Mathematical Induction

Algebra 2 Tutorial

Mathematical Induction


Mathematical induction is a form of proof that takes place in two parts.
It is surprisingly useful.

Part 1) Prove that the statement is true for the number 1.
Part 2) Prove that if it is true for any number n then it is also true for the number n+1.

Think about it… we first show what we want to prove is true for 1.
Using step 2 we know if it’s true for 1 then it is true for 2.
If it is true for 2 then it is true for 3.
If it is true for 3 then it is true for 4.
Et cetera, et cetera, et cetera

The logic goes on forever, so what we want to prove is true for ANY integer.

Sample Problem

Let’s prove that 1 + 2 + 3 ... n = \frac{n(n+1)}{2}}


The first part is simple… n=1
Is it true that 1=\frac{n(n+1)}{2}}=\frac{1*2}{2}}… Of course it is!

Part 2 is the tricky part.
We take, as an assumption, that
1 + 2 + 3 ... n = \frac{n(n+1)}{2}}

And we show that if this is true then
1 + 2 + 3 ... n + (n+1) = \frac{(n+1)(n+2)}{2}}

1 + 2 + 3 ... n = \frac{n(n+1)}{2}} …. our assumption
1 + 2 + 3 ... n + (n+1)= \frac{n(n+1)}{2}} + (n+1) … add (n+1) to both sides

1 + 2 + 3 ... n + (n+1)= \frac{n(n+1)+2(n+1)}{2}}… multiply numerator and denominator of (n+1) by 2

Now notice that the two terms on the right hand side each have a common factor of (n+1). We can factor that out and get what we want.

1 + 2 + 3 ... n + (n+1)= \frac{(n+1)(n+2)}{2}}

Our proof is complete. We have show that our equation is true when n=1 and we have shown that if it is true for any value n, then it is also true for n+1.

That is to say, our equation is true for all integers.

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