This deck is a from-scratch guide to STARTING an induction proof, not merely following one. It covers recognizing when a claim needs induction, writing the base case, the inductive hypothesis, and the inductive step, and choosing between weak and strong induction. The worked proofs cover a summation identity, an inequality, a divisibility claim, a strong-induction postage argument, and two recursive-structure proofs - a recursive sum function and binary-tree leaf counts - which connect induction to recursive algorithms. It targets six real mistakes: asserting the conclusion instead of deriving it, picking the wrong base case, writing a step that never uses the hypothesis, misstating the hypothesis as the whole universal claim, proving something weaker than what was required, and using weak induction where strong induction is needed.
Subject: CS3000 Algorithms · 132 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
This deck is about STARTING an induction proof, not just following one someone else wrote. By the end you can:
Warm-up
Discussion prompt
Before we open Proof by Induction: without looking back, what was the main idea of Asymptotic Notation (Big-O, Big-Omega, Big-Theta), 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 Big-O, Big-Omega, and Big-Theta notation as CS3000 uses them. It explains why we measure growth rather than exact time, gives the formal definitions in terms of a constant and a threshold, shows how to construct a Big-O proof by exhibiting a specific c and n0, and lays out the standard growth-rate hierarchy from constant to factorial. It targets four real misconceptions: treating Big-O as a synonym for worst case, treating constant-factor multiples as different classes, reporting a loose Big-O when a tight Big-Theta is actually known, and assuming that an algorithm with a larger Big-O is always slower on the small inputs you actually run.
Concept
Before any new material: cover the screen.
You have named 4 reusable moves so far. Say as many as you can out loud, by number, from memory.
Do not advance until you have actually tried. Getting four of eight is information; skipping the exercise is not.
Here they are. Score yourself.
Today adds 2 moves to this list. Everything else you will need is already above.
The question that starts every proof from here on is not how do I begin. It is which of these applies here?
Counterexample
Discussion prompt
You have named 4 reusable moves so far. Say as many as you can out loud, by number, from memory.
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:
Do not advance until you have actually tried. Getting four of eight is information; skipping the exercise is not.
Picture it
Animation
Shows: Loop invariants have three obligations — a rendered Manim animation.
Rendered with Manim.
Takeaway: The first two are induction; the third is why you bothered.
Concept
Some mathematical claims are not about one number - they are about an entire, endless family of numbers at once. Look for a letter, almost always n, standing for any natural number: one, two, three, and so on forever.
If a claim reads something like 'for every n' or 'for all positive integers n', that n is your signal. The claim is really shorthand for infinitely many separate statements, one for each value the letter can take.
P(n) — A shorthand name for the statement, whatever it says, with the number n plugged in. The claim at n equals one is called P at one, the claim at n equals seven is called P at seven, and so on - all from the same template.
Concept
Picture the claim as a machine: feed it a natural number, and it hands back one true-or-false statement about that number. Saying the claim holds for all n means every single output of that machine, forever, is true.
That is what makes these claims hard to prove directly: you cannot list infinitely many numbers and check each one by hand. You need an argument that covers all of them at once - that argument is induction.
Concept
Plugging in a few small numbers and watching the claim come out true is tempting, and it feels like progress. It is not a proof.
Checking three cases only tells you about those three cases. The claim could fail much later, and nothing you checked would have warned you. A real proof must rule that out for every case, without checking every case by hand.
counterexample — A single value where the claim turns out false. Finding one destroys the whole claim; finding none by testing a few cases proves nothing about the rest.
Definition probe
Sort into buckets
Every line below is part of the definition of P(n) or of counterexample — one or the other, never both. Put each where it belongs.
Intuition
Picture an endless line of dominoes, one for each natural number. A domino falls exactly when the claim is true at that number.
You do two things, and only two: push over the first domino, and set up every domino so that if it falls, it knocks over the very next one. Do both, and every domino in the infinite line falls - you never touch the later dominoes yourself.
Analogy
Discussion prompt
Explain Dominoes: knock the first, guarantee the next by analogy to something with no CS3000 Algorithms 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 an endless line of dominoes, one for each natural number. A domino falls exactly when the claim is true at that number.
Intuition
Now picture an infinite ladder. You prove you can stand on the first rung. Then you prove one general fact: standing on any rung, you can always step up to the next one.
Those two facts together mean you can reach every rung, no matter how high, without ever climbing the whole ladder yourself. Induction is that second fact, proved once, doing infinite work.
Concept
Before writing anything, ask three questions about the claim in front of you.
One: is it about a natural number? Two: is it claimed for every such number from some starting point onward, not just one value? Three: can you describe what changes between the claim at one number and the claim at the very next number?
If the answer to all three is yes, induction is very likely your tool. If the claim is about a single fixed number, or about a variable with no clear 'next' value, induction does not apply the way you would expect.
Explain it
Discussion prompt
Explain The induction checklist: three questions 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 writing anything, ask three questions about the claim in front of you.
Concept
Induction never requires you to check every value by hand, and it never requires you to already know why the claim is true for some huge specific number.
It also does not require cleverness at every step - only at two moments: proving the smallest case, and describing the one general move from any case to the next. Everything else follows automatically.
Socratic
Discussion prompt
Induction never requires you to check every value by hand, and it never requires you to already know why the claim is true for some huge specific number.
Suppose that were not true. What is the first thing in Proof by Induction that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Picture it
Animation
Shows: Induction is exactly two obligations — a rendered Manim animation.
Rendered with Manim.
Takeaway: Skip either and the ladder has no bottom rung, or no rungs.
Concept
The base case proves the claim directly, by plain computation, at the smallest number the claim covers - no assumptions, no shortcuts, just arithmetic or direct substitution.
This is the first domino you personally push over. If it does not actually fall - if the base case is false - nothing later in the proof can rescue the claim.
Picture it
Animation
Shows: One argument, every case — a rendered Manim animation.
Rendered with Manim.
Takeaway: The step is written once and applies at every rung.
Concept
Call the starting value the base value. It is not always the number one - read the claim itself to find where it starts, and if the claim does not say, test the smallest few values by hand until you find where it first becomes true.
A wrong choice of starting value is one of the most common beginner mistakes: picking a start where the claim is actually false, then not noticing because the base case was never truly checked.
Concept
The inductive hypothesis is the single assumption you are allowed to make in the middle of the proof: fix one particular number - call it k - that is at least as large as the starting value, and assume the claim is true at that one number.
You do not get to assume the claim is true for every number - that would be assuming the very thing you are trying to prove. You only get one fixed case, handed to you for free.
Picture it
Animation
Shows: Point at where you used the hypothesis — a rendered Manim animation.
Rendered with Manim.
Takeaway: A step that never invokes the hypothesis is proving something else.
Concept
The phrase 'fixed but arbitrary' describes this k exactly. Fixed means k is one specific number, not a range - inside the step, k never changes value partway through.
Arbitrary means you never used any special property of k to assume the claim there - so whatever you prove about the next number after k will work no matter which k you had picked. That is precisely how one step ends up covering every step at once.
