# {Edited} Factorials.

okay...I love math and i enjoy math a lot, mainly for its perfection and the aspect of not contradicting itself. But lately i have been runnig into contradictions and things that should not exist...and they piss me off.

The worst of them is solving 0!

0!=1

why? i asked my math teacher and he worte the following (which indeed makes sense)

5!=5*4*3*2*1 or 5!=5*4!

so 4!=4*3!
3!=3*2!
2!=2*1!
1!=1*0!

but this is where it lost me. 1! should simply be 1, not 1*0! maybe i'm worng. aniway, let's take a look at 0!

0!= all the natural numbers larger than zero (5! = 5*4*3*2*1...all of them larger than zero) but also no bigger than zero. (5*4*3*2*1 has no 6 in it.)

So...let's see...0!=0*Whatever is larger than zero and smaller (or equal) than zero at the same time.

so 0!=0*all natural numbers that belong to (0,0]. that makes the smallest allowed value bigger than the largest. I can't picture a REAL such a number. Let's call the whatever number W. So...0!=0*W

Not only we have a number in that bizzare scenario with a min > max, but it is also multiplied with zero. LAst time i multiplied something by zero it did not turn out to equal one.

so...any good reason why 0!=1?

plz...have mercy, i'm only in calc 1. no crazy explanations that i can't even begin to understand.

Thank you.
~Robokapp

I looked this up on mathworld. It says that 0! is simply defined to be 1 because there is exactly one way to organize zero objects (the empty set), and factorials are usually used for finding how many ways there are to organize things.

but in how many ways can you give zero apples to zero kids? that's infinity...because 0/0 is undefined. dividing by zeor always causes asymptotes...

i'm not saying i need help figuring out why you can't divide by zero. i know the 3/0=3-0-0-0-0-0-0-0-0....-0=0 demonstration.

but still...i think zero objects can be organised in many ways. or in no way...but i'll go with your answer :D

thank you.

Just do what i do, teach yourself the math because your math teacher is a drunk who comes to class eveyrday hung over. Its funny to watch he starts talking then forgets whats he alking about so he puts an overhead on shuts of the lights puts his head down and taps the glass.

The first term is 2. 0 isn't included in n! And 1 is negligtable. Also 1! = 1, this is was defined to match the binominal theorem.

Well, you seem to have noticed a pattern that (n+1)! = (n+1)*n!. So, if we put n = 0, then we get 1! = 1*0!, and 1! = 0! = 1. There are other ways of justifiying it, but that way is quite simple.

Keep in mind that 0! = 1 is a definition, not any sort of theorem.

If we define n! to be equal to gamma(n+1)

However I doubt that will be of much help to someone at the level of the topic poster.

It is only to sastisfy the binominal theorem.

I believe understanding that the $\Gamma(n+1)=n!$ is defined as an integral and then seeing that it works for integer n will make the poster realise that the supposedly arbitrary definition of 0! as 1 makes sense.

It's the same as realising that $e^{i\theta}=\cos\theta +i\sin\theta$ is consistent due to the Taylor expansions of these various functions.

I encourage the original poster to look up gamma functions on the internet, and even if he/she doesn't understand the integrals etc. to realise that if this general function works for all kinds of n, then perhaps it's a good idea to take it's definition of 0!

matt grime
Homework Helper
What is your definition of the factorial function, Robokapp? Mine is that it is a function from the set {0,1,2,3,...} of natural numbers to the integers satisfying 0!=1 and n!=n(n-1)!

Now, why is that a problem? It is, as people have said a definiton.

Don't bother with gamma functions as a justification. 0! was defined to be 1 well before gamma functions; it is disingenuous to use those to justify this definition. (warning that assertion is purely speculation but MUST be correct)

And we really cannot emphasize it enough that it IS a definition, one that is incredibly useful, consistent and practical.

Why do you say "solve for 0!" it isn't an equation; one cannot solve it.

Now, given we've made this or ANY other definition, the only valuation we can place on it is its usefulness, and this is a very useful one.

ooh. i think i get what you're talknig about. math is created to justify the real life situations not the other way around, so 0! has to =1 so the system given by the definition works.

right?

Tom Mattson
Staff Emeritus
Gold Member
No, it's not right. While it's true that "real life situations" are not created to justify mathematical statements, it is nonetheless not true that mathematical statements are created to justify said real life situations. Mathematicians develop their subject quite independently of whether it turns out to be descriptive of the physical world.

CRGreathouse
Homework Helper
4! = 4 * 3!
3! = 3 * 2!
2! = 2 * 1!
1! = 1 * 0!

This just shows that if there is a 0!, it 'should' be 1 to follow the same rules as the other numbers.

Here's another method you can consider. 4! is the product of 4 numbers (1,2,3,4). 3! is the product of 3 numbers (3, 2, 1). 2! is the product of 2 numbers (1,2). 1! is the product of 1 number (1). For this to make sense we have to have a way to take the product of just one number, and I think you'll find that the only way that 'makes sense' is to take the number itself:

$$k=\prod_{i=1}^1k$$

In the same way, the only way to make a product of 0 elements work is to define it as 1:

$$1=\prod_{i=1}^0k$$

This is because 1 is the multiplicative identity. Does this make any more sense?

A final argument for 0!=1 is that it is needed to define binomial coefficients and many other mathematical objects cleanly.

masudr said:
I believe understanding that the $\Gamma(n+1)=n!$ is defined as an integral and then seeing that it works for integer n will make the poster realise that the supposedly arbitrary definition of 0! as 1 makes sense.

It's the same as realising that $e^{i\theta}=\cos\theta +i\sin\theta$ is consistent due to the Taylor expansions of these various functions.

