Overview

Induction is a strategy (rather than a theorem) that gets a reccursion on a particular cases and automatically prove the statement.

Definition

As you might know, induction has two parts: base case and inductive case. Let P(n)P(n) be a statement where nn is a positive integer.

  • The base case, where you show that P(1)P(1) is true. This part is often trivial, but don't forget to write it down. I recommend you to write it in the very first of the proof.
  • Now for the inductive case, assume that P(n)P(n) is true, and show that P(n+1)P(n+1) is also true. This will complete the problem.

Thus, it can be written as

P(0)  ⟹  P(1)  ⟹  P(2)  ⟹  …P(0)\implies P(1)\implies P(2)\implies \dots

showing that P(n)P(n) is true in general.

Example 1: the classic

We only outline the solution, and leave it as an exercise for the readers.

problem: Show that for any positive integer nn, the following equality holds:

P(n) ⁣:(1+2+⋯+n)2=13+23+⋯+n3.P(n)\colon (1+2+\dots +n)^2=1^3+2^3+\dots +n^3.

We proceed via induction. The base case is trivial as n=1n=1.

Now for the inductive step, assume that for n=kn=k the equality holds. Now take the difference of P(k)P(k) and P(k+1)P(k+1) on both sides, as it looks like a lot of things cancel on the RHS. We have

((k+1)(k+2)2)2−(k(k+1)2)2=(k+1)3\left(\frac{(k+1)(k+2)}{2}\right)^2-\left(\frac{k(k+1)}{2}\right)^2=(k+1)^3

and the rest is an easy calculation.

Advanced induction

As some might notice, induction is not only for n↦n+1n\mapsto n+1 yet! We will state a few example.

  • One of the famous one is strong induction, where you assume that all of the P(1)P(1), P(2)P(2), …\dots, P(n)P(n) is true, instead of P(n)P(n) alone. (Actually you can assume this anytime when you induct).
  • Maybe for some integer hh the base case is P(1)P(1), P(2)P(2), …\dots, P(h)P(h) and the inductive step is P(n)↦P(n+h)P(n)\mapsto P(n+h).
  • Induction with prime factors.
  • Sometimes we induct with an interval; Let P(I)P(I) be a statement where II is some interval. With the base case I=[0,M)I=[0, M) and inductive step P([kM,(k+1)M))  ⟹  P([(k+1)M,(k+2)M))P([kM, (k+1)M))\implies P([(k+1)M, (k+2)M)) then you can prove over nonnegative reals that P(I)P(I) is true.
  • It might get a reccursion with multiple variables. This is called dynamic programming in computer science.

Example 2: Induction on NT

TODO

Example 3: Induction on integer FE

Induction is especially common in integer FE as well, as it should be not too surprising.

When should I induct?

So when to use induction? Some advice is that

  • If you can get a nice reccursion when lot of things cancel then you'll have fun with induction. It makes a lot easier to work with.
  • It might be useful to use induction if a lot of things preserved while the inductive case, even if doesn't actually cancel out.

More advice on induction:

  • Keep in mind that induction is just a strategy. It's a basic ``idea'' rather than a theorem, and it's really useful in many problems.
  • Be careful of the base case! For example, it might be possible but you proved P(1)P(1) is true and P(n)P(n) is true where n≥3n\ge 3 by induction, but missing the case n=2n=2. I recommend you to confirm until about first six cases if it's easy enough.
  • Think about which varients are fixed and which do you induct with. This is sometimes not so clear.

Practice Problems

StatusSourceProblem NameDifficultyTags
Canadaeasy
Show TagsInduction
IMOmedium
Show TagsInduction
USAMOhard
Show TagsInduction
Berkeley Math Circle Take-Home Contest #1Hard
Show TagsDivisibility / Factorization, Induction / smoothing
Berkeley Math Circle Take-Home ContestHard
Show TagsInduction / smoothing, Linear and quadratic inequalities, Vectors
Berkeley Math CircleHard
Show TagsFunctional Equations, Induction / smoothing
Berkeley Math CircleHard
Show TagsAlgebraic properties of binomial coefficients, Induction / smoothing, Modular Arithmetic
Berkeley Math Circle Take-Home Contest #6Hard
Show TagsInduction / smoothing, Polynomial operations, Sums and products, Telescoping series
Berkeley Math Circle Monthly Contest 3Hard
Show TagsConstructions and loci, Induction / smoothing
Berkeley Math Circle Monthly Contest 1Hard
Show TagsInduction / smoothing, Recursion, bijection
Berkeley Math Circle Monthly Contest 4Hard
Show TagsFactorization techniques, Induction / smoothing
Berkeley Math Circle Take-Home Contest #6Hard
Show TagsGreatest common divisors (gcd), Induction / smoothing, Polynomial operations, Recurrence relations
MathNetHard
Show TagsColoring schemes, extremal arguments, Induction / smoothing
Berkeley Math Circle Monthly Contest 8Hard
Show TagsInduction / smoothing, Modular Arithmetic, Recurrence relations, Recursion, bijection
Berkeley Math CircleHard
Show TagsColoring schemes, extremal arguments, Induction / smoothing
Berkeley Math Circle Monthly Contest 6Hard
Show TagsInduction / smoothing, Recurrence relations, Sums and products
Berkeley Math Circle Monthly Contest 1Hard
Show TagsFunctional Equations, Induction / smoothing, Injectivity / surjectivity
Berkeley Math Circle: Monthly Contest 7Hard
Show TagsInduction / smoothing, Integers
Berkeley Math Circle Monthly Contest 2Hard
Show TagsInduction / smoothing, Techniques: modulo, size analysis, order analysis, inequalities
Berkeley Math Circle Monthly Contest 5Hard
Show TagsInduction / smoothing, Polynomial operations
Berkeley Math CircleHard
Show TagsAlgebraic properties of binomial coefficients, Coloring schemes, extremal arguments, Induction / smoothing, Invariants / monovariants
Berkeley Math CircleHard
Show TagsColoring schemes, extremal arguments, Induction / smoothing
Berkeley Math CircleHard
Show TagsColoring schemes, extremal arguments, Induction / smoothing, Invariants / monovariants
Berkeley Math Circle Monthly Contest 4Hard
Show TagsFunctional Equations, Induction / smoothing
Berkeley Math Circle: Monthly Contest 2Hard
Show TagsColoring schemes, extremal arguments, Expected values, Induction / smoothing, Sums and products

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