Concept
Before writing any equation, say the hypothesis out loud in plain words: assume the claim is true when the number equals some fixed value k. Only after that sentence is clear should you translate it into symbols.
This habit catches the single most common setup mistake: writing a hypothesis that quietly claims more than one fixed case - for example, accidentally writing it for every number at once, which smuggles in the whole theorem before you have earned it.
Concept
The inductive step proves one general implication: if the claim is true at the fixed number k, then it is also true at the very next number after k.
Nothing else counts as the step. Not restating the claim at the next number, not checking a few more numbers by hand - only a derivation that starts from the assumption at k and ends at the conclusion at the next number.
Picture it
Animation
Shows: The step, done carefully — a rendered Manim animation.
Rendered with Manim.
Takeaway: Line two is where the hypothesis is used — the only place it may be.
Concept
Somewhere in the middle of your derivation, the assumption at k has to actually appear - substituted into an equation, plugged into an inequality, used to name a quantity. If it never appears, you have not used it.
A step that reaches the right answer without ever touching the hypothesis has proved something true, perhaps, but it has not completed an inductive step. That distinction is the difference between real induction and something that only looks like it.
Intuition
New students often feel uneasy assuming the claim is true at k - isn't that assuming the very thing you're trying to prove? It is not, and here is why.
You are never assuming the claim for every number. You are assuming it for one fixed number, in exchange for proving one narrow, honest fact: that truth at that one number forces truth at the very next number. The base case supplies the first truth for free; the step then cascades it forward, one link at a time, exactly like the domino guarantee.
Socratic
Discussion prompt
New students often feel uneasy assuming the claim is true at k - isn't that assuming the very thing you're trying to prove? It is not, and here is why.
Suppose that were not true. What is the first thing in Proof by Induction that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Concept
Every induction proof has exactly this shape: state the claim precisely, prove it directly at the starting value, assume it at one fixed k, and derive it at the next number using that assumption.
\[ \big[P(n_0)\ \text{true}\big] \ \text{and} \ \big[P(k) \Rightarrow P(k+1) \ \text{for every } k \ge n_0\big] \ \Longrightarrow \ P(n) \ \text{true for every } n \ge n_0 \]
Notice what is doing the work: the base case supplies exactly one true statement, and the step is a single general rule. Together, dominoes fall and rungs get climbed forever - you never touch a single value past the start yourself.
Concept
Every proof of this kind has the same five or six moves in the same order. The order is not something you rediscover each time.
It is on the right. It will stay on the right through the worked examples that follow.
Why this matters: the structure is now handled. You are not spending working memory on what comes next — you are spending all of it on the one hard step.
Steps 1, 2, 3 and 6 are bookkeeping. Step 4 is the only place you have to think — and step 5 is automatic once step 4 is done right.
Picture it
Animation
Shows: Each case knocks over the next — a rendered Manim animation.
Rendered with Manim.
Takeaway: The step is one argument that works at every rung.
Concept
Consider the claim: adding up every whole number from one through n always equals a particular formula in n. Run it through the checklist.
It is indexed by a natural number. It is claimed for every such number starting from one. And moving from one number to the next changes the sum in a very specific, describable way - you just add one more term. All three boxes are checked: this is an induction claim.
Concept
Before proving anything, write down the exact statement the claim asserts, with n as the free variable.
\[ P(n):\quad 1+2+\cdots+n = \frac{n(n+1)}{2} \]
Everything that follows is just this statement applied at different numbers: applied at one is the base case, applied at k is the hypothesis, and applied at the number after k is the target of the step.
Concept
The claim describes the sum of the first n positive integers, and the smallest positive integer is one - so the natural place to start is with n equal to one.
Read the claim again to confirm: it never mentions starting at zero, so there is no reason to start there. The base value is one.
Intuition
What move should we make next?
We have assumed the hypothesis, and we have to reach the goal:
\[ \text{assume } \; 1 + 2 + \cdots + k = \frac{k(k+1)}{2} \]
\[ \text{goal } \; 1 + 2 + \cdots + k + (k+1) = \frac{(k+1)(k+2)}{2} \]
The goal is a sum of k+1 things. The hypothesis is about a sum of k things. Something has to bridge that gap.
_Look at your toolkit. Say a move number out loud before this slide advances._ A wrong guess is useful. A silent guess is not.
Socratic
Discussion prompt
We have assumed the hypothesis, and we have to reach the goal:
Suppose that were not true. What is the first thing in Proof by Induction that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
The goal is a sum of k+1 things. The hypothesis is about a sum of k things. Something has to bridge that gap.
Ranking
Put in order
Put the moves of Worked proof: the sum of the first n positive integers 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. Plug the value one directly into both sides and check they match - no assumptions are allowed here.
Worked example
Claim: for every positive integer n, one plus two plus all the way up through n equals n times n plus one, divided by two.
\[ P(n):\quad 1+2+\cdots+n = \frac{n(n+1)}{2} \]
Prove the base case at n equal to one
Why: Plug the value one directly into both sides and check they match - no assumptions are allowed here.
\[ n=1:\quad \text{LHS}=1,\quad \frac{1(2)}{2}=1\ \checkmark \]
State the inductive hypothesis
Why: Fix one particular integer k that is at least one, and assume the sum formula holds there. This is the one thing we are allowed to assume.
\[ 1+2+\cdots+k = \frac{k(k+1)}{2} \]
Add the next term to both sides
Why: The sum up to k plus one is the sum up to k, plus the new term k plus one. Substituting the hypothesis for the sum up to k is exactly where it gets used.
\[ 1+2+\cdots+k+(k+1) = \frac{k(k+1)}{2}+(k+1) \]
Simplify the right side
Why: Factor k plus one out of both terms and combine over a common denominator to reach a single fraction.
\[ \frac{k(k+1)}{2}+(k+1) = \frac{(k+1)(k+2)}{2} \]
Verify the result matches the target statement
Why: The simplified expression is exactly the original formula with n replaced by k plus one, so the step is complete. Checking n equal to five by direct arithmetic confirms the formula independently.
\[ n=5:\ 1+2+3+4+5=15=\frac{5\cdot 6}{2}\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Worked proof: the sum of the first n positive integers", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The simplified expression is exactly the original formula with n replaced by k plus one, so the step is complete. Checking n equal to five by direct arithmetic confirms the formula independently.
Fill the middle
Fill in the blanks
From The move we just made, named — finish the line. Write what belongs on the right of the equals sign before you look.
\frac\frac{(k+1)(k+2)}{2}___ + (k+1) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The sum of the first k+1 numbers is literally the sum of the first k numbers, plus one more number.
Concept
The move: #5 (Peel one off), then #6 (Substitute the hypothesis).
