Greetings. This is my first post and I would like to share a proof that I recently put on youtube. It hasn't gotten many views yet so I figured I'd share it here .
It is kind of like when people say something that goes away if you know what I mean.
Greetings. This is my first post and I would like to share a proof that I recently put on youtube. It hasn't gotten many views yet so I figured I'd share it here .
It is kind of like when people say something that goes away if you know what I mean.
If you write your proof in text form, then not only it would take about 10^4 times less memory, but it would be more clearly seen and people would probably be more willing to check it out. You could use LaTeX ([TEX]2^{n+1}<(n+1)![/TEX] gives ), the superscript tags (2[sup]n+1[/sup] gives 2^{n+1}) or just plain text writing 2^(n+1) for 2^{n+1}.
Sorry, I will try and capture it more clearly but here it is line by line:
Prove that 2^n < n! for all n greateroreaqual to 4, for n in N
Base Case: Let n = 4 . 2^4 < 4! is true since 16 < 24
Let n = k, for k in N and k > 4
2^k = 2^(k-1) x 2 and
k! = (k-1)! * k
and 2^4 = 16
4! = 24
2^k < (k)! for all k greatororequal to 4
since 2 < k
Thanks for the input, I updated the video so it can be read. I am half-Polish living in the USA and I do feel my method is different. If you watch my other videos you will observe a difference but you see it works https://www.youtube.com/user/danbabb . Thanks.
Guys, on a tangent, never divides , nor do any prime number to the power of . It is because there are not enough exponents in , for any prime number .
Salahuddin
Maths online
More exactly, p^n does not divide ((p-1)*n)! for any prime number p.
Salahuddin
Maths online