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Thread: answer this

  1. January 22nd, 2004, 09:23 AM #1
    saintakaagni
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    answer this

    does anybody have the proof or know where the answer can be found for the following question
    0!=1
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  2. January 22nd, 2004, 11:38 AM #2
    fourdc
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    I wasn't a math major but 0!=1 couldn't happen.

    1!= 1 2!=2x1 3!=3x2x1 etc etc so since there is no subset of positive integers below zero, 0! like the square root of -1 would fall into the imaginary number category.
    ddddc

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  3. January 22nd, 2004, 01:39 PM #3
    cgkanchi
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    AFAIK, I don't think it has any proof. It's just an assumption to ensure that while calculating permutations and combinations (among other things), you don't get stuck with a 0 in the denominator.
    Cheers,
    cgkanchi
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  4. January 23rd, 2004, 09:07 AM #4
    saintakaagni
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    please see this site
    www.newdream.net/~sage/old/numbers/fact.htm
    it has the value for 0! but how it came i dont know
    Life is a shipwreck but we must not forget to sing in the lifeboats. ~Voltaire
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  5. January 24th, 2004, 03:56 AM #5
    jinxy
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    I'm sorry i'm no maths expert. 0=0 to me, to think otherwisemakes no sense.
    By the way x=un-known quantity
    Spurt=Drip under pressure
    What happens if a big asteroid hits the Earth? Judging from realistic simulations involving a sledge hammer and a common laboratory frog, we can assume it will be pretty bad. - Dave Barry
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  6. January 26th, 2004, 07:20 PM #6
    The Mormon
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    Of course 0!=1

    Code:
    #include <stdio.h>

main()
{
if(0!=1)
printf("Hey! I'm right!");
else
printf("You Suck!");

return 0;
}
    (joke)
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