Peel the last term off the goal's left side
Why: The sum of the first k+1 numbers is literally the sum of the first k numbers, plus one more number. Splitting it that way is a choice you make, not something the algebra hands you.
\[ \underbrace{1 + 2 + \cdots + k}_{\text{this is } P(k)} + (k+1) \]
Substitute the hypothesis for the peeled part
Why: The underbraced piece is exactly what you assumed. Swap it for the closed form. This step is automatic — the thinking already happened when you chose where to peel.
\[ \frac{k(k+1)}{2} + (k+1) = \frac{(k+1)(k+2)}{2} \]
Every induction proof you write this term is these two moves. If a step of yours never touches the hypothesis, you skipped move #6 and you have not proved anything.
Notation
Annotate
From The move we just made, named — read this one piece at a time. What is each part doing?
On: \( \underbrace{1 + 2 + \cdots + k}_{\text{this is } P(k)} + (k+1) \)
Anomaly
Predict first
A student writes this, and it looks reasonable:
For the same sum claim, a student writes the inductive step by starting from the k-plus-one statement itself.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This treats the very thing that needs to be proved as already given, then just displays it - no hypothesis at k was ever used to reach it.
Start only from the hypothesis at k, and derive the k-plus-one statement by algebra.
Why: This treats the very thing that needs to be proved as already given, then just displays it - no hypothesis at k was ever used to reach it. That is circular: assuming what you want and calling it shown.
Trap
For the same sum claim, a student writes the inductive step by starting from the k-plus-one statement itself.
\[ \text{Target: } 1+2+\cdots+(k+1) = \frac{(k+1)(k+2)}{2} \]
Write down the target equation as the starting point
Why: This treats the very thing that needs to be proved as already given, then just displays it - no hypothesis at k was ever used to reach it. That is circular: assuming what you want and calling it shown.
\[ \text{Wrong 'step': assume } 1+2+\cdots+(k+1) = \frac{(k+1)(k+2)}{2}\ \text{directly, no derivation} \]
Start only from the hypothesis at k, and derive the k-plus-one statement by algebra.
\[ \text{Hypothesis: } 1+2+\cdots+k = \frac{k(k+1)}{2} \]
Add the next term to the hypothesis and simplify
Why: This is the real derivation: begin at what you are allowed to assume, and arrive at the target through valid steps, never assuming the target itself.
\[ \frac{k(k+1)}{2}+(k+1) = \frac{(k+1)(k+2)}{2}\ \text{(derived, not assumed)} \]
Concept
Consider the claim: two raised to the power n is always bigger than n itself, for every positive integer n. It looks different from an equation, but check it against the same three questions.
It is indexed by n, claimed for every n from one onward, and moving from n to the next number changes both sides in a describable way - the left side doubles. Inequalities fit the induction template exactly as equations do.
Concept
State the exact claim first.
\[ P(n):\quad 2^{n} > n \]
The hypothesis fixes one particular k and assumes the inequality there - nothing more. Writing it as 'greater than' instead of 'equals' does not change how the hypothesis works; it is still one fixed case, assumed once.
\[ \text{Hypothesis: fix } k \ge 1,\ \text{assume } 2^{k} > k \]
Intuition
Watch me not know the answer. This is what the first two minutes actually look like.
The claim, for every whole number n starting at 1:
\[ 2^{n} > n \]
First instinct: just compare the two sides directly
Why: Two to the k+1 is bigger than k+1 because exponentials grow faster. That sentence feels true, so let us see whether it can be made into a step.
It cannot
Why: Exponentials grow faster is the thing we are being asked to prove. Using it as a reason is circular. The step also never touched the hypothesis, which is the other tell.
Dead end. Not a mistake — a move that was worth trying and did not pay off. This happens in most proofs.
Back up. Peel one off instead
Why: Rewrite the k+1 case so the k case physically appears inside it. Two to the k+1 is two times two to the k — and two to the k is what the hypothesis is about.
\[ 2^{k+1} = 2 \cdot 2^{k} > 2 \cdot k = k + k \ge k + 1 \]
That is #5 (Peel one off) followed by #6 (Substitute the hypothesis) again — the same two moves as the sum proof, on a claim that looks nothing like it.
The expert does not see the whole path in advance. The expert tries something, reads the result, and adjusts. That is the skill.
Step zero
Discussion prompt
Worked proof: 2 to the n exceeds n — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Prove the base case at n equal to one
Answer:
Worked example
Claim: for every positive integer n, two raised to the power n is strictly greater than n.
\[ P(n):\quad 2^{n} > n, \quad n \ge 1 \]
Prove the base case at n equal to one
Why: Two to the power one is two, which is bigger than one. The base case holds by direct computation.
\[ 2^{1} = 2 > 1\ \checkmark \]
State the inductive hypothesis
Why: Fix one integer k that is at least one, and assume two to the power k exceeds k.
\[ 2^{k} > k \]
Double both sides of the hypothesis
Why: Multiplying an assumed true inequality by two, a positive number, preserves the inequality, and the left side becomes exactly two to the power k plus one.
\[ 2^{k+1} = 2\cdot 2^{k} > 2k \]
Bound the right side below by k plus one
Why: Since k is at least one, two times k is at least k plus one - a small, separate fact about numbers that connects the doubled bound to the target.
\[ 2k \ge k+1 \quad (\text{since } k\ge 1) \]
Verify the chain proves the target statement
Why: Combining the two inequalities gives two to the power k plus one exceeds k plus one exactly, completing the step. Checking n equal to six directly confirms the formula independently.
\[ 2^{k+1} > 2k \ge k+1 \ \Rightarrow\ 2^{k+1} > k+1; \quad n=6:\ 2^{6}=64>6\ \checkmark \]
Picture it
Animation
Shows: The base case is a real proof — a rendered Manim animation.
Rendered with Manim.
Takeaway: Skip it and the ladder has no bottom rung.
Anomaly
Predict first
A student writes this, and it looks reasonable:
For the claim n squared is at least three times n plus four, true only for large enough n, a student assumes the base case must start at n equal to one, out of habit.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Plugging in one gives one on the left and seven on the right - false.
Test small values in order to find where the claim genuinely becomes true.
Why: Plugging in one gives one on the left and seven on the right - false. Since the base case is not actually verified, everything that follows is built on nothing, even if the algebra afterward looks fine.
Trap
For the claim n squared is at least three times n plus four, true only for large enough n, a student assumes the base case must start at n equal to one, out of habit.
\[ P(n):\quad n^{2} \ge 3n+4 \]
Test n equal to one and press on anyway
Why: Plugging in one gives one on the left and seven on the right - false. Since the base case is not actually verified, everything that follows is built on nothing, even if the algebra afterward looks fine.
\[ n=1:\ 1^{2}=1,\quad 3(1)+4=7,\quad 1 \ge 7 \ \text{is false} \]
Test small values in order to find where the claim genuinely becomes true.
\[ P(n):\quad n^{2} \ge 3n+4 \]
Check n equal to one, two, three, then four in order
Why: One gives one versus seven, false. Two gives four versus ten, false. Three gives nine versus thirteen, false. Four gives sixteen versus sixteen, true. The smallest value that works is four.
\[ n=1,2,3:\ \text{false}; \qquad n=4:\ 16 \ge 16\ \checkmark \]
Use four as the base case
Why: Only start the induction where the claim is actually true. The base case must be verified, never assumed out of habit.
