6
$\begingroup$

This is from Discrete Mathematics and its Applications enter image description here

My question is on 22c. From the book, enter image description here

I inferred from the question that all of the listed mathematical expressions are functions. Is the mathematical expression $(x + 1)/(x + 2)$ even a function though? What I got from my other question, Why is this not a function? was that a function needs to have exactly one output from every possible input in the domain. However this function doesn't have an output for $x = -2$, which is the domain of $\mathbb{R}$.

$\endgroup$
3
  • $\begingroup$ the misconception here is that the function is a triple; it has domain, it has range and it has a rule that says what happens to each element in the domain. usual one only sees the rule given prominent place and other two are not mentioned at all. $\endgroup$
    – abel
    Jan 17, 2015 at 12:40
  • $\begingroup$ There are three parts to the definition of a function? I thought it was just map one element of a set to one element of another set. $\endgroup$ Jan 27, 2015 at 20:04
  • $\begingroup$ @committedandroider—which sets? $\endgroup$
    – wchargin
    Jul 19, 2015 at 22:44

2 Answers 2

7
$\begingroup$

You are essentially correct. While $f(x)=\frac{x+1}{x+2}$ is actually a function, its domain is not $\mathbb{R}$ since $f$ is not defined for $x=-2$. Thus it cannot be a function from $\mathbb{R}$ to $\mathbb{R}$, much less a bijection.

$\endgroup$
4
  • $\begingroup$ I am so confused. The book's definition of function from A to B is an assignment of exactly one element of B to each element of B. Basically, each element of A can be assigned exactly one element of B. If it doesn't meet this definition of function, how can it be a function? $\endgroup$ Jan 17, 2015 at 3:25
  • 2
    $\begingroup$ If you assume that $A=B=\mathbb{R}$, then $f$ is not a function from $A$ to $B$, which is what is important for your problem. But $f$ is still a function on domains not containing $-2$, since $f$ takes on a exactly one value for every other real number. $\endgroup$
    – AMPerrine
    Jan 17, 2015 at 3:31
  • $\begingroup$ so overall definition of function is one output for one input. But on a specific domain, function must be able to produce an output for every input in the domain? $\endgroup$ Jan 17, 2015 at 3:34
  • 4
    $\begingroup$ Yes, that's right. If you ask, "Is this a function?" without specifying a domain, we assume the domain is everything for which the expression is defined and only consider the uniqueness condition. Once you include the domain $A$ in the question, you can ask whether $f(x)$ even makes sense for every $x$ in $A$. A couple more examples: $f(x)=\pm x$ is not a function on $\mathbb{R}$. But it is a function on the set $\{0\}$. $g(x)=\frac{1}{x}$ is not a function on $\mathbb{R}$. But it is a function on $\mathbb{R}\setminus \{0\}$. $\endgroup$
    – AMPerrine
    Jan 17, 2015 at 3:47
4
$\begingroup$

You're correct that $f$ is not defined at $-2$. It is still a function, though; its domain is $\mathbb R \setminus \{ -2 \}$, and its range is $\mathbb R \setminus \{ 1 \}$.

So we have, say, $f : \mathbb R \setminus \{ -2 \} \to \mathbb R \setminus \{ 1 \}$. Is this function a bijection? Yes, it is; its inverse is $$f^{-1}(y) = \frac{1 - 2y}{y - 1},$$ which is defined everywhere on the codomain of $f$. But is this function a bijection from $\mathbb R$ to $\mathbb R$, as the question asks? Clearly not.

$\endgroup$
4
  • $\begingroup$ What definition of function are you using? The one I am using says that a function from A to B means that every element of A must be assigned exactly one element of B. If it doesn't meet this requirement, how can this still be a function? $\endgroup$ Jan 17, 2015 at 3:31
  • 4
    $\begingroup$ @committedandroider But in this answer, $A = \mathbb R\setminus \{-2\}$. So why doesn't the function meet the requirements? You would have a fair point if you meant that we shouldn't use $f$ to refer to a function on a different domain than the one originally described, but this kind of equivocation is pretty common in undergraduate math. (And it probably contributes to misconceptions that students have about what a function is.) $\endgroup$
    – Erick Wong
    Jan 17, 2015 at 3:52
  • $\begingroup$ oh so definition of function depends on what domain you define $\endgroup$ Jan 17, 2015 at 3:58
  • 2
    $\begingroup$ @committedandroider Right. Part of the definition of any function is its domain and its codomain. If I let $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{Z} \to \mathbb{Z}$, the functions are different, even if, say, $f(x) = x$ and $g(x) = x$. $\endgroup$
    – wchargin
    Jan 17, 2015 at 4:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.