February 24, 2017

Homework Help: Abstract Algebra

Posted by Dina on Friday, September 21, 2007 at 2:00pm.

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^2-2n-1>0, and if I add 2 then n^2-2n+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.

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