Break the constraint
Discussion prompt
The rule this trap just fixed:
One gives one versus seven, false. Two gives four versus ten, false. Three gives nine versus thirteen, false. Four gives sixteen versus sixteen, true. The smallest value that works is four.
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.
Answer:
Plugging in one gives one on the left and seven on the right - false. Since the base case is not actually verified, everything that follows is built on nothing, even if the algebra afterward looks fine.
Concept
Consider the claim: n cubed minus n is always divisible by three, for every positive integer n. Divisible by three means the quantity equals three times some whole number.
\[ P(n):\quad n^{3}-n = 3\cdot(\text{some integer}) \]
Writing the claim this way, as an equation with an unnamed integer multiplier, is exactly what will let the hypothesis be substituted into an equation later.
Concept
Before doing any algebra, decide in advance: the hypothesis says k cubed minus k equals three times some integer, call it m. The step must expand the next case and show that same quantity, three times m, sitting inside it.
Planning this in advance - expecting to see the hypothesis appear as a substitution - is exactly the habit that prevents accidentally proving the fact by a different route that skips the hypothesis entirely.
Intuition
What move should we make next?
Hypothesis and goal for the divisibility claim:
\[ \text{assume } \; 3 \mid (k^{3} - k) \]
\[ \text{goal } \; 3 \mid \big((k+1)^{3} - (k+1)\big) \]
Do not expand anything yet. First answer this: where inside the goal do you expect the hypothesis to show up, and what will be left over?
_Look at your toolkit. Say a move number out loud before this slide advances._ A wrong guess is useful. A silent guess is not.
Step zero
Discussion prompt
Worked proof: 3 divides n cubed minus n — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Prove the base case at n equal to one
Answer:
Worked example
Claim: for every positive integer n, three divides n cubed minus n.
\[ P(n):\quad 3 \mid (n^{3}-n) \]
Prove the base case at n equal to one
Why: One cubed minus one is zero, and zero is three times zero - divisible by three by direct computation.
\[ n=1:\ 1^{3}-1 = 0 = 3\cdot 0\ \checkmark \]
State the inductive hypothesis
Why: Fix one integer k that is at least one, and assume k cubed minus k equals three times some integer m - m is just a name for whatever whole number makes that equation true.
\[ k^{3}-k = 3m \quad \text{for some integer } m \]
Expand the next case
Why: Multiply out the cube for k plus one and collect terms; the result splits naturally into the exact quantity from the hypothesis, plus something extra.
\[ (k+1)^{3}-(k+1) = k^{3}+3k^{2}+3k+1-k-1 = (k^{3}-k) + 3(k^{2}+k) \]
Substitute the hypothesis into the expression
Why: This is the moment the hypothesis is used: replace k cubed minus k with three m, then factor out the shared three.
\[ = 3m + 3(k^{2}+k) = 3\big(m+k^{2}+k\big) \]
Verify the result is 3 times an integer
Why: The expression is now three times a whole number, matching the target statement exactly. Checking n equal to five directly confirms the divisibility independently.
\[ n=5:\ 5^{3}-5 = 120 = 3\cdot 40\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Worked proof: 3 divides n cubed minus n", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The expression is now three times a whole number, matching the target statement exactly. Checking n equal to five directly confirms the divisibility independently.
Fill the middle
Fill in the blanks
From Trap: a step that never uses the hypothesis — finish the line. Write what belongs on the right of the equals sign before you look.
(k+1)^k(k+1)(k+2)\ (\text{true, but the hypothesis never appears})-(k+1) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. The expression factors as k times k plus one times k plus two - three consecutive whole numbers - and among any three consecutive integers, one is always divisible by three.
Trap
For the same divisibility claim, a student factors the expanded expression a completely different way.
\[ P(n):\quad 3 \mid (n^{3}-n) \]
Factor into three consecutive integers instead
Why: The expression factors as k times k plus one times k plus two - three consecutive whole numbers - and among any three consecutive integers, one is always divisible by three. That is a true and correct fact, but the hypothesis, that k cubed minus k equals three m, was never substituted anywhere. Whatever this argument is, it is not the inductive step.
\[ (k+1)^{3}-(k+1) = k(k+1)(k+2)\ (\text{true, but the hypothesis never appears}) \]
Redo the step so the hypothesis is explicitly substituted.
\[ \text{Hypothesis: } k^{3}-k = 3m \]
Expand, then substitute three m from the hypothesis
Why: This is the same computation as the actual worked proof: expand the cube, recognize the k cubed minus k piece, and replace it using the assumed equation. Now the hypothesis is genuinely load-bearing.
\[ (k+1)^{3}-(k+1) = 3m + 3(k^{2}+k) = 3(m+k^{2}+k) \]
Translation
\( P(n):\quad 3 \mid (n^{3}-n) \)
Draw it
Translate both ways. First write the expression above as a sentence with no symbols in it at all. Then cover it, and write your sentence back as notation. If the two versions disagree, the disagreement is the thing to fix.
Concept
In every proof so far, the hypothesis handed you exactly one fact: the claim is true at k, the single number immediately before the target. This is ordinary, or weak, induction.
Weak induction is enough whenever reaching the target at the next number only ever requires knowing the truth at that one immediately prior number - nothing further back.
Socratic
Discussion prompt
In every proof so far, the hypothesis handed you exactly one fact: the claim is true at k, the single number immediately before the target. This is ordinary, or weak, induction.
Suppose that were not true. What is the first thing in Proof by Induction that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
Weak induction is enough whenever reaching the target at the next number only ever requires knowing the truth at that one immediately prior number - nothing further back.
Picture it
Animation
Shows: Induction on an inequality — a rendered Manim animation.
Rendered with Manim.
Takeaway: The claim only starts holding past a threshold, so the base case moves.
Concept
Sometimes, to build the claim at the next number, the natural argument needs a fact about a much earlier number - not k, but maybe a couple of numbers before that, depending on the situation.
When that happens, weak induction leaves you stuck: it only licenses using the fact at k, and nowhere else. That mismatch between what the step needs and what weak induction grants is the warning sign that you need a stronger hypothesis.
Intuition
Ordinary induction lets each rung push you up to only the very next one - a strict, local handoff. Strong induction changes the rule: you may look down at every rung below you, not just the last one, and use whichever earlier rung actually helps.
This matters exactly when climbing to the next rung genuinely depends on a rung further down, not the one directly beneath your feet.
Concept
The strong inductive hypothesis assumes the claim holds at every value from the starting number all the way up through k, not just at k alone.
\[ \text{Strong hypothesis: } P(n_0),\ P(n_0+1),\ \ldots,\ P(k) \ \text{all assumed true} \]
The step then may use whichever of those earlier cases it actually needs to reach the target at the next number - you are not forced to use all of them, only permitted to reach for any of them.
Picture it
Animation
Shows: Strong induction assumes more — a rendered Manim animation.
Rendered with Manim.
Takeaway: Needed whenever the recursion splits into pieces of unequal size.
