# Mathematical Induction

Mathematical Induction is a special way of proving things. It has only 2 steps:

- Step 1. Show it is true for the
**first one** - Step 2. Show that if
**any one**is true then the**next one**is true

Then **all** will be true

Have you heard of the "Domino Effect"? - Step 1. The first domino falls
- Step 2. If any domino falls, the next domino falls
So ... That is how it works. |

In the world of numbers we say:

- Step 1. Show it is true for
**n=1** - Step 2. Show that if
**n=k**is true then**n=k+1**is also true

## How to Do it

Step 1 is usually easy, you just have to prove it is true for **n=1**

Step 2 is best done this way:

**Assume**it is true for**n=k****Prove**it is true for**n=k+1**(you can use the**n=k**case as a**fact**.)

Step 2 can often be * tricky* ... because you may need to use imaginative tricks to make it work!

Like in this example:

### Example: 3^{n}-1 is a multiple of 2

**Is that true? Let us find out.**

**1.** Show it is true for **n=1**

**3 ^{1}-1 = 3-1 = 2 **

Yes 2 is a multiple of 2. That was easy.

3^{1}-1 is true

**2.** Assume it is true for **n=k**

3^{k}-1 is true

(Hang on! How do we know that?
We don't!

It is an **assumption** ... that we treat

**as a fact** for the rest of this example)

Now, prove that **3 ^{k+1}-1** is a multiple of 2

3×3^{k}
And then split
And each of these are multiples of 2 |

Because:

**2·3**is a multiple of 2 (you are multiplying by 2)^{k}**3**(we said that in the assumption above)^{k}-1 is true

So:

3^{k+1}-1 is true

DONE!

Did you see how we used the **3 ^{k}-1** case as being

**true**, even though we had not proved it? That is OK, because we are relying on the

**Domino Effect**...

... we are asking **if** any domino falls will the **next one** fall?

So we take it as a fact (temporarily) that the "**n=k**" domino falls (i.e. **3 ^{k}-1** is true), and see if that means the "

**n=k+1**" domino will also fall.

## Tricks

I said before that you often need to use imaginative tricks.

A common trick is to rewrite the **n=k+1** case into 2 parts:

- one part being the
**n=k**case (which is assumed to be true) - the other part can then be checked to see if it is also true

We did that in the example above, and here is another one:

### Example: Adding up Odd Numbers

1 + 3 + 5 + ... + (2n-1) = n^{2}

**1.** Show it is true for **n=1**

1 = 1^{2} is True

**2.** Assume it is true for **n=k**

1 + 3 + 5 + ... + (2k-1) = k^{2} is True

Now, prove it is true for "k+1"

1 + 3 + 5 + ... + (2k-1) + (2(k+1)-1) = (k+1)^{2} ... ?

We know that **1 + 3 + 5 + ... + (2k-1) = k ^{2}** (the assumption above), so we can do a replacement for all but the last term:

**k ^{2}** + (2(k+1)-1) = (k+1)

^{2}

Now expand all terms:

k^{2} + 2k + 2 - 1 = k^{2} + 2k+1

And simplify:

k^{2} + 2k + 1 = k^{2} + 2k + 1

**They are the same! So it is true.**

So:

1 + 3 + 5 + ... + (2(k+1)-1) = (k+1)^{2} is True

DONE!

So there you have it!