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prove by induction that 2^n is less or equal to n!
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a set of positive numbers whose square is 25
what?
mmmm im not sure but i wrote a test about it eash it dealt with me
if n = 1 then 2 < = 2!=2. If we have for n = k that 2^k <= k!. Then for n = k+1 we will have 2^(k+1) = (2^k)*2 <=k!*(k+1) = k!. what we wanted.
true!!!!
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basis step: for n>=4 , 2^n<=n! Inductive step: assume that if k=4, p(k) is true. {2^4<=4!} for k>=0, we show that if 2^k<=k!, then 2^(k+1)<=(n+1)! 2^(k+1) = 2*2^k <=2*k! <(k+1)k! = (k+1)!
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