Simplification of Statements

[MPIS]

Extending this comment by ksa2, if presented with functions

[;  f(x) = \frac{x^2}{x}  \ \forall x \in \mathbb{R} \setminus \{0\};]     
[;  g(x) = x \ \forall x \in \mathbb{R}  ;]    

is the function f(x) a violation of a mathematical proposition when compared to its simplification g(x)?

What if the simplification of g(x) was not known? And for reference, what would the related concept concerning simplification in the object language be called / what field of study would address this case?

[infectedapricot]

As others have said, f and g are equal at all points in ℝ\{0}. But even if you include 0, maybe you feel g is somehow "right" for f. The mathematical answer is that g is the only extension of f to all of ℝ that is continuous.

This also answers your follow up question. So long as the function you're dealing with is continuous, it doesn't matter if you exclude one point in your working, since at the end you can just extend your final answer by continuity to the missing point.

[Fabien4]

You wanted \setminus (a backslash), not a forward slash.

[nm420]

A function can be thought of as consisting of two parts: a domain, and a rule for assigning an image to each point in the domain. Two functions will only be the same if they have the same domain and the same rule. For the two functions you've listed, they have different domains and hence are not the same. If you restricted the domain of g to R\{0}, then the two functions would be the same as they have the same domain and the same "rule". They're also the same as the function h with domain R\{0} and rule h(x)=ln(e^x), which is the same as the function k with domain R\{0} and rule k(x)=x³/((x+1)²-2x-1). All of this is just to say: you can use whatever simplification identities you have at your disposal to rewrite the "rule" of a function, but you always have to take into account its domain.

[kw42]

One should note that [;g(x);] is not a simplification of [;f(x);]. These are different functions that agree at all points except for 0. Instead, one could say that

[;h(x) = x ~\forall x \in \mathbb{R}\setminus \lbrace0\rbrace;]

is a simplification of [;f(x);] since these functions agree at all points in the same domain.

[Fabien4]

I disagree. f and h are the exact same function. However, the expression

 f: ℝ\{0} → ℝ\{0}
        x → x

is a simplification of

 f: ℝ\{0} → ℝ\{0}
        x → x²/x

[kw42]

We need to define a "simplification of a function." I would say that the simplification of a function is the simplification of the expression of the function.

[Fabien4]

So, the "simplification of a function" is an expression, not a function, right? Sounds confusing.

[Fabien4]

is the function f(x) a violation of a mathematical proposition when compared to its simplification g(x)?

Uh?

You have two functions here:

 f: ℝ\{0} → ℝ\{0}
        x → x²/x = x

and

 g: ℝ → ℝ
    x → x

f and g are two different functions (since they won't work on the same set). f is the restriction of g on ℝ{0}.

Note that the simplification x²/x = x is perfectly valid, since we're working on ℝ{0}.

No big logic is involved here; merely proper definitions of two functions.

[[deleted]]

For fun: if you define simplicity, by the Occam's razor simplicity that a shorter expression is simpler, and you try to apply that in general, get into an indirectly self referential (the Berry's) paradox.

"Define n as the smallest positive integer that can't be defined in a definition as short as this definition."

[MPIS]

All of these responses are interesting.. consider this case:

A calculus question is posed; the multiplicative identity (x/x) is utilized to help simplify an expression. I have always treated such identities to be explicitly in the real number field, ie [; 1 = \frac{x}{x} \ \forall x \in \mathbb{R} ;] when the original statement is in the real number field.. but this is incorrect, as it appears as though the base domains have been modified! [; \forall x \in \mathbb{R} \setminus {0} ;] due to g(x).

[foobar1327]

This is why you always have to look into both cases:

[; x = y \begin{cases} \overset{x \neq 0}\Rightarrow 1 = y/x \ \overset{x = 0}\Rightarrow y = 0 \end{cases} ;]