Concept
If the step for strong induction needs to reach back several numbers, then the very first few applications of the step have nothing far enough back to reach. Those first few values must be verified directly instead, as extra base cases.
So a strong induction proof sometimes needs a handful of base cases checked by hand, not just one, before the general step can take over.
Concept
Ask: does the natural argument for the next number only ever need the fact at k? If yes, weak induction suffices.
Ask instead: does the natural argument need a fact about some earlier number that is not always k - a number whose distance back can vary? If yes, use strong induction, and expect to verify a few extra base cases by hand.
Concept
Consider the claim: every amount of postage worth eight cents or more can be made using only three-cent and five-cent stamps.
To build the next amount, the natural move is to remove one three-cent stamp and check whether the remaining amount can be made - but that remaining amount is two less than the target, not the number directly before it. That mismatch is exactly the warning sign for strong induction.
Picture it
Animation
Shows: When ordinary induction is not enough — a rendered Manim animation.
Rendered with Manim.
Takeaway: Any recursion that splits unevenly needs the stronger hypothesis.
Concept
\[ \text{Strong hypothesis: for every } j \text{ with } 8 \le j \le k,\ j \text{ can be made from 3-cent and 5-cent stamps} \]
This hypothesis grants access to every amount from eight up through k, so whichever earlier amount the step actually needs is guaranteed to be covered.
Concept
The step will reach back to an amount two less than the target. For that reduced amount to land on a value the hypothesis actually covers, the target must be at least eleven cents - so the step only takes over from eleven onward.
That leaves the amounts eight, nine, and ten with nothing earlier to lean on: they must each be verified directly, by hand, as base cases.
Intuition
What feels wrong about this?
The claim is that every postage amount from 8 cents up can be made from 3-cent and 5-cent stamps. The natural step is to build the k+1 amount from a smaller one.
\[ \text{but } \; (k+1) - 3 = k - 2, \quad \text{not } \; k \]
_Plain English only. No notation, no algebra. Just say what bothers you._
The feeling: you land three cents back, not one, so the rung you need is not the rung you assumed. Weak induction only hands you the rung directly behind you.
That feeling is the proof. It is not a substitute for the proof — it is the thing the proof writes down.
So the fix is not a new technique — it is a bigger hypothesis. Assume every case up to k, and the rung you land on is covered whichever one it turns out to be.
Ranking
Put in order
Put the moves of Worked proof: every amount from 8 cents up is 3s and 5s 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. Eight, nine, and ten cents each need to be checked by hand, since the step below cannot reach back far enough to cover them.
Worked example
Claim: for every integer n of at least eight, n cents of postage can be made using only three-cent and five-cent stamps.
\[ P(n):\quad \forall n \ge 8,\ \exists\, a,b \ge 0:\ n = 3a+5b \]
Verify the three base cases directly
Why: Eight, nine, and ten cents each need to be checked by hand, since the step below cannot reach back far enough to cover them.
\[ 8=3(1)+5(1),\quad 9=3(3)+5(0),\quad 10=3(0)+5(2) \]
State the strong inductive hypothesis
Why: Assume every amount from eight up through some fixed k, at least ten, can be made from three-cent and five-cent stamps - not just the amount k alone.
\[ \text{for every } j,\ 8\le j \le k\ (k\ge 10):\ j = 3a+5b\ \text{for some } a,b\ge 0 \]
Reduce the target by one 3-cent stamp
Why: The target is k plus one; removing three cents leaves k minus two, which sits between eight and k because k is at least ten - exactly the range the hypothesis covers.
\[ (k+1)-3 = k-2, \qquad 8 \le k-2 \le k \]
Apply the hypothesis to the reduced amount
Why: By the hypothesis, k minus two can be made from some number of three-cent and five-cent stamps; add one more three-cent stamp to that combination to reach the target.
\[ k-2 = 3a'+5b' \ \Rightarrow\ k+1 = 3(a'+1)+5b' \]
Verify the construction reaches the target
Why: The combination totals exactly k plus one cents, completing the step. Checking n equal to fourteen directly confirms it: eleven cents is two threes plus one five, and one more three-cent stamp reaches fourteen.
\[ n=14:\ 11=3(2)+5(1) \ \Rightarrow\ 14 = 3(3)+5(1)=9+5=14\ \checkmark \]
Picture it
Animation
Shows: Each line of the worked example "Worked proof: every amount from 8 cents up is 3s and 5s", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Eight, nine, and ten cents each need to be checked by hand, since the step below cannot reach back far enough to cover them.
Anomaly
Predict first
A student writes this, and it looks reasonable:
A student tries to prove the same postage claim using only weak induction - assuming just the single amount k, not the whole range up to k.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Suppose the hypothesis only grants that nine cents can be made, using three three-cent stamps.
Use strong induction so the hypothesis covers every earlier amount, not just one.
Why: Suppose the hypothesis only grants that nine cents can be made, using three three-cent stamps. There is no one-cent stamp to bridge from nine to ten - the representation for nine gives no way to construct ten. The single fact at k simply is not enough; the natural argument needs a fact about a different, earlier number, which weak induction never grants.
Trap
A student tries to prove the same postage claim using only weak induction - assuming just the single amount k, not the whole range up to k.
\[ \text{Weak hypothesis (insufficient): only } k = 3a+5b \text{ is assumed} \]
Try to reach the next amount directly from k alone
Why: Suppose the hypothesis only grants that nine cents can be made, using three three-cent stamps. There is no one-cent stamp to bridge from nine to ten - the representation for nine gives no way to construct ten. The single fact at k simply is not enough; the natural argument needs a fact about a different, earlier number, which weak induction never grants.
\[ 9=3(3)+5(0) \ \not\Rightarrow\ 10 = \text{(no bridge exists from this one fact)} \]
Use strong induction so the hypothesis covers every earlier amount, not just one.
\[ \text{Strong hypothesis: every } j,\ 8\le j\le k,\ \text{is } 3a+5b \]
Reach back to a different earlier amount instead
Why: With the strong hypothesis, the amount two less than the target - a different number from k - is guaranteed to be constructible, and adding one three-cent stamp to that construction reaches the target. This is the actual fact the step needs, and only strong induction supplies it.
\[ k-2 = 3a'+5b' \ \Rightarrow\ k+1 = 3(a'+1)+5b' \]
Concept
A recursive function is defined by exactly two pieces: a base case that returns an answer directly, with no further calls, and a recursive case that computes the answer for n by calling the same function on a smaller input and combining the result.
That is precisely the base-case-plus-step shape. It is not a coincidence - recursion and induction are the same idea, one written as a proof and the other written as a program.
Concept
The three parts of an induction proof are already sitting in any correct recursive procedure. Line for line, the base case is the base case and the recursive call is the inductive hypothesis.
SUM-TO(n)
if n == 0
return 0
return n + SUM-TO(n - 1)The recursive call on line 4 is where you use the hypothesis. You do not trace it, and you do not re-derive it. You assume it returns the right answer for a smaller input, exactly as induction lets you assume the claim for k.
