Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I teach math and have had this debate with a fellow teacher. I believe the domain of a function is really not important to determine if a function is or is not injective if there is no "real life" context.

What do you think?

would you find the domain of a function like

$f(x)=1-\ln(2-e^{2x})$

to show if the function is injective? If so, why?

share|improve this question
9  
It is extremely important to know specifically what the domain is. For example, the function $g:\mathbb{R}\rightarrow \mathbb{R}$ given by $g(x)=x^2$ is not injective, but the function $h:[0,\infty)\rightarrow \mathbb{R}$ given by $h(x)=x^2$. The only difference in these functions is their domains. Giving the rule alone is not enough. A function is not defined by a rule--it needs a rule, a domain, and a range before it is a function. –  MPW Mar 1 at 12:37
    
I meant exactly that when I mentioned the "real life" context. Of course we can change the domain and make a non-injective function into an injective one but if no restriction is given to the domain, shouldn't we consider the biggest domain possible? And in this case should indicating it be a step in the correct resolution of the exercise? Or is it enough to say: take $x, y \in D_f$? –  Concept7 Mar 1 at 13:09

3 Answers 3

up vote 6 down vote accepted

By definition functions have domains, proving injectivity without considering the domain is something devoided of sense, the domain of a function is part of its essence.

In practice there can be times in which whatever the domain is, the proofs will look the same (if the functions are in fact injective), but they are necessarily different because the 'first' step in proving injectivy of a function is taking 'two' arbitrary elements on its domain. You can't do this if you don't focus on the domain.

This doesn't mean that you have to 'find' the domain of a function to prove it is injective. Taking your example and letting $D_f$ denote the domain of $f$, the aforementioned first step would just be: let $x,y\in D_f\,\ldots$

share|improve this answer
    
Yes, I also think that the domain is important but if no context is given then you don't have to analyse the domain... just say, as you put it: $x,y \in D_f$. The debate was based on this: is finding the domain a necessary step in answering if a function is injective? Do you believe it is? –  Concept7 Mar 1 at 13:05
1  
No, as I've hinted in my answer, I don't think it is necessary. A function necessarily has a domain, if you're given a function, you're given a domain. If this was not the case, then maybe I'd need to change my answer, but it is the case. To further exemplify, taking your example, let $x,y\in D_f$ and assume that $1-\ln (2-e^{2x})=1-\ln (2-e^{2y})$. Then $\ln (2-e^{2x})=\ln (2-e^{2y})$ from where one infers that $2-e^{2x}=2-e^{2y}$ (because $\ln$ is injective). It follows that $e^{2x}=e^{2y}$ and consequently $x=y$. This completes the proof in a way that makes sense without 'finding' $D_f$. –  Git Gud Mar 1 at 13:10
3  
To stress the importance of having an explicit domain in mind, note that the exponential function is not injective if you consider it as having the larger domain $\mathbb{C}$. In fact, it is periodic with period $2\pi i$. It is only the restriction to the real line you had in mind that lets you have injectivity. –  MPW Mar 1 at 13:28
    
oh.. that is true! I wasn't even considering that because the students to which the exercise was presented don't even know the existence of $\mathbb{C}$ and it is said that we are working in $\mathbb R$. But that is a very good point! –  Concept7 Mar 1 at 13:35

A function $f:\ A\to B$ is tantamount to a subset $G_f\subset A\times B$ having certain properties. In most cases $A$ and $B$ are clearly specified in advance, and one can then start right away to investigate whether $f$ is injective or not.

Very often $A$ and/or $B$ have to be surmised from the context. While the exact envisaged range $B$ (e.g., ${\mathbb R}$ or ${\mathbb C}$) is not relevant for injectivity, the exact domain $A$ certainly is.

In cases where the function $f$ is given as an expression, e.g., $$f(x):=\qquad {x\over 1+x^2},\quad \cos x, \quad 1-\log\bigl(2-e^{2x}\bigr),\quad {\rm etc.},$$ the tacit understanding is the following: Consider as domain the set of all $x$ envisaged in the context (integer, real, complex, or otherwise) for which the expression can be evaluated without asking questions. In this sense $0$ would not belong to the domain of $x\to{\sin x\over x}$. If one wants to restrict $f$ to a smaller domain $A$, say in order to make it injective, one has to specify this domain explicitly.

share|improve this answer

I agree with Christians and Gits point, that the domain is an essential part of a function. But the problem here is also one of terminology. Especially in education before university, no clear distinction is made between the term, which is used to describe the combination of more elementary function to a new one, and the function itself.

The term $x⋅x$ (or $x^2$ by convention) can describe a lot of functions depending on the domain one chooses for the free variable $x$. It is valid for any set $M$ endowed with an operation $M×M →M$.

In school math the default domain is usually taken to be $ℝ$ or its biggest subset on which the function is well-defined. This particular choice of domain does however not make the domain irrelevant!

Like MPW mentioned in his comments, the exponential function is only injective on $ℝ$, but not on the bigger domain $ℂ$ (or even on an arbitrary manifold with affine connection, in general), while $x^2$, although not injective on $ℝ$, is injective when restricted to a suitable subset.

So it is clear, that the domain is vital information to determine the injectivity of a function. The only way in which it is not important is the following:

If a function with domain $M$ (e.g. $ℝ$) is injective, it is also injective after being restricted to any subset of $M$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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