Proof the induction 2^n > n+4, n>=3
pka Elite Member Joined Jan 29, 2005 Messages 11,991 May 16, 2020 #2 @arz, can you tell us what steps are require in order to prove a statement is true by induction? If so then post what you know.
@arz, can you tell us what steps are require in order to prove a statement is true by induction? If so then post what you know.
Steven G Elite Member Joined Dec 30, 2014 Messages 14,598 May 16, 2020 #3 n=3: Is 2^n > n+4? 2^3 =8 and 3+4 =7 so yes! Now you assume something is true for n=k. What do you assume and what is the restriction on k? Now you show that something is true for n=k+1. What is it that you want to show?
n=3: Is 2^n > n+4? 2^3 =8 and 3+4 =7 so yes! Now you assume something is true for n=k. What do you assume and what is the restriction on k? Now you show that something is true for n=k+1. What is it that you want to show?
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 May 18, 2020 #4 Assume that, for some k, 2k>k+4\displaystyle 2^k> k+ 42k>k+4. Then 2k+1=2(2k)>2(k+4)=2k+8\displaystyle 2^{k+1}= 2(2^k)> 2(k+ 4)= 2k+ 82k+1=2(2k)>2(k+4)=2k+8. Now you need to show that 2k+8>(k+1)+4=k+5\displaystyle 2k+ 8> (k+1)+4= k+ 52k+8>(k+1)+4=k+5. That looks very easy.
Assume that, for some k, 2k>k+4\displaystyle 2^k> k+ 42k>k+4. Then 2k+1=2(2k)>2(k+4)=2k+8\displaystyle 2^{k+1}= 2(2^k)> 2(k+ 4)= 2k+ 82k+1=2(2k)>2(k+4)=2k+8. Now you need to show that 2k+8>(k+1)+4=k+5\displaystyle 2k+ 8> (k+1)+4= k+ 52k+8>(k+1)+4=k+5. That looks very easy.