contrapositive not Q⇒not P\text{not } Q \Rightarrow \text{not } P. The contrapositive always has the same truth value as the original. The converse and inverse need not — they are each other's contrapositives, so they stand or fall together.

Example
A true statement with a false converse

"If nn is divisible by 66, then nn is divisible by 33." True: n=6k=3(2k)n = 6k = 3(2k). Converse: "if 3∣n3 \mid n then 6∣n6 \mid n" — false, because n=9n = 9 has 3∣93 \mid 9 and 6∤96 \nmid 9. Contrapositive: "if 3∤n3 \nmid n then 6∤n6 \nmid n" — true, as it must be, since it says the same thing as the original.

Example
Repairing a false statement

"If x2=9x^2 = 9 then x=3x = 3" is false: x=−3x = -3 makes the hypothesis true and the conclusion false, which is exactly the shape a counterexample must have. There are two honest repairs. Strengthen the hypothesis: "if x>0x > 0 and x2=9x^2 = 9 then x=3x = 3". Or weaken the conclusion: "if x2=9x^2 = 9 then x=3x = 3 or x=−3x = -3". Both are true, and the second is even an "if and only if".

Example
Hidden quantifiers

"If a number is prime, then it is odd" really means "for every integer nn, if nn is prime then nn is odd". A counterexample needs one nn that is prime and not odd: n=2n = 2. The numbers 99 (odd, not prime) and 44 (even, not prime) are not counterexamples — their hypothesis is false, so the implication makes no claim about them. Finding the right kind of counterexample is half the skill of reading a statement.

Side note
Where it goes wrong

Using a converse as if it had been proved. Inside a longer argument this is usually silent: "4∣n4 \mid n and 6∣n6 \mid n, so 24∣n24 \mid n" treats "if 24∣n24 \mid n then 4∣n4 \mid n and 6∣n6 \mid n" as reversible. It is not: n=12n = 12 is divisible by 44 and by 66 but not by 2424. The step is valid only when the two divisors share no common factor.

Try it
Try it: four versions

Statement: "If a quadrilateral is a square, then its diagonals are equal." Write the converse, the inverse and the contrapositive, and say which of the four are true.

Show the answer

Original: true. Converse: "if a quadrilateral's diagonals are equal, it is a square" — false; a 2×12 \times 1 rectangle has equal diagonals. Inverse: "if a quadrilateral is not a square, its diagonals are not equal" — false, by the same rectangle (the inverse is the converse's contrapositive, so they fail together). Contrapositive: "if the diagonals are not equal, it is not a square" — true, because it is equivalent to the original.

Reading a Proof

Concept
How to read

Read a proof three times. First, find the claim and the assumptions. Second, at each line, ask "why is this true?" and do not move on until you can say. Third, ask "where was each assumption used?" — if some assumption was never used, either the theorem is stronger than stated or the proof is wrong.

Mathematicians read slowly. A page of proof may take an hour, most of it spent with a pencil checking a line the author called "clearly". That checking is the reading; a proof skimmed is a proof not read. The third pass — where were the hypotheses used? — is the most instructive one, because it shows what the proof is really about and what would break without each assumption.

Example
Find where the hypothesis is used

Theorem: if n2n^2 is even then nn is even. Proof by contrapositive: suppose nn is odd, n=2k+1n = 2k + 1. Then n2=4k2+4k+1=2(2k2+2k)+1n^2 = 4k^2 + 4k + 1 = 2(2k^2 + 2k) + 1 is odd. So if n2n^2 is even, nn cannot be odd. Where is "n2n^2 even" used? Only at the end, as the thing being contradicted. The real work is the algebra showing odd squares are odd.

Example
Read with a pencil

Claim: the product of four consecutive integers, plus 11, is a perfect square. The printed proof says: "n(n+1)(n+2)(n+3)+1=(n2+3n+1)2n(n + 1)(n + 2)(n + 3) + 1 = (n^2 + 3n + 1)^2." A reader does not accept this; a reader checks it. Pair the outer and inner factors: n(n+3)=n2+3nn(n + 3) = n^2 + 3n and (n+1)(n+2)=n2+3n+2(n + 1)(n + 2) = n^2 + 3n + 2. With m=n2+3nm = n^2 + 3n the left side is m(m+2)+1=m2+2m+1=(m+1)2m(m + 2) + 1 = m^2 + 2m + 1 = (m + 1)^2, which is the claim. A number test catches a misprint before the algebra does: n=1n = 1 gives 1⋅2⋅3⋅4+1=251\cdot2\cdot3\cdot4 + 1 = 25 and n2+3n+1=5n^2 + 3n + 1 = 5.

Side note
Where it goes wrong

Reading for plausibility instead of for reasons. A line like "since a2=b2a^2 = b^2, a=ba = b" looks routine and is false in general (a=3a = 3, b=−3b = -3). It is correct only if the proof has already established a,b≥0a, b \ge 0 — so the reader's job is to find where that was established, or to flag the gap.

Try it
Try it: spot the gap

Find the error: "Claim: if a∣bca \mid bc, then a∣ba \mid b or a∣ca \mid c. Proof: since aa divides bcbc, aa shares a factor with bb or with cc, so it divides one of them."

Show the answer

The claim is false: 6∣4⋅3=126 \mid 4\cdot3 = 12, but 6∤46 \nmid 4 and 6∤36 \nmid 3. The broken step is "shares a factor, so it divides": 66 shares the factor 22 with 44, which does not make 66 a divisor of 44. The statement becomes true when aa is a prime pp (Euclid's lemma), and its proof needs more than this sentence.

Putting It Together

Claim: for every integer nn, n2−nn^2 - n is even.

Draft A. "1−1=01 - 1 = 0, 4−2=24 - 2 = 2, 9−3=69 - 3 = 6, 16−4=1216 - 4 = 12: all even, so n2−nn^2 - n is always even." Flaw: this checks four cases and says nothing about n=1001n = 1001. No reason is given that covers every nn.

Draft B. "Let nn be an integer. Then n2−n=n(n−1)n^2 - n = n(n - 1), a product of two consecutive integers. Of any two consecutive integers one is even: if nn is even it is nn, and if nn is odd then n−1n - 1 is even. A product with an even factor is even. ■\blacksquare"

Draft B is a proof because every sentence is a definition, a line of algebra, or a fact with its reason attached, and because nn is arbitrary — nothing about it was assumed except that it is an integer.

Tip
The checklist behind every proof

Before writing: state the claim with its quantifiers and unpack every definition. While writing: each line is an assumption, a definition, a proved fact, or follows from earlier lines — and says which. After writing: the last line is the claim, word for word.

Try it
Try it: from checking to proving

Prove that n2+3nn^2 + 3n is even for every integer nn.

Show the answer

n2+3n=n(n+3)n^2 + 3n = n(n + 3). If nn is even, the product has an even factor. If nn is odd, n=2k+1n = 2k + 1 and n+3=2k+4=2(k+2)n + 3 = 2k + 4 = 2(k + 2) is even. Either way the product is even. ■\blacksquare (Alternatively: n2+3n=n(n+1)+2nn^2 + 3n = n(n + 1) + 2n, a product of consecutive integers plus an even number.)