Attempt at proving result from intro to pure mathematics by liebeck

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  1. Lukedavidhopkins1's Avatar
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    Attempt at proving result from intro to pure mathematics by liebeck
    If n=m^3-m for some integer m, then n is a multiple of 6.

    I assume that any integer m can be written in the form x+1 for some value of x

    n=(x+1)^3-(x+1)
    n=x(x+1)(x+2)

    From this result it can be seen that whatever the value of the integer x, n will have a factor which is a multiple of 2 and a factor which is a multiple of 3. From this n must be divisible by 6 as it is divisible by both 2 and 3. Thus, it is a multiple of 6

    This is my first real attempt at proving anything so please do not be too harsh. Is it valid?

    And for anyone well versed in LaTex, why has my second line of x's and n's changed like that?
    Last edited by Lukedavidhopkins1; 25-06-2012 at 00:38. Reason: Putting it in LaTex, and fixing it
  2. Tla's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    Looks valid to me. And when you wanna do some more latex on a new line you should end the first line (/latex) and start a new (latex)(/latex) on the next line.
    n=(x+1)^3-(x+1)
    n = x(x+1)(x+2)
  3. Lukedavidhopkins1's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by Tla)
    Looks valid to me. And when you wanna do some more latex on a new line you should end the first line (/latex) and start a new (latex)(/latex) on the next line.
    n=(x+1)^3-(x+1)
    n = x(x+1)(x+2)
    Cheers, much appreciated on both accounts. Have now fixed the mistake and will remember for the future.
  4. notnek's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    It is valid but you don't really need the substitution:

    m^3-m = m(m^2-1) = (m-1)(m)(m+1)

    This is also the product of three consecutive integers.
  5. Lukedavidhopkins1's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by notnek)
    It is valid but you don't really need the substitution:

    m^3-m = m(m^2-1) = (m-1)(m)(m+1)

    This is also the product of three consecutive integers.
    Damn, you're right. I had tried factorizing it but I hadn't noticed the connection until I made the substitution. Thanks for the input
  6. SubAtomic's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by Tla)
    Looks valid to me. And when you wanna do some more latex on a new line you should end the first line (/latex) and start a new (latex)(/latex) on the next line.
    Or use these like this \\ sometimes it required \\ \\ to look neat but today it is just one set weird n and x on second line, what's going on? Does some funny stuff does tex, the different n is because I pressed enter after the \\.

    n=(x+1)^3-(x+1) \\ 

n = x(x+1)(x+2)
    Last edited by SubAtomic; 26-06-2012 at 15:50.
  7. DFranklin's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by SubAtomic)
    Or use these like this \\ sometimes it required \\ \\ to look neat but today it is just one set weird n and x on second line, what's going on?
    Other things being equal, it's better to use separate [latex]...[/latex] blocks, because then each line is treated separately by the vBulletin system. It makes it easier for other people to put comments between lines etc.
  8. SubAtomic's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by DFranklin)
    Other things being equal, it's better to use separate [latex]...[/latex] blocks, because then each line is treated separately by the vBulletin system. It makes it easier for other people to put comments between lines etc.
    I will bear that in mind in the future, cheers:cool: How come the n and x'es changed in the second line of my reply? Never seen that happen before.
  9. DFranklin's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by SubAtomic)
    I will bear that in mind in the future, cheers:cool: How come the n and x'es changed in the second line of my reply? Never seen that happen before.
    Looks like it's changed to "text mode". I don't ever do multi-line LaTeX on here (and it's 20 years since I've done it at all), so I don't think I can suggest anything beyond that!
  10. f1mad's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by SubAtomic)
    I will bear that in mind in the future, cheers:cool: How come the n and x'es changed in the second line of my reply? Never seen that happen before.
    I'd suggest writing "\displaystyle" on each spaced line. That should do the trick.

    



\displaystyle n=(x+1)^3-(x+1)



\displaystyle n = x(x+1)(x+2)

    I'm no expert mind.
    Last edited by f1mad; 26-06-2012 at 03:38.
  11. raheem94's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by SubAtomic)
    Or use these like this \\ sometimes it required \\ \\ to look neat but today it is just one set weird n and x on second line, what's going on?

    n=(x+1)^3-(x+1) \\ 

n = x(x+1)(x+2)
     n=(x+1)^3-(x+1) \\  n = x(x+1)(x+2)

    See my LaTex.

    To start a new line, just write "\\" but then don't press the enter key, write everything in the same line.

    NB: I might be wrong because i am not an expert at it.
  12. SubAtomic's Avatar
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    Re: Attempt at proving result from intro to pure mathematics by liebeck
    (Original post by f1mad)
    I'd suggest writing "\displaystyle" on each spaced line. That should do the trick.

    



\displaystyle n=(x+1)^3-(x+1)



\displaystyle n = x(x+1)(x+2)

    I'm no expert mind.

    (Original post by raheem94)
     n=(x+1)^3-(x+1) \\  n = x(x+1)(x+2)

    See my LaTex.

    To start a new line, just write "\\" but then don't press the enter key, write everything in the same line.

    NB: I might be wrong because i am not an expert at it.
    I prefer it to look like this though, with a bit of a space between lines, so it is either

    \displaystyle n=(x+1)^3-(x+1) \\ \\ n=x(x+1)(x+2)

    Or like DFranklin said,

    \displaystyle n=(x+1)^3-(x+1) press enter twice and start the second line.

    \displaystyle n=x(x+1)(x+2)


    Just a case of having the (tex)\displaystyle copied in the second way.
    Last edited by SubAtomic; 26-06-2012 at 15:48.
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