Notation
Every line of SUM-TO says one thing. Read the line, then read what it does — not the other way round.
Annotate
Invariant
The stack winds down to the base case, then unwinds. The unwinding is the proof direction: each return value is built from the one below it, which is what the inductive step promises.
Step through it
At the deepest call, say out loud which line of an induction proof you are standing in.
Picture it
Animation
Shows: SUM-TO executing: the current line of pseudocode is highlighted while the data it touches changes.
Rendered with Manim.
Takeaway: The stopping condition is the base case and the recursive call is the inductive hypothesis — a correct recursion and a correct induction are the same object.
Picture it
Animation
Shows: The classic induction target — a rendered Manim animation.
Rendered with Manim.
Takeaway: Two obligations, and nothing else is required.
Intuition
When a recursive function calls itself on a smaller input and then trusts the result without re-deriving it from scratch, that trust is exactly the inductive hypothesis.
You are allowed to assume the smaller call already works correctly - the same permission an inductive step gets to assume the claim at k - and your job is only to show that the current call builds the correct answer out of that trusted, smaller result.
Concept
For a recursive algorithm, the claim is not an equation about numbers in the abstract - it is the statement that the function returns the correct answer when called on an input of size n.
Stating the claim this way turns 'prove the algorithm works' into an ordinary induction claim: indexed by n, claimed for every n, with a clear notion of what changes between size n and a smaller size.
Socratic
Discussion prompt
For a recursive algorithm, the claim is not an equation about numbers in the abstract - it is the statement that the function returns the correct answer when called on an input of size n.
Suppose that were not true. What is the first thing in Proof by Induction that would stop working?
Hint: Follow it one step downstream. The answer is whatever was quietly relying on it.
Answer:
Stating the claim this way turns 'prove the algorithm works' into an ordinary induction claim: indexed by n, claimed for every n, with a clear notion of what changes between size n and a smaller size.
Concept
The base case of the induction proof is exactly the function's stopping condition - the smallest input where the function returns an answer directly, without calling itself.
Just as with any base case, this must be checked directly: does the function actually return the correct answer on that smallest input, with no assumptions borrowed from anywhere else?
Explain it
Discussion prompt
Explain Setup decision: the base case is the stopping condition 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:
The base case of the induction proof is exactly the function's stopping condition - the smallest input where the function returns an answer directly, without calling itself.
Step zero
Discussion prompt
Worked proof: a recursive sum function is correct — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Verify the base case at n equal to zero
Answer:
Worked example
Define a function that adds up the whole numbers from zero to n by calling itself on a smaller input.
\[ S(0)=0; \qquad S(n) = n + S(n-1)\ \text{for } n \ge 1 \]
Claim: this function always returns the closed-form sum formula.
\[ P(n):\quad S(n) = \frac{n(n+1)}{2} \]
Verify the base case at n equal to zero
Why: The stopping condition returns zero directly; the formula at n equal to zero also gives zero - they match with no assumptions used.
\[ S(0)=0=\frac{0(1)}{2}\ \checkmark \]
State the inductive hypothesis
Why: Fix one integer k at least zero and assume the smaller call, on input k, already returns the correct formula. This is the recursive call's return value, trusted rather than re-derived.
\[ S(k) = \frac{k(k+1)}{2} \]
Unfold the recursive case at k plus one
Why: By the function's own definition, the call on k plus one returns k plus one, plus whatever the smaller call on k returns - and the hypothesis tells us exactly what that smaller call returns.
\[ S(k+1) = (k+1) + S(k) \]
Simplify using the hypothesis
Why: Substitute the assumed value for the smaller call and combine into a single fraction, matching the target formula at k plus one.
\[ S(k+1) = (k+1) + \frac{k(k+1)}{2} = \frac{(k+1)(k+2)}{2} \]
Verify against a direct unrolling
Why: The simplified result matches the target formula exactly. Unrolling the recursion by hand for n equal to four gives four plus three plus two plus one plus zero, which is ten, matching the formula's four times five over two.
\[ S(4) = 4+3+2+1+0 = 10 = \frac{4\cdot 5}{2}\ \checkmark \]
Picture it
Animation
Shows: Proving a recursive algorithm correct — a rendered Manim animation.
Rendered with Manim.
Takeaway: Assuming the recursion works is legitimate here, not circular.
Explain it to yourself
Discussion prompt
In Same two moves, now on code this move is made:
The inductive hypothesis IS trusting the call
Why is that legal? Name the rule or definition it rests on before you read on.
Hint: If you can only say "because that is what you do", the rule is the thing to go and find.
Answer:
Assuming the call on n-1 returns the right answer is exactly assuming P(k). You are not tracing the call — you are substituting the hypothesis.
Concept
The move: #5 (Peel one off), then #6 (Substitute the hypothesis).
The recursive call IS the peel
Why: A function that returns n plus a call on n-1 has already done move #5 for you. The code peels one off; you did not have to choose where.
The inductive hypothesis IS trusting the call
Why: Assuming the call on n-1 returns the right answer is exactly assuming P(k). You are not tracing the call — you are substituting the hypothesis.
\[ \texttt{sum}(k+1) = (k+1) + \underbrace{\texttt{sum}(k)}_{\text{hypothesis}} = (k+1) + \frac{k(k+1)}{2} \]
This is why you can trust recursion without tracing every call: the proof template and the code have the same shape. Recognizing that is worth more than any single proof today.
Concept
Once the base case and the inductive step are both proved, you never again need to trace through every recursive call by hand to believe the function works on some huge input.
The induction proof is doing exactly the same job the domino guarantee did: prove the smallest case, prove that trusting one smaller call is enough to build the next case correctly, and correctness cascades to every input size at once.
Analogy
Discussion prompt
Explain Why this justifies trusting recursion instead of tracing every call by analogy to something with no CS3000 Algorithms 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:
Once the base case and the inductive step are both proved, you never again need to trace through every recursive call by hand to believe the function works on some huge input.
Concept
Not every claim starts out as an equation. Consider: a certain kind of tree, built by repeatedly turning a single leaf into a branching point with two new leaves, always ends up with one more leaf than it has branching points.
Before proving anything, translate the words into a precise statement: after n such steps, the number of branching points equals n, and the number of leaves equals n plus one. Only once the claim is written this precisely can a base case or hypothesis be stated about it.
Ranking
Put in order
Put the moves of Worked proof: counting leaves in a full binary tree 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 any steps, there is one leaf and zero branching points - the formula gives zero plus one equals one, which matches.
