The contradiction comes from the wrong assumption: m!=n! implies m=n.

So we can modify this rule as follows:

For m, n > 1, m!=n! implies m=n.

The contradiction comes from the wrong assumption: m!=n! implies m=n.

So we can modify this rule as follows:

For m, n > 1, m!=n! implies m=n.

The contradiction comes from the wrong assumption: m!=n! implies m=n.

So we can modify this rule as follows:

For m, n > 1, m!=n! implies m=n.

Regards

m! = n! no imply m=n how 1! =0! bu t no 1=0

@ Cenap Özel

If n!=7! , what is n?

n=7

So, m!=n! implies m=n

0! = 1 for convenience. It does not follow the same pattern as the others and so should not be considered among others, as Tareq pointed out.

Dear Professor,

In the combinations concern m!=n! is true only when m,n>0. So the question contradicts the basic condition .

@Jack Son

Who put the basic condition ?

"m!=n! is true only when m,n>0".

Thus if m,n =0 then m!=n! does not imply m=n.

Hence 0!=0! does not imply 0=0 .

@Tareq Al-shami

@Colton Magnant

Similarly, if you assert that "For m, n > 1, m!=n! implies m=n"

Then For m, n ≤ 1, m!=n! does not imply m=n.

Thus

1!=0! , does not imply 1=0.

But also

1!=1! , does not imply 1=1

0!=0! , does not imply 0=0

To overcome the problem, we restate it as follows:

m!=n! implies m=n for all (m, n)∈ N ×N \{(1, 0),(0, 1)}

Regards

To overcome the problem, we restate it as follows:

m!=n! implies m=n, for all (m, n)∈ N ×N \{(1, 0),(0, 1)}

Regards

m!=n! is true only when m,n>0.

Ahlam H. Tolba

"m!=n! is true only when m,n>0"

Thus if m,n =0 then m!=n! does not imply m=n.

Hence 0!=0! does not imply 0=0

inconsistent.

Tareq Al-shami

"To overcome the problem"

Thus, we fled to face the problem.

@Salah A. Mabkhout

Who prove that m!=n! implies m=n???!!!

This mathematical expression is not correct, because 1!=0! implies 1=0 !!!

So the logically mathematical expression is that:

m!=n! implies m=n, for all (m, n)∈ N ×N \{(1, 0),(0, 1)}

Dear all

Precisely, the question about why?

Why 1!=0! does not follow the rule " m!=n! implies m=n" ?

Obviously, if we ignore the case 1!=0!, there would be no contradiction.

Well,

n!=г(n+1)=nг(n)

1=1!=г(1+1)=1г(1)=г(1)

1=0!=г(0+1)=0г(0)

One may said the expression

n!=г(n+1)

is not defined when n=0 : 0!=г(0+1)=0г(0)

Hence, so does its consequence 0!

0! is not defined

There is no paradox.

The factorial function is defined to be n! = n*(n-1)*(n-2)*...*(2)*1, for integers n greater than or equal to 1. The recurrence relation corresponding to this is: (n+1)! = (n+1)*(n!), with initial condition 0! = 1.

Thus 0! = 1! = 1.

As to the original poster's question....

In general, [f(x) = f(y) implies that x = y] if and only if f is a one--to-one function. Here, the factorial function is not a one-to-one function, since 0! = 1! = 1.

Richard M. Low

"The factorial function is defined for integers n greater than or equal to 1". That is true. Hence, the factorial function n! is not defined for n=0. That is, 0! is not defined. Now you define 0!=1, a contradiction.

The problem relies behind 0!=1.

From where you brought this initial condition

"with initial condition 0! = 1" ???

How could you justify this initial condition?

The only way to define 0! is to set 0!=г(1)

and evaluate г(1) from the integral form of Gamma Function.