Proofs

Lecture 10, DSCI 220, 2026W1

Announcements

• EX2 is next week, Thu 10/8 through Sun 10/11. It covers through today.
• Practice problems are on PrairieLearn.
• HW2 is due Sun Oct 4.

What is a proof?

You Have Been Writing Them

A proof is a finite chain of statements that

• starts from assumptions and definitions
• moves by valid rules
• ends with the claim

You’ve already done each of those pieces. The rules are the ten on your Formula Sheet, together with notions of equivalence. What is new today is the story that you tell with the arguments.

Say plainly that nothing new arrives today except vocabulary and habit. They have had the rules since Friday of week 2, the quantifiers since week 3, and they spent Monday deciding whether nested statements were true.

The one genuinely new idea is on the next two slides: what “for all” does to the START of a proof.

First Proof

Theorem: For any integers \(a\) and \(b\), if \(ab\) is even, then \(a\) is even or \(b\) is even.

Important definitions:

   An integer \(a\) is even iff ______________________

   For integers \(d \neq 0\) and \(n\), \(d\) divides \(n\) (written \(d \mid n\)) iff __________________

An integer a is even iff there is an integer k with a = 2k.

Write it as an EXISTENTIAL. They will need it as one on the next slide, and it is the first time a definition has been a quantified statement rather than a word.

Then d | n iff there is an integer k with n = dk – the same existential shape, and worth saying out loud that EVEN IS THE d = 2 CASE. “a is even” and “2 | a” are the same claim. Two definitions, one shape.

Divisibility is needed twice more today, on the Proof by Cases slides, where the claim is that n^3 - n is divisible by 3 and by 2. The bar notation is also all over the practice set, so introduce it here rather than leaving them to guess.

Technical Rewrite

\(\forall a,b\in \mathbb{Z}\)
    \(( (\exists k\in\mathbb{Z}, ab=2k) \to ((\exists j\in\mathbb{Z}, a = 2j) \lor (\exists j\in\mathbb{Z}, b=2j)))\)

You cannot check every pair. So how does a proof of a \(\forall\) statement start?

____________________________________________

“Proof: Consider arbitrary integers a and b …”

THIS IS THE SLIDE. Nothing about the domain is exhaustible, so you cannot verify a for-all by checking. Instead you take ONE, chosen arbitrarily, and argue about it in a way that uses nothing particular about it – and then the argument applies to every one.

That move has a name: universal instantiation going in, universal generalization coming out. Between them sits ordinary propositional inference, which they have had for three weeks.

So the whole template is:

take arbitrary x  ->  assume P(x)  ->  derive Q(x)  ->  conclude forall x (P(x) -> Q(x))

Say that out loud, then say that this is what a proof IS. They have had every piece since week 2 and have never seen them assembled.

Which are valid arguments?

Four attempts at the theorem above.

• Pf1: Suppose \(a\) and \(b\) are odd. Then there are integers \(k\) and \(j\) so \(a=2k+1\) and \(b=2j+1\). Then \(ab=(2k+1)(2j+1) = 2(2jk + j + k) +1\). Since \(2jk+j+k\) is an integer, \(ab\) is odd.

• Pf2: Assume \(a\) or \(b\) is even. Suppose \(a\) is even – the argument is the same if it is \(b\). Then \(ab=2kb\) for some \(k\), and thus \(ab\) is even.

• Pf3: Suppose that \(ab\) is even but \(a\) and \(b\) are both odd. Namely, \(ab=2n\), \(a=2k+1\), and \(b=2j+1\), for some integers \(n\), \(j\), and \(k\). Then \(2n=(2k+1)(2j+1)\) and \(n = 2jk + k + j + 1/2\) which is a contradiction.

• Pf4: Let \(ab\) be an even number, say \(ab=2n\), and \(a\) be an odd number, say \(a=2k+1\). Then \(2n = (2k+1)b\) and thus \(b=2(n-kb)\) and \(b\) is even.