Worked example
Start from a single leaf with no branching points. Each step picks any existing leaf and gives it two brand-new leaf children, turning that leaf into a branching point.
\[ T_0:\ 0\ \text{branch points},\ 1\ \text{leaf} \]
Claim: after n such growth steps, the tree has exactly n branch points and n plus one leaves.
\[ P(n):\ \text{after } n \text{ steps: branch points} = n,\ \text{leaves} = n+1 \]
Verify the base case at n equal to zero
Why: Before any steps, there is one leaf and zero branching points - the formula gives zero plus one equals one, which matches.
\[ n=0:\ \text{branch points}=0,\ \text{leaves}=1=0+1\ \checkmark \]
State the inductive hypothesis
Why: Fix one integer k at least zero and assume that after k steps, there are exactly k branching points and k plus one leaves.
\[ \text{after } k \text{ steps: branch points}=k,\ \text{leaves}=k+1 \]
Apply one more step
Why: Pick any current leaf and turn it into a branching point with two new leaf children. This removes one leaf and adds two, and adds exactly one branching point.
\[ \text{one leaf} \to \text{one branch point} + \text{two new leaves} \]
Count branch points and leaves using the hypothesis
Why: Branch points go from k to k plus one directly. Leaves go from k plus one, the hypothesis's count, down by one for the leaf that changed, then up by two for its new children.
\[ \text{branch points} = k+1; \qquad \text{leaves} = (k+1)-1+2 = k+2 \]
Verify the counts match the target statement
Why: Branch points equal k plus one and leaves equal k plus two, which is exactly k plus one, plus one - matching the target formula. Building the tree by hand through three steps gives three branch points and four leaves, confirming it directly.
\[ k+2 = (k+1)+1\ \checkmark; \qquad n=3:\ 3\ \text{branch points}, 4\ \text{leaves}\ \checkmark \]
Picture it
Animation
Shows: Structural induction on a tree — a rendered Manim animation.
Rendered with Manim.
Takeaway: Same shape, but the ordering is by structure rather than by number.
Anomaly
Predict first
A student writes this, and it looks reasonable:
For the inequality claim that two to the n exceeds n, a student writes the inductive hypothesis like this.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: This states the entire universal claim as the assumption, rather than fixing one particular k.
Fix exactly one number and assume the claim only there.
Why: This states the entire universal claim as the assumption, rather than fixing one particular k. Since the whole claim - true for every n - is exactly what the proof is supposed to establish, assuming it here makes the argument circular before the step even begins.
Trap
For the inequality claim that two to the n exceeds n, a student writes the inductive hypothesis like this.
\[ P(n):\quad 2^{n} > n \]
Write the hypothesis as true for every n
Why: This states the entire universal claim as the assumption, rather than fixing one particular k. Since the whole claim - true for every n - is exactly what the proof is supposed to establish, assuming it here makes the argument circular before the step even begins.
\[ \text{Wrong hypothesis: } 2^{n} > n \ \text{for every } n \ge 1 \]
Fix exactly one number and assume the claim only there.
\[ P(n):\quad 2^{n} > n \]
Fix one particular k and assume the claim only at k
Why: This assumes a single, narrow fact - true at one arbitrary number - which is a legitimate premise, and it is exactly what licenses deriving the claim at the next number without circularity.
\[ \text{Correct hypothesis: fix } k \ge 1,\ \text{assume } 2^{k} > k \]
Notation
Annotate
From Trap: writing the hypothesis as the whole claim — read this one piece at a time. What is each part doing?
On: \( \text{Correct hypothesis: fix } k \ge 1,\ \text{assume } 2^{k} > k \)
Anomaly
Predict first
A student writes this, and it looks reasonable:
Continuing the same inequality proof, after doubling the hypothesis a student stops one move too early.
It is wrong. Say what breaks — and say it before you turn the page.
Correct: Since two k is certainly bigger than k, this conclusion is technically true - but it is not what the step needs to show.
Push one more inequality to land exactly on the required target.
Why: Since two k is certainly bigger than k, this conclusion is technically true - but it is not what the step needs to show. The target requires exceeding k plus one, not just k. Stopping here leaves the actual step unfinished.
Trap
Continuing the same inequality proof, after doubling the hypothesis a student stops one move too early.
\[ 2^{k+1} = 2\cdot 2^{k} > 2k \]
Conclude only that the next term exceeds k
Why: Since two k is certainly bigger than k, this conclusion is technically true - but it is not what the step needs to show. The target requires exceeding k plus one, not just k. Stopping here leaves the actual step unfinished.
\[ 2^{k+1} > 2k \ \Rightarrow\ 2^{k+1} > k\quad(\text{not the required target}) \]
Push one more inequality to land exactly on the required target.
\[ 2^{k+1} = 2\cdot 2^{k} > 2k \]
Bound two k below by k plus one, then chain the inequalities
Why: Since k is at least one, two k is at least k plus one. Chaining this with the doubled hypothesis lands exactly on the statement that the next term exceeds k plus one - precisely the target, nothing weaker.
\[ 2k \ge k+1\ (k\ge 1) \ \Rightarrow\ 2^{k+1} > 2k \ge k+1 \ \Rightarrow\ 2^{k+1} > k+1 \]
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.
Concept
Look back at every worked proof so far: a summation identity, an inequality, a divisibility claim, a postage-stamp existence claim, a recursive function, and a tree-growth claim. The subject matter was completely different every time.
The setup decisions were identical every time: recognize the claim as indexed by n, state it precisely, choose the right base case, fix one k for the hypothesis - or a whole range, for strong induction - and derive the next case by actually using what was assumed. That repeating shape is the entire skill.
Explain it to yourself
Discussion prompt
In The induction setup checklist this move is made:
4. Write the inductive hypothesis
Why is that legal? Name the rule or definition it rests on before you read on.
Hint: If you can only say "because that is what you do", the rule is the thing to go and find.
Answer:
Fix one arbitrary k at least as large as the start, and assume the claim at k alone - or, for strong induction, the claim at every value from the start through k.
Pattern
1. Recognize the claim
Why: Is it indexed by a natural number, and claimed for every such number from some starting point onward? If yes, induction is very likely the right tool.
2. State the claim precisely
Why: Write the exact statement, in symbols, before doing anything else - every later part of the proof is just this statement applied at a different number.
3. Prove the base case directly
Why: Find the correct starting value by testing small numbers if needed, then verify the claim there by plain computation, no assumptions.
4. Write the inductive hypothesis
Why: Fix one arbitrary k at least as large as the start, and assume the claim at k alone - or, for strong induction, the claim at every value from the start through k.
5. Derive the next case, using the hypothesis
Why: The assumption must appear somewhere in the algebra or reasoning - not just the conclusion. This is where most real proofs are won or lost.
6. Verify the result
Why: Confirm the derived statement matches the target exactly, then sanity-check the whole formula on one concrete number.
Real world
Discussion prompt
Outside this lesson: where does Proof by Induction 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 The induction setup checklist 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:
A from-scratch guide to STARTING an induction proof, not just following one: recognizing when a claim needs induction, writing the base case, the inductive hypothesis, and the inductive step, and choosing weak versus strong induction. Worked proofs cover a summation identity, an inequality, a divisibility claim, a strong-induction postage argument, and two recursive-structure proofs (a recursive sum function and binary-tree leaf counts) connecting induction to recursive algorithms.
Picture it
Animation
Shows: A loop invariant is induction in disguise — a rendered Manim animation.
Rendered with Manim.
Takeaway: Which is why proving a loop correct feels so familiar.
