Abstract Algebra
posted by Dina .
If 2^n>n^2 and n>5, then 2^n+1>(n+1)^2
Proof: Assuming that 2^n>n^2 then I can say that 2*2^2>2*n^2 = 2^n+1>2n^2
If I can show that 2n^2>(n+1)^2 then I will be done by transitivity. So 2n^2>(n+1)^2? then 2n^2>n^2+2n+1? then n^2>2n+1, hence n^22n1>0, and if I add 2 then n^22n+1>2... after this I do not know what to do. If you can see that I did something wrong somewhere along the proof please let me know.

Proof: Assuming that 2^n>n^2 then I can say that 2*2^n>2*n^2 = 2^n+1>2n^2
yes, correct
If I can show that 2n^2>(n+1)^2 then I will be done by transitivity.
So 2n^2>(n+1)^2?
then 2n^2>n^2+2n+1?
then n^2>2n+1, now, n>5, so n= 5+m, where m is any positive integer.
n^2 >2n+1
(5+m)^2>2(5+m)+1
25+2m + m^2>10+2m+1
m^2>14, and since m is any postive integer, it is proved.
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