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Thread: Check out my new math proof

  1. #1
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    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.
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  2. #2
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    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 2n+1) or just plain text writing 2^(n+1) for 2n+1.
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  3. #3
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    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
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  4. #4
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    Re: Check out my new math proof

    Quote Originally Posted by danbabb View Post
    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.
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  5. #5
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    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.
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  6. #6
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    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.
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  7. #7
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    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
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  8. #8
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    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.
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  9. #9
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    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
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