We knew n!=n(n-1)!
Hence 1!=1(1-1)!
1!=0!
But m!=n! implies m=n
So, 1!=0! implies 1=0
How to resolve such paradox?
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
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.
@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
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.
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