0!

0!=1!=!, and the question to you is...why?! (! is factorial, i.e. teh product of all the numbers up to that nubmer, like 5!=1*2*3*4*5 - but you all knew that, didn't you?)

n! is defined as follows:
0!=1
for n>0, n!=n*(n-1)!

which is basically the same thing as teh product of all the numbers up to that nubmer, but it explains the 0 case

Originally posted by Mouse2
n! is defined as follows:
0!=1
for n>0, n!=n*(n-1)!

which is basically the same thing as teh product of all the numbers up to that nubmer, but it explains the 0 case

don't you hate it when you post a puzzle that isn't so much a puzzle as a bit of knowledge that people'll get first time anyway? ah well-i've gotta find some more puzzle, or dig up some old uns...ah well-how about, for the mean time, integrate lnx?

You'll just get x*lnx - x plus a constant. How about do it other than a simple way though.

Genius's modified puzzle:

Integrate ln x in the most convoluted way possible...

Originally posted by genius
0!=1!=!, and the question to you is...why?! (! is factorial, i.e. teh product of all the numbers up to that nubmer, like 5!=1*2*3*4*5 - but you all knew that, didn't you?)

I love this argument for 0!

0! = 1! /1 = 1/1 = 1

Why not just define k! as the number of permutations of k objects? If you have zero objects, then you have one possible permutation, so 0! = 1 with no special case definition.

I think those reasons are all made up after it was defined this way. That was probably because it fits best that 0!=1 and not that 0!=0, so people looked for 'logical' explanations. But that's the mathematicians way.

Originally posted by piderman
I think those reasons are all made up after it was defined this way. That was probably because it fits best that 0!=1 and not that 0!=0, so people looked for 'logical' explanations. But that's the mathematicians way.

well-how about "prove that 0!=0"? 😛

Originally posted by piderman
I think those reasons are all made up after it was defined this way. That was probably because it fits best that 0!=1 and not that 0!=0, so people looked for 'logical' explanations. But that's the mathematicians way.

Even sillier is the following definition: n! := Gamma(n+1)

i think another reason 0!=1 is because if you end up with a 0! in the denominator of a fraction, then the whole sentence becomes undefined. and sometimes the are solutions. like n choose r -
5 choose 0-
5!/(5-0)!*0!
there is one way to choose 0 out of 5, so it is possible. if 0!=0, then it would be undefined, whereas now it is 1.

by the way, im gonna be a freshman in highschool (but im in advanced math), so if that doesnt make any sense, its my fault, not urs.

Originally posted by royalchicken
Why not just define k! as the number of permutations of k objects? If you have zero objects, then you have one possible permutation, so 0! = 1 with no special case definition.

Yep.

Originally posted by Bowmann
Yep.

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