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When does 0!=1?

ummmmmm....... 0! = 1 is true. Zero factorial equals one.

Apparently it's possible for 0 prime to equal 1? Any ideas?
Click to expand...

I have no idea wtf you are talking about here. Are you confusing prime with factorial???
 
There's a fake proof out there to show that 0=1, but you have to divide by zero to do it. If you could divide by zero, you could have Jesus powers and make whatever you wanted. Say you want an orange, but you don't have one, you just do the proof, divide by zero, and make the 0 oranges you have turn into 1 orange which you can eat. Oranges are awesome.
 
If the question is "When does 0! = 1" then the answer is always.
If the question is "When does 0 != 1" then the answer is still always.

I'm confused.
 
0! is defined as 1 simply because it makes a lot of mathematics easier (look at Taylor series, for example). There is really no other fundamental reason for it to be defined that way.
 
Originally posted by: notfred
If the question is "When does 0! = 1" then the answer is always.
If the question is "When does 0 != 1" then the answer is still always.

I'm confused.
Click to expand...

I don't know what language you use but I know at least PHP PERL and C/C++ would st@b you in your sleep for using syntax like that ;p

Now $0! = 1; might pass with PERL or C/C++ (kinda doubt it in PERL, not sure about C) 🙂
And $0 != 1; Is only true so long as its not proveeded by $0 = 1; 🙂


Hahaha, this is so sad....
But actually now that I think about it, I don't think you are allowed to start a PHP variable with a number 🙂

Carry on.
 
Originally posted by: Balthazar
Originally posted by: notfred
If the question is "When does 0! = 1" then the answer is always.
If the question is "When does 0 != 1" then the answer is still always.

I'm confused.
Click to expand...

I don't know what language you use but I know at least PHP PERL and C/C++ would st@b you in your sleep for using syntax like that ;p

Now $0! = 1; might pass with PERL or C/C++ (kinda doubt it in PERL, not sure about C) 🙂
And $0 != 1; Is only true so long as its not proveeded by $0 = 1; 🙂


Hahaha, this is so sad....
But actually now that I think about it, I don't think you are allowed to start a PHP variable with a number 🙂

Carry on.
Click to expand...

in C:
#include <stdio.h>

int main(){
if(0 != 1){
printf("0 isn't 1\n");
}
}
Click to expand...

in perl:
#!/usr/bin/perl

if(0 != 1){
print "0 ins't 1\n";
}
Click to expand...

Go ahead and compile and run them, they both work.
 
By definition:
n! = n*(n-1)*...*1
(n-1)! = (n-1)*...*1
From the above two lines:
(n-1)! = n! / n

let n = 4 ---> (n - 1) = 3

by definition we have
4! = 4 * 3 * 2 * 1
3! = 3 * 2 * 1
by above: (n-1)! = n! / n
3! = (4 * 3 * 2 * 1 ) / 4 = 3 * 2 * 1

makes sense, follows the definition of n! even.

how about now: n = 1
(n-1)! = n! / n
(1-1)! = 1!/1
0! = 1

we often think of factorials as multiplying by increasing "n's". if you think of factorialization backwards, as n decreases, you're dividing. so, by definition, and "backwards" thinking of n! and (n-1)!, we have it clearly stated that 0! = 1

if you look at this analysis, you see clearly why (n-1)! is not defined for n = (-1) or n=0. And by recursion, you see that factorials are not defined for ANY negative number, simply because you get a division by 0 error.

please note that even though we "define" 0! as 1, it makes sense that 0! is 1, and that factorial is truly undefined for negative numbers.
 
Originally posted by: Mday
By definition:
n! = n*(n-1)*...*1
(n-1)! = (n-1)*...*1
From the above two lines:
(n-1)! = n! / n

let n = 4 ---> (n - 1) = 3

by definition we have
4! = 4 * 3 * 2 * 1
3! = 3 * 2 * 1
by above: (n-1)! = n! / n
3! = (4 * 3 * 2 * 1 ) / 4 = 3 * 2 * 1

makes sense, follows the definition of n! even.

how about now: n = 1
(n-1)! = n! / n
(1-1)! = 1!/1
0! = 1

we often think of factorials as multiplying by increasing "n's". if you think of factorialization backwards, as n decreases, you're dividing. so, by definition, and "backwards" thinking of n! and (n-1)!, we have it clearly stated that 0! = 1

if you look at this analysis, you see clearly why (n-1)! is not defined for n = (-1) or n=0. And by recursion, you see that factorials are not defined for ANY negative number, simply because you get a division by 0 error.

please note that even though we "define" 0! as 1, it makes sense that 0! is 1, and that factorial is truly undefined for negative numbers.
Click to expand...

beat me. nice work too.
 
