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Edit: Man, I actually had a bachelor's completed when I asked this dumb question, like even more than the terry tao dumb thing. Don't judge me! But to be fair even when I was in calculus we're always asked to 'find the domain' of single variable functions or 'sketch the domain' bivariate functions.

Suppose we have a function, say, $f(x) = x+2$. Its domain is $\mathbb{R}$. How do you prove this? Or is this something not needed to be proven since it is "defined" $\forall$ x $\in \mathbb{R}$?

If to be proven (ignore if not needed): Induction seems to do the trick but that would only cover positive integers. I guess I could cover negative integers using a similar argument. Maybe I could even extend to all rational numbers. What about irrational numbers then?

If not to be proven (ignore if not needed): So highschool teachers should say the domain of $f(x) = x+2$ is $\mathbb{R}$ by definition?

BCLC
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    If you have a function, the definition of the function has to contain the domain of the function, otherwise it is not reasonable to call it a function. However, in school it is handled a bit sloppy. If pupils are asked for the "domain of a function", it is often meant as somehow the "maximal domain", where we can define the function. But this is strongly dependent on what you know. For instance, your example could also be defined on the complex numbers or only over a finite field etc... – Alex Jul 04 '14 at 12:21
  • So is the question "find the domain of each function" actually wrong? – BCLC Jul 04 '14 at 12:28
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    It is not reasonable if the teacher has not defined what he means with that kind of question...:-). And mathematically I would stick with naslundx answer. – Alex Jul 04 '14 at 12:33
  • @Alex relevant comment? https://math.stackexchange.com/questions/2115693/how-to-prove-the-domain-of-a-function#comment4351201_2115693 – BCLC Dec 01 '20 at 00:11

2 Answers2

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You do not and can not prove the domain of a function, you specify it (either explicitly or implicitly) for the function you're discussing.

The function $f(x)=x+2$ is defined for all $x \in \Bbb R$ and hence its domain may be $\Bbb R$. However, you may also define the function $g(x) = x+2$ for $g: [0,1] \to [2,3]$ and then the domain of $g$ is simply $[0,1]$, even though it can be extended to $\Bbb R$.

Also, note that the function $f$ is also defined for all $x \in \Bbb C$ and hence its domain could also be said to be $\Bbb C$.

naslundx
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  • In highschool textbooks, students are asked "find the domain of each function", that is find all values for which a given function is defined. If one cannot prove one's answer, isn't the question actually wrong? – BCLC Jul 04 '14 at 12:26
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    @BCLC Terminology is usually simplified for pre-university studies. A better question would be "find all the (real) values for which the function is defined". In high school, I suppose one does not deal with restricted functions, so that definition of domain works. – naslundx Jul 04 '14 at 12:29
  • Suppose a student (highschool or o/w) asks why x+2 is defined $\forall \mathbb{R}$...? – BCLC Jul 04 '14 at 12:31
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    It is an axiom of the real numbers: there is an operation called "addition" $(x,y)\in\mathbb{R}^2\mapsto x+y\in\mathbb{R}$. Since $2\in\mathbb{R}$ and if we assume $x\in\mathbb{R}$ the axiom states that $x+2\in\mathbb{R}$. –  Jul 04 '14 at 12:50
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    To a high school student I would use the geometric explanation of addition: $x$ is a point on the number line, move it 2 units to the right, and you get $x+2$. To a college student in mathematics, I would start with the natural numbers and their addition operation that satisfies the Peano axioms; then I would construct the integers as equivalence classes $(m,n)~(p,q) \iff m+q=n+p$, and define their addition; then I would construct the rational numbers as equivalence classes $(a,b) \sim (c,d) \iff ad = bc$, and define their addition; ... – Lee Mosher Jul 04 '14 at 12:52
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    then I would construct the real numbers as equivalence classes of Cauchy sequences of real numbers, and define their addition. Or I would take the shortcut of @Leonhard and go straight to the axioms for the real numbers themselves. – Lee Mosher Jul 04 '14 at 12:52
  • @LeeMosher http://math.stackexchange.com/questions/856299/how-do-you-prove-a-function-is-defined-for-a-a-certain-set – BCLC Jul 04 '14 at 13:01
  • @Leonhard http://math.stackexchange.com/questions/856299/how-do-you-prove-a-function-is-defined-for-a-a-certain-set – BCLC Jul 04 '14 at 13:01
  • @naslundx relevant comment? https://math.stackexchange.com/questions/2115693/how-to-prove-the-domain-of-a-function#comment4351201_2115693 – BCLC Dec 01 '20 at 00:11
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Yes, absolutely, it should be defined. I had the same experience with my high-school teacher who actually got mad. I do not remember exactly, but apparently it is assumed (without stating) that the domain is the largest possible set on which the (rational, real, complex?) function is defined. I mean, something like $\sqrt{x-4}$ could als be defined on $[2014,\infty)$. So always ask for a specification of the domain.

Nicky Hekster
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  • In highschool textbooks, students are asked "find the domain of each function", that is find all values for which a given function is defined. Is the question actually wrong? – BCLC Jul 04 '14 at 12:31
  • @BCLC That question is not wrong, but that is not the statement of the post here. – Nicky Hekster Jul 04 '14 at 12:34
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    @NickyHekster Could it also be defined for $x=3$ in real numbers even thought it would be discontinuous? – user599310 Aug 19 '20 at 21:53
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    @user599310 Yes you can do that. – Nicky Hekster Aug 20 '20 at 06:49
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    @NickyHekster Doesn't this violate the definition of function where every element of the domain must associate with exactly one element of the range? I mean if we have the domain for all real numbers for $x=3$ the function is undefined therefore we don't have an association of an element of the domain with an element of the range. – user599310 Aug 20 '20 at 11:02
  • @user599310 in complex analysis i think you would just say $\frac 1z$ is like 'meromorphic' at $z=0$ and therefore include $z=0$ in the domain of $\frac 1z$. however for reals... maybe the idea is $f(x): = \sqrt{x-4}$, $x \ge a$, for some $a \ge 4$ and then say $f(x):= -78 \pi^2$, $x=3$. Therefore we extend domain of $f$ from $[4,\infty)$ to ${3} \cup [4,\infty)$ – BCLC Nov 17 '20 at 22:12