prove 0!=1?

Answer 1

n!=n(n-1)!
1!=1*(1-1)!
1!=1*0!
1!/1=0!=1

Note:You can't use this formula with negative or franctional numbers.n means natural number.

Answer 2

0+1=1

Answer 3

0 = 0 Therefore 0 != 1

Answer 4

This is not a proovable thing, it is a definition introduced to make several formulae look and work better in combinatorics.
I.e. the newton symbol "n over k"=n!/(n!*(n-k)!) indicates how many ways there are to choose k elements in a set of n elements. If you put k=n, there is only one way, so 1=n!/(0!*n!) for this to be true, we would like 0! to be =1
Also, you can define the factorial by using the Euler's gamma function
n!=Gamma(n+1)
so 0!= Gamma(1)
and the Gamma function is defined as an integral
Gamma(z)=Integral_0^(infinity) (t^(z-1)* exp(-t) dt)
0!=Gamma(1)=Integral_0^(infinity) (exp(-t) dt)=1
Enough?

Answer 5

PHX is absolutely correct.

Due to its recursive nature,

n! = n(n - 1)!

Plugging in n = 1, we get

1! = 1(1 - 1)!

Simplifying

1! = 1(0)!

Showing that 1! = 0! = 1.

I was trying to come up with a way to find the factorial of negative numbers using PHX's method but couldn't.

Answer 6

in the field R 0 is the unique neutral element of addition
if 1 = 0 then 0 isnt unique .

Answer 7

this is an axiom and it cannot be proved just like any number to the power 0 is equal to 1 :)

Answer 8

1*0 = 0
0*0 = 0
divide by 0!
1 = 0!!!!!!!!!!!!!

Answer 9

n! = n * (n - 1) * (n - 2) * ... * 2 * 1
n! = n*(n - 1)!

Using first equation 1! = 1
Using second equation 1! = 1 * 0!

So 1 * 0! = 1
So 0! = 1

Answer 10

n!=n*n-1*n-2*........1;
that means,
n!=n*n-1!
put n=1;
1!=1!*0!
or
0!=1!/1
but 1!=1
0!=1/1=1
...proved