Write E(x) for “x is even”. The theorem is E(ab) -> E(a) v E(b). Valid: 1, 3, 4. Not a proof of this theorem: 2.

Pf1 – CONTRAPOSITIVE. Valid. It proves ~E(a) ^ ~E(b) -> ~E(ab). By DM, ~E(a) ^ ~E(b) is ~(E(a) v E(b)), so this is ~(E(a) v E(b)) -> ~E(ab), the contrapositive of the theorem. It never says so: it stops at “ab is odd” and leaves the reader to see that this settles the theorem. Worth saying out loud.

Pf2 – CONVERSE. Not a proof of this theorem. It proves E(a) v E(b) -> E(ab). That is true, and the proof of it is fine (“the argument is the same if it is b” is a fair shortcut), but it is the converse. Proving the converse is not proving the theorem – lecture 4.

Pf3 – CONTRADICTION. Valid. It assumes E(ab) ^ ~E(a) ^ ~E(b) and gets n = 2jk + k + j + 1/2. That says n is not an integer, but n was introduced as an integer (ab = 2n). That is the contradiction. The proof just says “which is a contradiction” without naming what contradicts what – ask them to name it.

Pf4 – DIRECT, of an equivalent form. Valid. It assumes E(ab) and ~E(a), and derives E(b). So it proves E(ab) -> (~E(a) -> E(b)), and ~E(a) -> E(b) is E(a) v E(b) by IMP. This is the usual way to prove an OR: assume one side is false and show the other must be true. (It is also cases on a in disguise: if a is even, done; if a is odd, this shows b is even.) b appears on both sides of b = 2(n - kb). That is fine: it says b is twice the integer n - kb. Expect someone to call it circular.

Common to 1, 3 and 4: “a is odd, so a = 2k+1 for some integer k” – the definition is an existential, and we name the witness. That is existential instantiation. Each also turns “not even” into “odd”, which uses the fact that every integer is exactly one of even and odd.

Algebra checked: (2k+1)(2j+1) = 2(2jk + j + k) + 1, and b = 2(n - kb).

Proof Skeleton Cards

\(\to\) Direct Proof of \(P\to Q\)

1. Assume \(P\).
2. Reason – Unpack definitions / apply given rules to derive needed facts from \(P\).
3. … Therefore \(Q\).

\(\leftrightsquigarrow\) Contrapositive Proof of \(P\to Q\)

1. Assume \(\neg Q\)
2. Use rules/definitions that match \(\neg Q\) to drive toward \(\neg P\).
3. … Therefore \(\neg Q\to\neg P\), and thus \(P\to Q\).

↯ Contradiction Proof of \(P\to Q\)

1. Assume \(P\) and \(\neg Q\).
2. Reason – Unpack definitions / apply given rules to derive needed facts from \(P\land \neg Q\), watch for contradiction (\(a\land\neg a\)).
3. … Therefore False, and our assumption of \(\neg Q\) must be false, so \(P\to Q\).

⊞ Cases Proof of \(P\to Q\)

1. Partition \(P\) into cases \(P_1\), \(P_2\), etc. covering all possibilities.
2. For each \(P_k\), show that \(P_k\to Q\).
3. … Therefore, in each case \(Q\), and \(P\to Q\).

\(\to\) Pause and Try: Direct

Let \(T_n = \frac{n(n+1)}{2}\) be the \(n\)th triangular number: \(T_0 = 0\), \(T_1 = 1\), \(T_2 = 3\), etc.

A positive integer \(g\) is a glow number if there exists a positive integer \(k\) so that \(g = T_{k-1} + T_k\).

A positive integer \(n\) is a perfect square if there is a positive integer \(m\) with \(n = m^2\).

Theorem: every glow number is a perfect square.

Proof: Consider an arbitrary glow number \(g\)…

 

 

 

 

… therefore \(g\) is a perfect square.

There exists a positive integer k with g = T_{k-1} + T_k (name the k – existential instantiation). T_{k-1} + T_k = (k-1)k/2 + k(k+1)/2 = k(2k)/2 = k^2. So g = m^2 with m = k, which is a positive integer: the witness.

Aside: Is the converse true?

Converse: every perfect square is a glow number.

Come to class Monday with an answer!

Yes. A perfect square is m^2 for a positive integer m, and m^2 = T_{m-1} + T_m: the witness is k = m.

So the theorem and its converse together say the glow numbers are exactly the perfect squares, and the k-th glow number is k^2.

Nothing sits at the boundary: T_0 = 0 is defined, so 1 = T_0 + T_1 is a glow number, and both definitions are for positive integers, so 0 is not in question.

\(\leftrightsquigarrow\) Pause and Try: Contrapositive

Definitions:

• A number \(n\) is even if there is an integer \(k\) such that \(n=2k\).
• A number \(n\) is odd if there is an integer \(k\) such that \(n=2k+1\).

Theorem: If \(3n+2\) is odd, then \(n\) is odd.

Direct fails: from 3n+2 = 2k+1 you get n = (2k-1)/3, which says nothing useful. The contrapositive is the move – assume n is even, n = 2j, then 3n+2 = 6j+2 = 2(3j+1), which is even. Done.

Worth saying WHY direct stalls before showing the contrapositive. The point of having four techniques is that the first one you reach for does not always go through.

↯ Pause and Try: Contradiction

Claim: \(\sqrt 2\) is irrational.

Proof: Suppose, for contradiction, that \(\sqrt 2\) is rational. Then there are integers \(p\) and \(q\), sharing no common factor, with \(\sqrt 2 = p/q\).

Square both sides: 2 = p^2/q2, so p^2 = 2q^2, so p^2 is even, so p is even – that last step needs the theorem from the start of this lecture, ab even implies a or b even, with a = b = p. Say so; it is the callback.

Write p = 2r. Then 4r^2 = 2q^2, so q^2 = 2r^2, so q is even too. But p and q were chosen with no common factor, and both are even. Contradiction.

The shape to point at: we never showed sqrt 2 is irrational directly. We showed that supposing otherwise breaks. Same move as adding the negated conclusion and hunting for the empty clause – which they did all last week.

⊞ Proof by Cases

Claim: \(n^3 - n\) is divisible by ___ , for every integer \(n\).

First, factor it: \(n^3 - n =\) ____________________

Overview: ____________________________________

Divisible by 6. Factors as n(n-1)(n+1) – three consecutive integers.

Overview: show 3 divides it, show 2 divides it, and since 3 and 2 share no factor, 6 divides it. Two separate case analyses, which is why the next two slides exist.

Verified for every n from -20 to 20.

⊞ Divisible by 3

Consider an arbitrary \(n\). One of these must hold:

• Case A: \(n=3k\)    then __________ is divisible by 3

• Case B: \(n=3k+1\)   then __________ is divisible by 3

• Case C: \(n=3k+2\)   then __________ is divisible by 3

In all three cases, \(n^3-n\) is divisible by 3.

A: n itself. B: n-1 = 3k. C: n+1 = 3k+3.

The cases have to be EXHAUSTIVE, and that is the only thing that makes this work – every integer is one of those three. Worth saying that choosing the right partition is the whole skill.

⊞ Divisible by 2

Consider an arbitrary \(n\). One of these must hold:

• Case A: \(n=2k\)    then __________ is even

• Case B: \(n=2k+1\)   then __________ is even

So 2 and 3 both divide \(n^3-n\), and therefore _______ does.

A: n itself. B: both n-1 and n+1.

Final blank: 6. It needs 2 and 3 to share no common factor – worth saying out loud, because “divisible by 2 and by 3” does not in general give “divisible by 6” if the two numbers share a factor. Compare 4 and 6: divisible by both does not mean divisible by 24.

Words and Ideas

Proof

Arbitrary element

Universal instantiation, universal generalization

Direct proof

Proof by contrapositive

Proof by contradiction

Proof by cases, exhaustive cases