Does this proof make sense. prove n! n^2 for n=4. So we assume that k^2 1 is equivalent to 2k+1 = 4 we have 2k+1

Question

Does this proof make sense. prove n! > n^2 for n>=4. So we assume that k^2 < k! Notice that that k>1 is equivalent to 2k+1 < 3k It follows that k>= 4 we have 2k+1 < 3k < (k)! k Now using the assumption k^2 < k! we have k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)! So (k+1)^2 < (k+1)!

anonymous · Answer

i see eyes, lol

anonymous · Answer

this is a proof by induction, yes? seems ok. a proof by induction always needs verification of the base case, though.

so check 4! = 4*3*2 = 24, which is larger than 4^2.

anonymous · Answer

but what about the step, (k+1)!

anonymous · Answer

this is false 
k^2 +(2k+1) < k! +(2k+1) + k! +(k)! k=(k+1)!

anonymous · Answer

ah yes, you're right. that is incorrect.

anonymous · Answer

hehe

anonymous · Answer

hi pooch

anonymous · Answer

Cantor, does this qualify as a valid proof of induction?

Assume: k >= 4

From k >= 4, k > 1

From k >= 4 we also get the result k*k >= 4*k thus k^2 >= 4k

So, if we assume that k! > k^2, then:

(k + 1)! = (k + 1) * k! > (k + 1) * k^2 > (k + 1) * 4k (since k^2 >= 4k)

(k + 1) * 4k = 4k^2 + 4k = 4k^2 + 3k + k > 4k^2 + 3k + 1 (since k > 1)

4k^2 + 3k + 1 > k^2 + 2k + 1 (for all k)

And k^2 + 2k + 1 = (k + 1)^2

And thus (k + 1)! > (k + 1)^2

anonymous · Answer

one sec

anonymous · Answer

yes that works, a bit odd , but fine :)

anonymous · Answer

you proved a different question
2^n < n! , i was asking n^2 < n!, 
but i think your proof works for the former

anonymous · Answer

wait this is false

anonymous · Answer

sorrry:)

anonymous · Answer

2^(k+1) =2.2^k <2.k! <(k+1k! =(k+1)!

anonymous · Answer

you wrote 2.2^k < 2.k!, thats false

anonymous · Answer

B.S : 2^k<k!

anonymous · Answer

thats your inductive step, clean your proof up

anonymous · Answer

HAMED!!!!
your proof needs a bit of work and finesse. the logic is dodgy

anonymous · Answer