Mathematical Induction

Mathematical Induction is a method of proof typically used to establish that a given statement is true for all natural numbers. This is often the first type of proof statement seen by a high school student.

First Principle of Finite Induction
Let SS be a set of positive integers with the following properties:
(1) The integer 1 belongs to the set.
(2) Whenever the integer kk is in SS, then the next integer k+1k+1 must also be in SS.
Then SS is the set of all positive integers.

This statement seems almost immediately obvious, and it can be proved by contradiction. Often, an analogy is drawn with a sequence of falling dominos; if the first domino falls, and each subsequent domino pushes it's neighbor, then all dominos in this sequence must eventually fall. For now, we will focus on how to prove statements by induction. These proofs often have a standard format that can be followed, as shown below:

Statement: Let PnP_{n} be the proposition Induction Hypothesis, for nn in Domain.

Base Case: Consider Base Case.
LHS=LHS
RHS=RHS
Since LHS=RHS, thus Base Case is true.

Induction step: Assume PkP_{k} is true for some kk in Domain. Consider Pk+1P_{k+1}.
LHS Pk+1P_{k+1} =use Induction Hypothesis =RHS Pk+1P_{k+1}.
Hence PkP_{k} is true implies that Pk+1P_{k+1} is true.

Conclusion: By Mathematical Induction, since Base Case is true and PkP_{k} is true implies Pk+1P_{k+1} is true, thus PnP_{n} is true for all nn in Domain. \square

There are 4 main components in the proof, and you have to fill in the details of the various parts left in italics. Let us work through a specific example, to get a better sense of how this works.

Worked Examples

1. Show that for all positive integers nn,

i=1ni2=n(n+1)(2n+1)6. \sum_{i=1} ^n i^2 = \frac {n(n+1)(2n+1)} {6}.

Solution: This is a well known statement about the sum of the first nn squares, and we will show this via induction.

Statement: Let PnP_{n} be the proposition that i=1ni2=n(n+1)(2n+1)6 \sum_{i=1} ^n i^2 = \frac {n(n+1)(2n+1)} {6}, for all positive integers.

[We have stated what the Induction Hypothesis is, and specified the domain in which the statement is true.]

Base Case: Consider n=1n=1.
LHS= i=1112=1 \sum_{i=1} ^ 1 1^2 = 1 .
RHS= 1×(1+1)×(2+1)6=1 \frac{ 1 \times (1+1) \times (2 + 1) } {6} = 1 .
Since LHS=RHS, thus the Base Case is true.

[We have identified the Base Case, and shown that it is true.]

Induction Hypothesis: Assume PkP_k is true for some positive integer kk. Consider Pk+1P_{k+1} .
LHS Pk+1=i=1k+1i2=(i=1ki2)+(k+1)2=k(k+1)(2k+1)6+(k+1)2=(k+1)(2k2+k6+6(k+1)6)=(k+1)2k2+7k+66=(k+1)(k+2)(2k+3)6=RHS Pk+1\begin{aligned} \mbox{LHS } P_{k+1} & = \sum_{i=1}^{k+1} i^2 \\ & = \left( \sum_{i=1}^k i^2 \right) + (k+1) ^2 \\ & = \frac{ k(k+1)(2k+1) } { 6} + (k+1) ^2 \\ & = (k+1) \left( \frac{ 2k^2 + k } { 6} + \frac{ 6 (k+1) } { 6} \right) \\ & = (k+1) \frac{ 2k^2 + 7k + 6 } { 6} \\ & = \frac { (k+1) (k+2) ( 2k + 3) }{6} \\ & = \mbox{RHS } P_{k+1} \\ \end{aligned} Hence PkP_{k} is true implies that Pk+1P_{k+1} is true.

[We successfully used the induction hypothesis (in the 3rd equation) along with algebraic manipulation to show that the next statement is true.]

Conclusion: By Mathematical Induction, since P1P_1 is true and PkP_{k} is true implies Pk+1P_{k+1} is true, thus PnP_{n} is true for all positive integer nn. \square

[We now sit back and drink a cup of coffee.]

 

2. Show that for all positive integers nn, 32n+1+40n33^{2n+1}+40n-3 is a multiple of 64.

Solution: Statement: Let PnP_n be the proposition that 32n+1+40n33^{2n+1}+40n-3 is a multiple of 64, for all positive integers nn.

Base Case: Consider n=1n = 1 .
Since 33+403=64 3^{3} + 40 - 3 = 64 , it is a multiple of 64.
Hence, the Base case is true.

Induction Hypothesis: Assume PkP_k is true for some positive integer kk. Consider Pk+1P_{k+1} .
32k+3+40(k+1)3=9×(32k+1+40k3)320k+64 3^{2k+3} + 40(k+1) - 3 = 9 \times ( 3^{2k+1} + 40k - 3 ) -320k + 64 . Since 320k+64=64(5k+1) -320k+ 64 = 64 ( -5k + 1) is a multiple of 64, and the first term is a multiple of 64 by the Induction Hypothesis, hence the LHS is a multiple of 6464.
Hence PkP_{k} is true implies that Pk+1P_{k+1} is true.

Conclusion: By Mathematical Induction, since P(1)P(1) is true and PkP_{k} is true implies Pk+1P_{k+1} is true, thus PnP_{n} is true for all positive integer nn. \square

Note by Calvin Lin
5 years, 8 months ago

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Nice note!BTW how to add these white boxes or blockquotes in notes?

Calvin Edit. You just add a ">" at the start. thanks

Gautam Sharma - 4 years, 8 months ago

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Simply add a ">" to the start of the paragraph. I've edited your comment so you can use that as a reference. If you click on "Edit", you can see how it's done.

Calvin Lin Staff - 4 years, 8 months ago

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Thanks Sir! This is my first introduction to induction.

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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For more information, check out the Induction wikis

Calvin Lin Staff - 4 years, 8 months ago

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Going through them, Sir.

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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Sir, could I copy content from this Note to show the format of a proof by Induction here?

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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@Ishan Dasgupta Samarendra yes i think you can as you are filling a wiki.Good luck

Gautam Sharma - 4 years, 8 months ago

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@Gautam Sharma Did you made a new wiki or it existed previously? If you made a new one, then i have to say there is one here which contains your topic i think.

Gautam Sharma - 4 years, 8 months ago

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@Gautam Sharma Didn't understand. As in the topic was there as part of the constituent wikis under Induction, but it is empty.

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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@Ishan Dasgupta Samarendra So it existed previously then go ahead and complete your wiki. You can also write on Proof of mathematical induction wiki.

Gautam Sharma - 4 years, 8 months ago

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You didn't study it in 11?

Gautam Sharma - 4 years, 8 months ago

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I'll be going to class 11 this April.

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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@Ishan Dasgupta Samarendra Oh sorry i thought you were in 11 as your age shown here is 16.

Gautam Sharma - 4 years, 8 months ago

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@Gautam Sharma I am 16. But I grew up in Calcutta till Class 3 and there, when one is 17, one is in Class 11. When I came to Delhi, I obviously could not skip Class 4 and directly go to Class 5.

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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@Ishan Dasgupta Samarendra Haaha you will be surprised to know I skipped class 6 .I was admitted in school 1 yr later than other kids but i skipped 6 hence we were at same stage again.

Gautam Sharma - 4 years, 8 months ago

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@Gautam Sharma Wow!

Ishan Dasgupta Samarendra - 4 years, 8 months ago

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