Reply to Re: Some math
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Well, n! = 1 * 2 * ... * n.
One would think 0! therefore would be 0. Not so. Watch:
1! = 1
2! = 2 * 1! ; 2! = 2
3! = 3 * 2! ; 3! = 6
4! = 4 * 3! ; 4! = 24
.....
So, turn that around:
4! = 24
3! = 4!/4 ; 3! = 6
2! = 3!/3 ; 2! = 2
1! = 2!/2 ; 1! = 1
0! = 1!/1 ; 0! = 1
Now, how does that satisfy n!?
1! = 1
n! = n(n-1)! ; n[1, Infinity)
Take our answer from before and check to see that:
0! = 1
n! = n(n-1)! ; n[0, Infinity)
That's why 0! == 1.
One would think 0! therefore would be 0. Not so. Watch:
1! = 1
2! = 2 * 1! ; 2! = 2
3! = 3 * 2! ; 3! = 6
4! = 4 * 3! ; 4! = 24
.....
So, turn that around:
4! = 24
3! = 4!/4 ; 3! = 6
2! = 3!/3 ; 2! = 2
1! = 2!/2 ; 1! = 1
0! = 1!/1 ; 0! = 1
Now, how does that satisfy n!?
1! = 1
n! = n(n-1)! ; n[1, Infinity)
Take our answer from before and check to see that:
0! = 1
n! = n(n-1)! ; n[0, Infinity)
That's why 0! == 1.