Discrete Math for Data Science

DSCI 220, 2025 W1

September 24, 2025

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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?

05:00

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.

24:

25:

26:

point out universal quantifier, and also least value.

Also point out which values follow easily from these… sequences, covering integers.

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 \((n>1)\):

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.

Your turn!

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

10:00

Last Demo

This is a tromino:

This is a \(n\) by \(n\) deficient grid:

Claim: Every \(2^n\) by \(2^n\) deficient grid can be tiled with trominos.

Tile Work

Claim: Every \(2^n\) by \(2^n\) deficient grid can be tiled with trominos.

 

Proof: Line 1:

Line 2:

 

Small case(s):

Tile Work

Claim: Every \(2^n\) by \(2^n\) deficient grid can be tiled with trominos.

 

Proof continued… Inductive case: