1. Prove something stronger: 2^n > n (for some natural k n>k)
or: check to n=-3,-2,-1,0,1 and then you will see that: 2^n is natural number.
2. (3/2)^n >= (1+ n/2) n>=1
or:
(3/2)^n >={(n+2)/2} so...
Hello, I am having some trouble with the following proofs by induction. I understand how to get the base case and induction hypothesis but its the actual induction that's bothering me. Sorry that it looks ugly but this is my first post and I haven't been able to figure out the formatting codes yet.
1.) 2^n >= 1/8 n >= -3
2.) (3/2)^n >= (1+ n/2) n>=1
for 1 tried 2^(n+1) = 2*2^n >= 2*(1/8) > 1/8
This however does not preserve the >=, just >. This makes sense to me on an instinctual level because n>= -3 means that n+1 will always be greater than 1/8
for 2 I tried 1+(n+1)/2 = 1 + n/2 + 1/2 <= (3/2)^n + 1/2 < (3/2)^(n+1) but again this did not preserve <= but just <.
I'm afraid I don't see where you are coming from in both cases. I do not see how to connect 2^n >= n to 2^n >= 1/8 for n >=-3. Do you mean to check n from -3 to 0 and prove 2^n > n for n >=1? Similarly I don't see what re-expressing 1 + n/2 as (n+2)/2 gets you
That is not a concern.
The "equal" for question 1 occurs only when n is -3.
The "greater than" occurs for all n above -3.
Hence n+1 is above -3 so you need not be concerned about losing the "equal" part.
Same for the second question.
You can do the proof as ">" for n>-3 and so on.