Let S be a set with n elements. Show:

If an (n is subscript) equals the number of subsets of S then an+1 (n+1 is subscript) = 2an (n is subscript)

Use this to prove by induction that an (n is subscript) = 2^n

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- Jun 20th 2010, 11:56 AMhuntstyClueless on this proof, please help!!!
Let S be a set with n elements. Show:

If an (n is subscript) equals the number of subsets of S then an+1 (n+1 is subscript) = 2an (n is subscript)

Use this to prove by induction that an (n is subscript) = 2^n - Jun 20th 2010, 12:24 PMPlato
What you want to show that if $\displaystyle 2^n$ is the number of subsets of $\displaystyle \{1,2,3,\cdots,n\}$;

then $\displaystyle 2^{n+1}$ is the number of subsets of $\displaystyle \{1,2,3,\cdots,n,n+1\}$.

Think $\displaystyle \{1,2,3,\cdots,n,n+1\}=\{1,2,3,\cdots,n\}\cup\{n+1 \}$.

Any subset of $\displaystyle \{1,2,3,\cdots,n\}$ is also a subset of $\displaystyle \{1,2,3,\cdots,n+1\}$.

Take anyone of those and add $\displaystyle n+1$ to it to get a different subset of $\displaystyle \{1,2,3,\cdots,n+1\}$. - Jun 20th 2010, 12:53 PMhuntsty
thank you, very helpful for understanding. I just don't know how to symbolically show this. Will i need to introduce a new random variable in my proof?

- Jun 20th 2010, 01:03 PMPlato
- Jun 21st 2010, 11:02 AMHallsofIvy
If S contains n+ 1 members, choose one of them and call it "a". Removing that from S leaves you with set T that has n members and so, a(n) members.

Now note that every member of S either contains "a" or it doesn't. If it doesn't, it is one of the a(n) subsets of T. If it does, then it is one of the subsets of T with "a" added- there are still a(n) such subsets. Together there are a(n)+ a(n)= 2a(n) subsets of S.