# Being defined and existence.

1. Mar 9, 2012

### julypraise

In mathematics, people construct and define things. And along with these, I found something very confusing:

Statement 1: if a thing exists, then the thing is defined.

I think this statement is true in mathematics. But this kind of statement I can't translate to first order calculus. Anyway, if this is true, then it is also ture:

Statement 2: if a thing is not defined, then the thing does not exists.

Then here a confusion begins:

Example 1:

We know that 6/0 is undefined (at least, most people say so). So by Statement 2 we may derive that 6/0 does not exists. Well, I'm not sure if it is true... Suppose 6/0 is a number such that multyplying this with 9 produces 6. In a first thought, if I think 6/0 like this, then it treuly does not exists. But In this first thought, I kinda already defined what 6/0 is, which means, 6/0 is defined. If I don suppose 6/0 like that, I have no idea of what this is, thus I can't think of the existence of it at all.

Example 2:

Now, if you consider some strange symbol like (6,0)/0, then it is undefined. But in the first place, I don't think I can say of whether it exists or not at all, because I do not know what this is.

Question:

(1) Maybe, Statement 1 is wrong in the first place. Is it? Please correct me.

(2) If it is ture, where I am wrong? Is my notion of 'being defined' wrong? Please give me insight.

2. Mar 9, 2012

### I like Serena

Hi julypraise!

In mathematics we work with systems of definitions and axioms.
If something is not defined within that system, it does not exist in that system.
And if something exists in reality, or in another system, but it is not defined in the system we're currently working with, it is still not defined and does not exist within our system.

In your example 6/0 is usually not defined, because in the system that you are used to, the regular real numbers, we cannot make a consistent definition for 6/0, so it is explicitly undefined.

However, there are other mathematical systems in which 6/0 is defined, for instance, 6/0 might be defined as infinity (symbol ∞).
But this can only be done at the cost of making something else we take for granted undefined, so it does not exist within this system.

3. Mar 9, 2012

### SteveL27

Not sure where you saw that. Perhaps if you tell us, it would help. It's manifestly false. For example in ZFC set theory (the standard set theory in use by mathematicians) the collection of all subsets of the natural numbers exists. If N is the natural numbers, then P(N) is the power set of the natural numbers; and this exists by the power set axiom.

But most subsets of the natural numbers are not definable, in the technical sense.

http://en.wikipedia.org/wiki/Definable_real_number

So you have to tell us where you read your statements, and why someone is claiming they're true, when they're not.

What do you mean by "define?" I can define a unicorn as a beast with one horn; but that doesn't mean unicorns exist.

4. Mar 9, 2012

### julypraise

Okay, guys. The actual problem I have is on this post:

Anc actually both of you guys commented here.

Okay Rudin in his definition forms the quotient and

kinda construct the derivative by the limit of the quotient.

So in what point of ZFC axioms can this (these performance of forming and

defining) be justified?

I can't clarify how this works in the ground of ZFC,

especially when I use only predcate language.

Is my point clear by the way?

5. Mar 9, 2012

### julypraise

Okay, I've tried to clarify the Rudin's definition by using the first order language as much as I can. Please check if this is correct:

Definition (Rudin, p. 103; revised by me). Suppose $f$ is defined (and real-valued) on $[a,b]$. Suppose $x\in [a,a]$. Suppose there exists a function $\phi$ defined by ${\displaystyle \phi(t)=\frac{f(t)-f(x)}{t-x}\quad(a<t<b,t\neq x)}$. If $\lim_{t\to x}\phi(t)$ exists (in $\mathbb{R}$) then we say $f$ is differentiable at $x$. Suppose the limit exists for some $x \in [a,b]$. Then, in this case, we define a function $f'$, naming it the derivative of $f$, by $f'(x)=\lim_{t\to x}\phi(t)$ for all $x \in [a,b]$ such that the limit exists.

If this revised definition is correct (and the revision is correct), then we cannot use this definition for the function that I defined in #1 in the post

because such $\phi$ function does not exist (it can be proved by contradiction), and therefore it's not that the derivative does not exist but is undefined.

Last edited: Mar 10, 2012
6. Mar 10, 2012

### SteveL27

Oh I see you are the one asking if the derivative is defined on a singleton set by abuse of Rudin. No, it's not. An interval [a,b] requires that a < b where the inequality is strict. The discussion in that other thread really has nothing to do with Rudin or the real numbers.

7. Mar 10, 2012

### julypraise

Okay. But Rudin in his textbook (Baby Rudin) does not specify that it should be a<b when defining [a,b]. And anyway, if the interval should satisfy that requirement, then by his textbook, we cannot consider the derivative of a function defined at a isolated point at all. But he talks about this thing. Please refer to my recent post: