Proof by Induction

Lecture 11, DSCI 220, 2026W1

Announcements

• EX2 is Thu 10/8 through Sun 10/11, in ORCA. It covers lectures 6 through 10. Practice problems are on PrairieLearn.
• The lecture 10 video was re-recorded: the first upload had no audio.
• Tutorials meet this week.

Friday’s Question

Is the converse true? Is every perfect square a glow number?

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

Both definitions are for positive integers, so 0 never comes up, and T_0 = 0 is defined, so 1 = T_0 + T_1 is a glow number.

The theorem and its converse together: the glow numbers are exactly the perfect squares. Proving the converse means naming a k for a given m – an existential, proved by giving the witness.

Sudoku Warm Up

In a 4x4 Sudoku, each row, each column and each 2x2 box contains each of 1, 2, 3 and 4 exactly once.

Solve the Sudoku on the worksheet.

Speculate on an interesting feature of the corners.

Do all 4x4 Sudoku puzzles have this feature?

The puzzle has exactly one solution: 2 3 4 1 1 4 2 3 4 1 3 2 3 2 1 4 Its four corners (2, 1, 3, 4) are all different.

Do all 4x4 sudokus have that? NO – only 168 of the 288. Counterexample: 1 2 3 4 / 3 4 1 2 / 2 1 4 3 / 4 3 2 1 – corners 1, 4, 4, 1.

“THIS feature” (the a b c d corners against the * centre) is the next slide, Corners and Centre, with the proof in its notes.

A speculation checked on one example is not a theorem – that is the point to land before the proofs.

Corners and Centre

Do all 4x4 Sudoku puzzles have THIS feature?

YES, in all 288: the four corners (a, b, c, d) hold exactly the same four numbers as the four centre cells (*). Not cell by cell – the four together.

The proof never looks at a single cell. Rows and columns are the only collections whose contents we know (each is exactly 1, 2, 3, 4), so count with those, and treat each set of cells as a bag of numbers, duplicates kept.

The four rows: every cell once. = four copies of {1,2,3,4} = corners + border + centre Rows 1 and 4, plus columns 1 and 4: the four edges. The centre is not in them, and each corner is in one of those rows AND one of those columns, so the corners are counted twice. = four copies of {1,2,3,4} = corners + corners + border Same bag, so corners + border + centre = corners + corners + border. Take one copy of the corners and the border off both sides: centre = corners.

(border = the 8 edge cells that are not corners.)

Check on the worksheet’s solution 2341 / 1423 / 4132 / 3214: corners 2, 1, 3, 4; centre 4, 2, 1, 3. Same four numbers.

Not cell by cell: a centre cell is never equal to the corner in its own box, but it can equal any of the other three. In 1234 / 3412 / 2143 / 4321, r2c2 = 4 matches r1c4 and r4c1, not the far corner r4c4 = 1.

It never uses the boxes: true for all 576 grids whose rows and columns each hold 1-4. (A 9x9 version of the same argument is known as the Phistomefel ring.)

\(\to\) Sudoku: Direct

Prove that for any solution to this mini sudoku, if the solution has a 2 in the top-left square (r1c1), then it has a 2 in the bottom-right square (r4c4).

Scrutinize the form of the statement: for any solution, if P then Q.

Assume r1c1 = 2. Row 1 then holds 1, 2 and 3, so r1c4 = 4. Column 4 holds 4, 1 and 3, so r4c4 = 2. Done – two steps.

Checked against all 4x4 sudokus: the puzzle has 6 solutions, and in the 3 where r1c1 = 2, r4c4 = 2.

\(\leftrightsquigarrow\) Sudoku: Contrapositive

Theorem: For any valid Sudoku solution, if A and B and C are not 4, then D is 4.

Contrapositive:

If D is not 4, then A or B or C is 4.

The negation of “A and B and C are not 4” is “A or B or C is 4” – De Morgan, worth pointing at.

Proof: the box contains a 4 somewhere. It is not D, so it is A, B or C. The contrapositive is the easy direction here: it is just “the 4 has to be somewhere”.

↯ Sudoku: Contradiction

Prove that any solution to this mini sudoku has a 3 in the top-right corner (r1c4).

Scrutinize the form: “any solution has a 3 at r1c4”.

Suppose r1c4 is not 3. Row 1 needs a 3. Not c1 (column 1 has the 3 at r3c1), not c2 (that is the 2), not c4 (assumed). So r1c3 = 3.

The bottom-right box needs a 3. Not in row 3 (row 3 has its 3 at r3c1), and r4c4 is 1, so r4c3 = 3. Now column 3 has two 3s – r1c3 and r4c3. Contradiction.

Checked: the puzzle has 3 solutions and r1c4 = 3 in all of them; with r1c3 = 3 added there are none.

Proof by Induction

Today’s goals

• Understand role of components of an inductive proof.
• Write short inductive proofs.

Warm Up

A Tower is built by stacking 5m red panels, and 7m blue panels.

What tower heights are possible? Talk it through with the people around you: which heights can you build, and which can’t you?

Give this time – it took a while last year. Let them find the last height that cannot be built (23) themselves.

End up with a clear claim: For all n > 23, we can build a tower of height n using only 5m red and 7m blue panels.

Say “claim holds for smaller values”

Warm Up Continued

Claim: for all integers \(n > 23\), we can build a tower of height \(n\) using only 5m red and 7m blue panels.

Height	Red	Blue
24
25
26
27
28

Every height \(n \geq 29\): ____________________________

24 = 2 red + 2 blue. 25 = 5 red. 26 = 1 red + 3 blue. 27 = 4 red + 1 blue. 28 = 4 blue.

Every n >= 29: n - 5 is at least 24, so a tower of height n - 5 can be built. Add one red panel.

Why FIVE base cases: each height comes from the height 5 below it, so the first five have to be checked by hand. Three is not enough – 28 would need 23, which cannot be built. (27 would need 22, which can be built, but the claim says nothing about 22.) Worth asking before filling in the table.

Point out the universal quantifier, and the least value. This is proof by cases plus one new move: assume it for a smaller value (n - 5). That is induction, before it has a name.

Aside, only if it comes up: 23 = 5*7 - 5 - 7.

Inductive Proof Skeleton Card

Proof by Induction of \(\forall n, P(n)\)

1. First line: “Consider an arbitrary n.”
2. Second line (IH): “Assume \(\forall j < n, P(j)\).”
3. Determine exhaustive cases for \(n\), including one or more base cases and usually one inductive case.
4. Prove each case, base and inductive, independently. Base case is typically direct, and inductive case will require use of IH.

Proof by Induction

Claim: For any integer \(n > 0\), \(\sum\limits_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}\).

Proof: Consider an arbitrary integer \(n > 0\).

IH: Assume inductively that for any \(j<n\), \(\sum\limits_{k=1}^{j} k^2 = \frac{j(j+1)(2j+1)}{6}\).

Either \(n=1\), or \(n>1\).

Case 1 \((n=1)\):

Case 2 \((n>1)\):

Your turn!

Claim: For any \(n>0\), \(\sum\limits_{k=1}^n \frac{1}{(2k-1)(2k+1)} = \frac{n}{2n+1}\).

If this runs over, it opens Wednesday: the sum telescopes, which is L12’s Trick 2, so prove it by induction here and by telescoping there.

Words and Ideas

Induction, base case, inductive case

Inductive hypothesis (IH)