I encourage the original poster to look up gamma functions on the internet, and even if he/she doesn't understand the integrals etc. to realise that if this general function works for all kinds of n, then perhaps it's a good idea to take it's definition of 0!
This is a good explanation, but you should be carefull when you say that the integral works for integers.
The gamma-function is not defined for negative integers, it includes division by zero in some way. As you stated, n! = gamma(n+1), which means that (-1)! = gamma(0) which is undefined.
gamma(x) is defined for all reals except for the negative naturals including zero.

LeonhardEuler
Gold Member
I agree with you, Robokapp, that the explaination with
4! = 4 * 3!
3! = 3 * 2!
2! = 2 * 1!
1! = 1 * 0!
is not a good one. It is not true that 0!=0*(-1)! so this is no reason that 0! should be one. It is more like matt grime suggested: That 0! is defined to be one. While in general what Tom Mattson said (that "it is nonetheless not true that mathematical statements are created to justify said real life situations") is true, there are often motivations behind mathematical definitions in the real world. 0! being 1 is consistant with the idea of the n! being the number of ways to order n things. Rethink what you were thinking when you said "but in how many ways can you give zero apples to zero kids? that's infinity...because 0/0 is undefined. dividing by zeor always causes asymptotes". Actually there is no reason to be dividing by zero in this situation. There is in fact exactly one way to order zero objects: to put no object anywhere. There are no other ways to order nothing: this is the only one.

Well, solving any

[f(x+h)-f(x)]/h involves a division by zero. well 10^-99

but trough mathematical manipulation the division by zero is made possible.

Isn't it interesting how whole Calculus is based on dividing by zero?

Robokapp said:
Well, solving any
[f(x+h)-f(x)]/h involves a division by zero. well 10^-99
but trough mathematical manipulation the division by zero is made possible.
Isn't it interesting how whole Calculus is based on dividing by zero?
This is taking the analogy a bit far. While dividing by zero does not yield a real number, limits also do not necessarily yield a point in the image of the function being studied; they only yield limit points, which may or may not be equivalent to points in the graph of the function. In this case, we can see what value this new function is bounded by, but the bound is not a value of the function.

hypermorphism said:
This is taking the analogy a bit far. While dividing by zero does not yield a real number, limits also do not necessarily yield a point in the image of the function being studied; they only yield limit points, which may or may not be equivalent to points in the graph of the function. In this case, we can see what value this new function is bounded by, but the bound is not a value of the function.
well, you can have a limit but not a value if it's a hole in graph...but still you're manipulating a division by zero...are you not? i mean you're dividing by h and setting h = 0

lurflurf
Homework Helper
Robokapp said:
well, you can have a limit but not a value if it's a hole in graph...but still you're manipulating a division by zero...are you not? i mean you're dividing by h and setting h = 0
No. Limits do not set anything equal to anything. The theorem
for f a function continuous at x=a
$$\lim_{x\rightarrow a}f(x)=f(a)$$
often gives the impression limits are equivalent to function evaluation.
That is not true.
Limits explore what happens as a variable approches something.
There is a big difference between doing something and coming as close as possible to doing it, but not doing it.
Say you are defusing a bomb.
your safe if you defuse it with time left be it
100289340 s
10 s
1 s
.001 s
.0000000000000000001 s
but
0 s
boom you are in bad shape

the difference between
.0000000000000000001 s
and
0 s
may be small, but it is an important difference.

yea...yea i agree, h is the (0 value...meaning it's smallest value is the smallest positive number other than zero...meaning 10^-infinity but still...tangent at one point using the point itself and itself again is very precise in normal functions, but imagine f(x)=1/x and you're asked to find dy/dx at x=0.00000001. It will be slightly off becasue the small gap in points on x coordonate causes a massive increase in y-values due to proximity to an asymptote if your h is anything bigger than a couple billionths of a unit. I understand what you're saying though...by setting

lim=(h^2+4xh)/h and algebrating it into h/h * (h+4x), followed by reducement in h/h and setting h = 0 you get lim = 4x...but in fact you never set h=0. you set h=(0 or h=10^-999 so h/h is possible operation. this means your lim does not equal 4X but 4x+ a neglectable ammount...that is where the slight error comes form.

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You're making it so small that it dissapears, but it's still not equal to zero.
Your example, f(x) = 1/x, does not have a limit when x->0. Why? Because if you look at what you said, x = 0.000000001, you get a very big number, and x = 0.0000000000000001 gives you an even bigger number. It can look like it goes to infinity.
If you look at f(x)=1/x if x = -0.0000000001, you get a really small number, or a "big negative" number, if you wish.
So we have lim(x->0+) { 1/x } = +infinity and lim(x->0-) { 1/x } = -infinity. These two limits are not equal, hence the limit as x->0 does not exist.

Just remember that we never set the limit equal to zero, but very very close.

A fitting analogy is to look at limits that have values that are not in the image of the domain of the function as a being part of the boundary for values in the function set. Suppose the image of a set under a function is the open disc {x: $0\leq x < 1$}. The boundary of this disc is not in the image of the function, but studying limits within the domain will allow us to see what the boundary is (the unit circle).
Similarly, looking at the geometric picture of the derivative of a curve in the plane as the limit of the slope of a secant as it approaches the tangent to the curve, note that the final tangent line is a bound for secants to the curve with respect to the point in question (Attach a ray to one point in the graph and pivot it about that point. At some point, it no longer intersects the rest of the graph). Take the slopes of those secants as a set. Since they are continuous and are unbounded in one direction, you can put it in interval form. The interval is unbounded on one side, but is bounded on the other side. The question is, what is the least upper/greatest lower bound of your interval/set? What point is sitting there on the boundary? No mention of zero need be made.

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This is because 1 is the multiplicative identity. Does this make any more sense?