Well obviously, no is the answer. But why?
After all: If 5! = x! then x = 5.
let me define a function f(x) = 5
since f(5) = f(8), does that mean 5=8?
No, because the 5 in f(x) = 5 is a constant.
Does this have something to do with the fact that the value 0!=1 and 1!=1 are defined; and hence they are constants?
If so, how should I put it down to words?
The reason we define 0! to be equal to 1 is due to the combinatorial interpretation of the factorial function. This interpretation says the number of ways to arrange n objects is n! So there is exactly 1 way to arrange 0 objects, and there is also exactly 1 way to arrange 1 object (and 2 ways for 2 objects, etc.)
0!=1 by definition, yes.
Ok, if you don't accept that argument, then what about sin? If sin(x)=sin(y) does that mean x=y? Of course not (0, pi, 2pi, 3pi, etc..). Functions were we can say f(x)=f(y) implies x=y are called injective (or one-to-one) and it has nothing to do with 'constants' at all. It so happens that the factorial function is not injective.
James, there is no "inverse" factorial function that would return a when applied to a! as far as I know.
Take your equation:
What would you do to both sides to get 0=1?
(Now you might say that since sin has an invese function that sin(x)=sin(y) would imply that x=y if you took the arcsin of both sides, but remember that arcsin has a range of −π/2 ≤ y ≤ π/2 so all that you'll prove is that some value between -pi/2 and pi/2 is equal to itself.)
matt grime: if factorial function is not injective, then why is it that i can show by inspection: if 5! = x!, then x=5? My Math knowledge is limit, so my apology if I am mistaken, but surely, there is no other value that can satisfy x except 5?
Couldn't you also by inspection say that if x!=1! that x is either 1 or 0?
dav2008: yes, i could, which means the factorial function is also not injective. But how could a function be injective (assuming my argument is true) and not injective at the same time?
Because it happens to work for 5 (and many other numbers ), but not in general, so not always.
Who is saying that it's injective?
It can't, it's either injective or not
TD: Yes, it doesn't work for 1!, but why? I'm confused :(
What argument? You've not presented an argument. You do not have a valid argument that ! is injective since it is not. What's the problem? I fail to see the point of this discussion at all.
I think I need to clarify. I don't know if it is injective or not, but if it is not injective, and I'm sorry to repeat this, why if 5! = x! then x=5?
While it TD is right in saying that it wouldn't work for some numbers, I still don't get it. Possibly, there is some circular logic in my argument. I'm not sure...
Because for every value of x other than 0 or 1 x! determines x. So what's the issue?
Why should it work? It simply doesn't because it doesn't follow from the definition of the factorial.
As Matt showed, why should sin(x) = sin(y) imply x = y? It just doesn't.
If a function y=f(x) is non-injective that just means that there exists at least one value y for which f(x1)=f(x2)=y (Well I guess I said it backwards but I think it's clearer saying it that way for this purpose)
It doesn't mean that every single value of y has to have at least two corresponding values of x.
See this diagram of a non-injunctive function: http://upload.wikimedia.org/wikipedia/en/d/dd/Non-injective_and_surjective.png
it's clear that ! is not injective since 0!=1! and 0=/=1 which you know, so that makes me think you fail to understand what injective means.
however, if a function fails to be injective it simply means that there are two different inputs mapped to the same output. it doesn't say anything about the behaviour of the function anywhere else. It is perfectly possible that if we exclude the two points we know of that means it fails to be injective then the on the rest of the domain the function might or might not be injective. all things are possible.
again, what argument?
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