# Proof: Number of different subsets of A is equal to 2^n?

• SMA_01
In summary, the conversation discusses proving that the number of subsets of a set with n elements is equal to 2n using mathematical induction. The initial proposition is that a set with a single element has two subsets, and the induction step involves showing that adding another element to a set with k elements results in 2k more subsets. The conversation also touches on the use of binomial coefficients in this proof.

## Homework Statement

Prove that if a set a contains n elements, then the number of different subsets of A is equal to 2n.

## The Attempt at a Solution

I know how to prove with just combinatorics, where to construct a subset, each element is either in the set or not, leading to 2n subsets. I want to know how to prove it with mathematical induction though. How would I start?

I figured this using summation notation:
$\sum^{k=0}_{n}$ (n k)=2n

SMA_01 said:

## Homework Statement

Prove that if a set a contains n elements, then the number of different subsets of A is equal to 2n.

## The Attempt at a Solution

I know how to prove with just combinatorics, where to construct a subset, each element is either in the set or not, leading to 2n subsets. I want to know how to prove it with mathematical induction though. How would I start?

I figured this using summation notation:
$\sum^{k=0}_{n}$ (n k)=2n

So the proposition you want to prove is that the number of subsets of a set with n elements is ##2^n##. Start with ##n=1## A set with a single element, call it ##S_1 = \{a\}## has two subsets: ##\{a\}, \Phi## which is ##2^1##. Now you need to prove the induction step: Assume a set with ##k## elements has ##2^k## subsets and use that to show a set with ##k+1## elements has ##2^{k+1}## subsets. That's how you start. It's easy and you don't need binomial coefficients to do it.

Thank you. The part with k+1 elements is confusing me. I went through a long process of trying to make one side equal to the other, and it's not really working.

If you start with a set with ##k## elements and ##2^k##subsets, and add another element, all the subsets of the original set are subsets of the larger set. What additional subsets are there?

The additional subsets will be the ones formed using the new element? As if this new element is either in the subsets or not...

LCKurtz said:
If you start with a set with ##k## elements and ##2^k##subsets, and add another element, all the subsets of the original set are subsets of the larger set. What additional subsets are there?

SMA_01 said:
The additional subsets will be the ones formed using the new element? As if this new element is either in the subsets or not...

Yes, so...?

So, 2*2^k=2^(k+1), is that right?

SMA_01 said:
So, 2*2^k=2^(k+1), is that right?

Are you asking me if ##2\cdot 2^k = 2^{k+1}##? Of course, you know it does. Do you understand how to put this all together to make a well written induction proof of your theorem?

Yes, thank you. I just thought I would have to go through the long, tedious process of proving both sides are equal using the summation formula. Thanks for all your help