Mathematical induction
Mathematical Induction (MI) is an extremely important tool in Mathematics.
First of all you should never confuse MI with Inductive Attitude in Science. The
latter is just a process of establishing general principles from particular cases.
MI is a way of proving math statements for all integers (perhaps excluding a
finite number.) [1] says:
Statements proven by math induction all depend on an integer, say, n. For example,
(1) 1 + 3 + 5 + ... + (2n-1) = n2
(2) If x1, x2, ..., xn > 0 then (x1 + x2 + ... + xn)/n (x1·x2·...·xn)1/n
etc. n here is an "arbitrary" integer.
It's convenient to talk about a statement P(n). For (1), P(1) says that 1 = 12 which is incidently true. P(2) says that 1 + 3 = 22, P(3) means that 1 + 3 + 5 = 32. And so on. These particular cases are obtained by substituting specific values 1, 2, 3 for n into P(n).
Assume you want to prove that for some statement P, P(n) is true for all n starting with n = 1. The Principle (or Axiom) of Math Induction states that, to this end, one should accomplish just two steps:
- Prove that P(1) is true.
- Assume that P(k) is true for some k. Derive from here that P(k+1) is also true.
The idea of MI is that a finite number of steps may be needed to prove an infinite number of statements P(1), P(2), P(3), ....
Let's prove (1). We already saw that P(1) is true. Assume that, for an arbitrary k, P(k) is also
true, i.e. 1 + 3 + ... + (2k-1) = k2. Let's derive P(k+1) from this assumption. We have
|
1 + 3 + ... + (2k-1) + (2k+1) | = [1 + 3 + ... + (2k-1)] + (2k+1) |
| | = k2 + (2k+1) |
| | = (k+1)2 |
Which exactly means that P(k+1) holds. (For 2k+1 = 2(k+1)-1.) Therefore, P(n) is true for all n starting with 1.
Intuitively, the inductive (second) step allows one to say, look P(1) is true and implies P(2). Therefore P(2) is true. But P(2) implies P(3). Therefore P(3) is true which implies P(4) and so on. Math induction is just a shortcut that
collapses an infinite number of such steps into the two above.
In Science, inductive attitude would be to check a few first statements, say, P(1), P(2), P(3), P(4),
and then assert that P(n) holds for all n. The inductive step "P(k) implies P(k+1)" is missing. Needless to say nothing can be proved this way.
Remark
- Often it's impractical to start with n = 1. MI applies with any starting integer n0.
The result is then proven for all n from n0 on.
- Sometimes, instead of 2., one assumes 2':
Assume that P(m) is true for all m < (k+1).
Derive from here that P(k+1) is also true. The two approaches are equivalent, because one may consnider a statement Q: Q(n) = P(1) and P(2) and ... and P(n), so that Q(n) is true iff P(1), P(2), ..., P(n) are all true.
There are other examples proven by MI:
- A 1-1 correspondence
- An Extension of van Schooten's Theorem
- An infinite exponent
- Another pigeonhole problem
- A Problem of Divisibility by 5n
- Binary Euclid's algorithm
- Book Index Range
- Breaking Chocolate Bars
- Committee Chairs
- Constructible Numbers
- Construction Problem
- Continued Fractions
- Counting Triangles
- Counting Triangles II
- Cutting Squares
- Diagonal Count
- Difference of the Cantor Sets
- Euclid's Algorithm
- Farey series
- Fractions on a Binary Tree II
- Geometric Illustration of a Convergent Series
- Golomb's inductive proof of a tromino theorem
- Groups of Permutations
- Guessing Two Consecutive Integers
- Hamming and Levenshtein distance functions
- Inequality 1/2·3/4·5/6· ... ·99/100 < 1/10
- Inequality (1 + 1-3)(1 + 2-3)(1 + 2-3)...(1 + n-3) < 3
- Inequality 1 + 2-2 + 3-2 + 4-2 + ... + n-2 < 2
- Inequality between arithmetic and geometric means
- Infinite Latin Squares
- Infinitude of Primes Via Fermat Numbers
- Infinitude of Primes Via *-Sets
- Integers and Rectangles: a Proof by Induction
- Integral Domains: Remarks and Examples
- Josephus problem
- Linear Functions
- Marriage Problem
- Mathematical Induction
- Mathematicians Doze Off
- Morley's Pursuit of Incidence
- Pennies in Boxes
- Pigeonhole problem
- Pigeonhole principle complements math induction
- Poncelet Theorem
- Prim's and Kruskal's algorithms find a minimum spanning tree
- Problem on an Infinite Checkerboard
- Property of irriducible fractions on the Stern-Brocot tree
- Property of the Powers of 2
- Sierpinski Gasket
- Sierpinski Gasket and Tower of Hanoi
- Simple Cellular Automaton
- Solitaire on the Circle
- Splitting piles
- Stern-Brocot Tree
- Stern-Brocot Tree II
- Strange Integers: Divisors and Primes
- The Somos sequences
- Ways To Count
Reference
- D. Fomin, S. Genkin, I. Itenberg, Mathematical Circles (Russian Experience), AMS, 1996
- R. Graham, D. Knuth, O. Patashnik, Concrete Mathematics, 2nd edition, Addison-Wesley, 1994.
- R.Courant and H.Robbins, What is Mathematics?, Oxford University Press, 1996
On the Web
- An online and iPod video by Julio de la Yncera

Copyright © 1996-2008 Alexander Bogomolny
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