Originally posted by: DrPizza
Originally posted by: Mday
By definition:
n! = n*(n-1)*...*1
(n-1)! = (n-1)*...*1
From the above two lines:
(n-1)! = n! / n

let n = 4 ---> (n - 1) = 3

by definition we have
4! = 4 * 3 * 2 * 1
3! = 3 * 2 * 1
by above: (n-1)! = n! / n
3! = (4 * 3 * 2 * 1 ) / 4 = 3 * 2 * 1

makes sense, follows the definition of n! even.

how about now: n = 1
(n-1)! = n! / n
(1-1)! = 1!/1
0! = 1

we often think of factorials as multiplying by increasing "n's". if you think of factorialization backwards, as n decreases, you're dividing. so, by definition, and "backwards" thinking of n! and (n-1)!, we have it clearly stated that 0! = 1

if you look at this analysis, you see clearly why (n-1)! is not defined for n = (-1) or n=0. And by recursion, you see that factorials are not defined for ANY negative number, simply because you get a division by 0 error.

please note that even though we "define" 0! as 1, it makes sense that 0! is 1, and that factorial is truly undefined for negative numbers.
Click to expand...

beat me. nice work too.
Click to expand...

:beer: Never thought of it that way.
 
Originally posted by: notfred
Originally posted by: Balthazar
Originally posted by: notfred
If the question is "When does 0! = 1" then the answer is always.
If the question is "When does 0 != 1" then the answer is still always.

I'm confused.
Click to expand...

I don't know what language you use but I know at least PHP PERL and C/C++ would st@b you in your sleep for using syntax like that ;p

Now $0! = 1; might pass with PERL or C/C++ (kinda doubt it in PERL, not sure about C) 🙂
And $0 != 1; Is only true so long as its not proveeded by $0 = 1; 🙂


Hahaha, this is so sad....
But actually now that I think about it, I don't think you are allowed to start a PHP variable with a number 🙂

Carry on.
Click to expand...

in C:
#include <stdio.h>

int main(){
if(0 != 1){
printf("0 isn't 1\n");
}
}
Click to expand...

in perl:
#!/usr/bin/perl

if(0 != 1){
print "0 ins't 1\n";
}
Click to expand...

Go ahead and compile and run them, they both work.
Click to expand...


I'll be d@mned, learn something new every day 🙂
 
Originally posted by: Mday
By definition:
n! = n*(n-1)*...*1
(n-1)! = (n-1)*...*1
From the above two lines:
(n-1)! = n! / n

let n = 4 ---> (n - 1) = 3

by definition we have
4! = 4 * 3 * 2 * 1
3! = 3 * 2 * 1
by above: (n-1)! = n! / n
3! = (4 * 3 * 2 * 1 ) / 4 = 3 * 2 * 1

makes sense, follows the definition of n! even.

how about now: n = 1
(n-1)! = n! / n
(1-1)! = 1!/1
0! = 1

we often think of factorials as multiplying by increasing "n's". if you think of factorialization backwards, as n decreases, you're dividing. so, by definition, and "backwards" thinking of n! and (n-1)!, we have it clearly stated that 0! = 1

if you look at this analysis, you see clearly why (n-1)! is not defined for n = (-1) or n=0. And by recursion, you see that factorials are not defined for ANY negative number, simply because you get a division by 0 error.

please note that even though we "define" 0! as 1, it makes sense that 0! is 1, and that factorial is truly undefined for negative numbers.
Click to expand...

very nice

:beer:😎
 
0! = 1 for reasons that are similar to why
x^0 = 1. Both are defined that way. But there are reasons for these
definitions; they are not arbitrary.

You cannot reason that x^0 = 1 by thinking of the meaning of powers as
"repeated multiplications" because you cannot multiply x zero times.
Similarly, you cannot reason out 0! just in terms of the meaning of
factorial because you cannot multiply all the numbers from zero down
to 1 to get 1.

Mathematicians *define* x^0 = 1 in order to make the laws of exponents
work even when the exponents can no longer be thought of as repeated
multiplication. For example, (x^3)(x^5) = x^8 because you can add
exponents. In the same way (x^0)(x^2) should be equal to x^2 by
adding exponents. But that means that x^0 must be 1 because when you
multiply x^2 by it, the result is still x^2. Only x^0 = 1 makes sense
here.

In the same way, when thinking about combinations we can derive a
formula for "the number of ways of choosing k things from a collection
of n things." The formula to count out such problems is n!/k!(n-k)!.
For example, the number of handshakes that occur when everybody in a
group of 5 people shakes hands can be computed using n = 5 (five
people) and k = 2 (2 people per handshake) in this formula. (So the
answer is 5!/(2! 3!) = 10).

Now suppose that there are 2 people and "everybody shakes hands with
everybody else." Obviously there is only one handshake. But what
happens if we put n = 2 (2 people) and k = 2 (2 people per handshake)
in the formula? We get 2! / (2! 0!). This is 2/(2 x), where x is the
value of 0!. The fraction reduces to 1/x, which must equal 1 since
there is only 1 handshake. The only value of 0! that makes sense here
is 0! = 1.

And so we define 0! = 1.
 
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