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Alright, I've done research regarding this question have attained answers that identify $f$ as the function (which is obvious) and identify $\mathbb R \rightarrow \mathbb R$ as meaning "the domain is the set of all real numbers and the codomain is the set of all real numbers",

I've also read that, say $f : x \rightarrow x^2$ can be used as the notation for a function, and can otherwise be written as $f(x) = x^2$, which makes sense, but say for example, $f : 2\sqrt x \rightarrow \displaystyle \frac{x}{4}$, would this be equivalent to $f(x) = \displaystyle \frac{x^2}{64}$?

Furthermore, can the use of symbols indicating the sets of, for example, all real or rational numbers always be used to distinguish between when this notation is referring to a function and when it is referring to the domain and codomain of a function? All I would like is confirmation to these ideas so that I am aware of how to use them in the future. Thank you.

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  • $\begingroup$ a function $f:A\to B$ is a special subset of the set $A\times B$. In general the usual notation to describe a function is like $f:A\to B,\, x\mapsto f(x)$, where $f(x)$ is explicitly described, by example as $x^2$, $\lfloor x\rfloor$, $\ln x$, etc... $\endgroup$ – Masacroso Sep 6 '17 at 4:58
  • $\begingroup$ I will enfatize the answer of Mathematician42: you cannot define a function properly if you dont says what is it domain and what is it codomain, so $f(x)=x^2$ doesnt define correctly a function. $\endgroup$ – Masacroso Sep 6 '17 at 5:06
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The notation $f:A\to B$ is read "f is a function from $A$ to $B$." I always have imagined the colon to be "is a" or "has the type." (You could potentially also write $f\in B^A$ if you are comfortable with $B^A$ being the set of functions $A\to B$.)

I wouldn't use $f:x\to x^2$ as notation for an algebraic definition of a function since, if $x$ is a set, it could mean "$f$ is a function from $x$ to the set of pairs of values from $x$." Perhaps $f:x\mapsto x^2$, though I usually say "let $f$ be a function defined by $x\mapsto x^2$" or better "let $f:\mathbb{R}\to\mathbb{R}$ be the function defined by $x\mapsto x^2$." It is not unheard of to write $f=(x\mapsto x^2)$.

You may define a function implicitly, like $2\sqrt{x}\mapsto x/4$, but you must make sure it is well defined. For instance, it cannot be defined for negative numbers since $2\sqrt{x}$ is not negative. It is indeed equivalent to $f(x)=x^2/64$ if you make the domain of $f$ clear.

If I'm parsing it correctly, I think the answer to your last question is that you should use $\mapsto$ when you give an algebraic definition of a function and $\to$ when you give the type (domain and codomain) of a function.

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  • $\begingroup$ Thanks for the response, is there a way in which I can identify the domain and range and the function in one notation, or would it have to be written as $x \mapsto \displaystyle \frac{x^2}{64}$ and $Dom(f)=\mathbb R^+$ or $f : \mathbb R^+ \rightarrow \mathbb R^+$? $\endgroup$ – joshuaheckroodt Sep 6 '17 at 5:17
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    $\begingroup$ (Just in case: don't get the impression that learning math is about learning and using ever more cryptic notations. Clear communication is key.) If there is one, I have never used it. I'm pretty sure I only ever say something like "Let $f:\mathbb{R}^+\to\mathbb{R}$ be defined by $x\mapsto x^2/64$." or "Let $f:\mathbb{R}^+\to\mathbb{R}$ be defined by $f(x)=x^2/64$. If you insist, I could imagine $f(x:\mathbb{R}^+)=x^2:\mathbb{R}$, but don't do this unless you explain your notation to the reader. (Be careful with the word range: it could mean "codomain" or "image.") $\endgroup$ – Kyle Miller Sep 6 '17 at 5:26
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You are making this more complicated than it is.

The proper notation for a function is $$f:X\rightarrow Y:x\mapsto f(x).$$ The part $f:X\rightarrow Y$ means that $X$ is the domain of the function and $Y$ is the target/codomain. The part $x\mapsto f(x)$ explains what happens to elements of $x$. Here a particular $x$ is mapped to $f(x)\in Y$.

In case the domain and target are understood, people often use the bad notation $f(x)=\mbox{some expression}$. For example $f(x)=\sqrt{x}$, however this does not completely define the function as it is not clear what the domain and target are. Indeed comparing the functions $f:[0,1]\rightarrow [0,1]:x\mapsto \sqrt{x}$ and $g:\mathbb{R}^+\rightarrow \mathbb{R}^+:x\mapsto f(x)$, we see that $f$ has a maximum whereas $g$ does not, a huge difference.

The notation $x\mapsto f(x)$ is the same story as it really only specifies $f(x)$.

Writing $f:2\sqrt{x}\mapsto \frac{x}{4}$ is not done. The problem is that you do not specify the domain and a priori it's not clear that any element in $X$ can be written as $2\sqrt{x}$. Indeed, writing $\sqrt{x}$ implies that $x$ should be positive, but when you rewritre this function as $f(x)=\frac{x^2}{64}$, this restriction dissapears.

For sake of completeness, let me give some examples of functions which have weirder domains and codomains.

Consider the map $f:C([0,1])\rightarrow \mathbb{R}:h\mapsto \max_{x\in [0,1]}h(x)$, it's non-trivial that this map is well-defined, i.e. the maximum exists. Here $C([0,1])$ is the set of all real-valued continuous functions on $[0,1]$. Here's another one $g:\mathbb{R}[X]\rightarrow \mathbb{R}[X]:p(X)\mapsto p'(X)$. Here $\mathbb{R}[X]$ is the set of all polynomials in $X$ with real coefficients and $p'(X)$ denotes the derivative of $p(X)$.

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  • $\begingroup$ So, i.e. if I were to correctly write out the function in the question, would it be $f : \mathbb R^+ \rightarrow \mathbb R^+ : 2\sqrt{x} \mapsto \displaystyle \frac{x}{4}$, which means the domain and codomain of $f(x)$ has now been identified. $\endgroup$ – joshuaheckroodt Sep 6 '17 at 5:09
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    $\begingroup$ I have never seen the notation $f:X\to Y:x\mapsto f(x)$ before (in particular, the second colon). Where have you seen this? $\endgroup$ – Kyle Miller Sep 6 '17 at 5:13
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    $\begingroup$ @joshuaheckroodt: You could write that, but it's bad practice. In order for that to make sense you need to explain that any element in $\mathbb{R}^+$ has a unique representation as $2\sqrt{x}$. $\endgroup$ – Mathematician 42 Sep 6 '17 at 5:17
  • $\begingroup$ Thank you @Mathematician42 $\endgroup$ – joshuaheckroodt Sep 6 '17 at 5:21
  • $\begingroup$ @KyleMiller: Are you serious? (Non-offensive, I'm just surprised). I know certain countries shuffle the order around, but other than that, this is standard notation. How do you write it? $\endgroup$ – Mathematician 42 Sep 6 '17 at 5:25

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