How can I prove by induction that:

for all n>=4

I do:

n=k

P(k) = k^2 > 3k

P(k+1) = (k+1)^2 > 3(k+1)

How to solve

P(k) --> P(k+1) to prove that P(n) is truth ???

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- August 25th 2010, 03:30 PMApprentice123Induction
How can I prove by induction that:

for all n>=4

I do:

n=k

P(k) = k^2 > 3k

P(k+1) = (k+1)^2 > 3(k+1)

How to solve

P(k) --> P(k+1) to prove that P(n) is truth ??? - August 25th 2010, 03:38 PMDefunkt

Also, writing is syntactically wrong -- is a proposition, not a number. - August 25th 2010, 04:03 PMApprentice123
Why ??

- August 25th 2010, 04:17 PMDefunkt
Use the induction hypothesis -

- August 25th 2010, 04:33 PMApprentice123
OK. But I not understand why

K^2 > 3k

(k+1)^2 > 3(k+1) ---Why you used 3k+2k+1 ??? - August 25th 2010, 04:43 PMDefunkt

- August 25th 2010, 04:50 PMApprentice123
I not understand why 3k+2k+1 if 3(k+1) = 3k+3 not 3k+2k+1

- August 25th 2010, 05:10 PMDefunkt
It's an intermediate step..

Do you understand why ?

If yes, then my argument is

. - August 25th 2010, 05:39 PMApprentice123
I know that k^2+2k+1 > 3k+2k+1

From where did 3k+2k+1 ? - August 25th 2010, 05:43 PMDefunkt
Why does it matter? I don't really understand what's your difficulty. Can you tell me what part you don't understand of my last line in the last post?

- August 25th 2010, 06:00 PMApprentice123
What you did was take an any expression to prove that (k+1)^2 > 3k + 3 ?

For example

Prove that is divisible by 3

I do:

4m + 1 is divisible by 3 then also divisible by 3

Correct ? - August 25th 2010, 06:03 PMDefunkt
- August 25th 2010, 07:08 PMApprentice123
OK. Thank you