Check out my new math proof

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 Ultimate Math Proof - YouTube .

It is kind of like when people say something that goes away if you know what I mean.

Re: Check out my new math proof

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 $\displaystyle 2^{n+1}<(n+1)!$), the superscript tags (2[sup]n+1[/sup] gives 2^{n+1}) or just plain text writing 2^(n+1) for 2^{n+1}.

Re: Check out my new math proof

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

Re: Check out my new math proof

Quote:

Originally Posted by

**danbabb** 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

Here is how I would write this.

$\displaystyle \begin{align*}2^k &= 2^{k-1}\times 2\\&<(k-1)!\times 2&&\text{by inductive hypothesis}\\&<(k-1)!\times k&&\text{since }k>2\\&=k!\end{align*}$

The fact that 2^4 = 16 and 4! = 24 is not relevant to the inductive step.

Re: Check out my new math proof

I wrote that to show 2^n < n! in the case where n = 4 . Like how the stuff above the proof in the video is for demonstration also, but thanks for your input.

Re: Check out my new math proof

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.

Re: Check out my new math proof

Guys, on a tangent, $\displaystyle 2^n$ never divides $\displaystyle n!$, nor do any prime number to the power of $\displaystyle n$. It is because there are not enough exponents in $\displaystyle n!$, for any prime number $\displaystyle p$.

Salahuddin

Maths online

Re: Check out my new math proof

Salahuddin559, When n = 2, 2^n = 4 and n! = 2, so that would work for what you just said.

Re: Check out my new math proof

More exactly, p^n does not divide ((p-1)*n)! for any prime number p.

Salahuddin

Maths online