Question

I am having alot of trouble making sense of this proof from my professor. http://www.math.oregonstate.edu/~kovchegy/math231/231_lecture11.pdf
Can someone please look at page 5. n^2>n+1

Answer 1

Also, here is my proof that I feel is much more clear.

Answer 2

it looks right to me

Answer 3

His or mine?

Answer 4

his... not yours

Answer 5

Can you elaborate.

Answer 6

We don't know that \((k+1)^2>k+2\) .

we have to prove that it is, assuming that \(k^2>k+1\) for a certain number k.

Answer 7

ok. I am having trouble making sense of his algrebra where it seems (k+1)^2 become 3k+2

Answer 8

We start with \(k^2>k+1\) .... which we assume to be true ...

then we ad \(2k+1\) to both sides

Answer 9

so then we would have k^2 +2k +1>3k+2 right?

Answer 10

Assume that \(k^2>k+1\), we have \[(k+1)^2 = k
^2+2k+1\\~\\k
^2+\mathbf{2k+1 }> (k+1)+\mathbf{2k+1 }\]Because we assumed that \(k^2>k+1\)

and \((k+1)+2k+1 = 3k+2 > k+2\)

So \((k+1)^2>k+2\)

Answer 11

right, and since \(3k+2>k+2\) for any positive number k

... we can conclude: \((k+1)^2>k+2\)

if \(a>b\) and \(b>c\), then \(a>c\)