Elimination
Eliminate the wrong options
Which of these four claims is the best candidate for a proof by mathematical induction?
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: A is a single statement P(n) indexed by a natural number n, claimed for every n at once - exactly the shape induction handles: prove it at the smallest n, then show truth at k forces truth at k+1. B and D are single fixed facts about one number, with no index n and nothing to step from k to k+1. C is indexed by a real number, and reals have no 'next' value, so there is no k+1 to induct to.
Check
Read each of the four statements below and decide which is best suited to a proof by mathematical induction.
Check your understanding
Which of these four claims is the best candidate for a proof by mathematical induction?
Answer: A
Why: A is a single statement P(n) indexed by a natural number n, claimed for every n at once - exactly the shape induction handles: prove it at the smallest n, then show truth at k forces truth at k+1. B and D are single fixed facts about one number, with no index n and nothing to step from k to k+1. C is indexed by a real number, and reals have no 'next' value, so there is no k+1 to induct to.
Prediction
Predict first
Which is the correctly stated inductive hypothesis for the step of this proof?
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: Fix one integer k >= 1 and assume 2^k > k just for that k.
Why: The inductive hypothesis assumes the statement holds at exactly one fixed, arbitrary case k - not for every n (that would be assuming the whole theorem), and not the k+1 case itself (that is the conclusion you must derive, not assume).
Check
You are inducting on the claim below.
\[ P(n):\quad 2^{n} > n, \quad n \ge 1 \]
Check your understanding
Which is the correctly stated inductive hypothesis for the step of this proof?
Answer: A
Why: The inductive hypothesis assumes the statement holds at exactly one fixed, arbitrary case k - not for every n (that would be assuming the whole theorem), and not the k+1 case itself (that is the conclusion you must derive, not assume).
Check
Recall the claim that three divides n cubed minus n. A student writes the following as their entire inductive step.
\[ (k+1)^{3}-(k+1) = k(k+1)(k+2), \ \text{a product of three consecutive integers, so one of them is divisible by 3} \]
Check your understanding
Does this write-up of the inductive step complete a valid proof BY INDUCTION of the claim above?
Answer: A
Why: The factoring and the three-consecutive-integers fact are correct, but an inductive step must derive the claim at k+1 from the assumed claim at k. This argument proves the fact a completely different way and never substitutes the hypothesis that k cubed minus k equals 3m, so it does not complete an induction proof.
Commit first
Predict first
By testing small values, what is the correct smallest starting value to use as the base case?
Commit to an answer, then rate it — certain, fairly sure, or guessing — and write the rating down before you turn the page.
Correct: n = 4
Why: Checking directly: n=1 gives 1 >= 7, false; n=2 gives 4 >= 10, false; n=3 gives 9 >= 13, false; n=4 gives 16 >= 16, true. So n=4 is the smallest value where the claim actually holds, making it the correct base case.
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
Consider the claim that n squared is at least three times n plus four, said to hold for n sufficiently large.
\[ n^{2} \ge 3n+4 \]
Check your understanding
By testing small values, what is the correct smallest starting value to use as the base case?
Answer: A
Why: Checking directly: n=1 gives 1 >= 7, false; n=2 gives 4 >= 10, false; n=3 gives 9 >= 13, false; n=4 gives 16 >= 16, true. So n=4 is the smallest value where the claim actually holds, making it the correct base case.
Check
A sequence is defined by three starting values, and then every later term is the sum of the three terms directly before it.
\[ a_1=a_2=a_3=1, \qquad a_n = a_{n-1}+a_{n-2}+a_{n-3}\ (n\ge 4) \]
Check your understanding
To prove a property holds for every term of this sequence, which method should you use?
Answer: A
Why: Since each term depends on the three previous terms, a step at n needs facts about several earlier terms, not just the one immediately before it. That is exactly the signal for strong induction, which assumes the property for every earlier case, not just the last one.
Prediction
Predict first
Which student's write-up is a valid, non-circular inductive step?
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: Student A starts from 1+2+...+k = k(k+1)/2 (the hypothesis), adds (k+1) to both sides, and simplifies to (k+1)(k+2)/2.
Why: Student A begins from the assumed hypothesis at k and derives the k+1 statement through valid algebra - that is a real inductive step. Student B instead writes down the very statement to be proved (the k+1 case) as a starting assumption, which is circular: it assumes the conclusion rather than deriving it.
Check
Two students each write an inductive step for the sum-formula claim from earlier in this deck.
\[ P(n):\quad 1+2+\cdots+n = \frac{n(n+1)}{2} \]
Check your understanding
Which student's write-up is a valid, non-circular inductive step?
Answer: A
Why: Student A begins from the assumed hypothesis at k and derives the k+1 statement through valid algebra - that is a real inductive step. Student B instead writes down the very statement to be proved (the k+1 case) as a starting assumption, which is circular: it assumes the conclusion rather than deriving it.
Elimination
Eliminate the wrong options
What is wrong with this way of stating the inductive hypothesis, if anything?
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: Stating the hypothesis for every integer n assumes the full universal claim before it has been proved - the very thing induction is supposed to establish. The correct hypothesis fixes one specific, arbitrary k and assumes the statement only there, then uses that single case to reach the next number.
Check
Recall the claim that three divides n cubed minus n for every n of at least one. A student writes their inductive hypothesis this way.
\[ \text{Hypothesis: } 3 \mid (n^{3}-n) \ \text{for every integer } n \ge 1 \]
Check your understanding
What is wrong with this way of stating the inductive hypothesis, if anything?
Answer: A
Why: Stating the hypothesis for every integer n assumes the full universal claim before it has been proved - the very thing induction is supposed to establish. The correct hypothesis fixes one specific, arbitrary k and assumes the statement only there, then uses that single case to reach the next number.
Concept
Moves added today:
Moves you reused today:
Notice that move #2 came back today, in disguise: choosing the base case is choosing where the claim first becomes true, the same instinct as picking the n that breaks a bound.
Full toolkit so far: #1 through #6.
Next session opens with you naming every one of these from memory, before any new material.
Counterexample
Discussion prompt
Notice that move #2 came back today, in disguise: choosing the base case is choosing where the claim first becomes true, the same instinct as picking the n that breaks a bound.
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:
Next session opens with you naming every one of these from memory, before any new material.
Picture it
Animation
Shows: The classic broken induction — a rendered Manim animation.
Rendered with Manim.
Takeaway: The step must hold for EVERY k, including the smallest.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The induction setup checklist · Toolkit check-in: name them before you look · The tell: a claim indexed by a natural number n · One claim, infinitely many statements · Testing examples is not a proof. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
You can now START an induction proof, not just follow one. The setup decisions are the whole skill:
| Part | What it must do |
|---|---|
| Base case | Prove the claim directly at the correct starting value, no assumptions |
| Inductive hypothesis | Fix one arbitrary k (or a full range, for strong induction) and assume the claim there only |
| Inductive step | Derive the claim at the next number, using the hypothesis somewhere in the argument |
| Verify | Confirm the derived line matches the target exactly, then sanity-check one concrete number |
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