0
$\begingroup$

I have these functions which I am supposed to analyse on surjectivity and injectivity. My problem is that I understood what both mean and seeing a graph, I can tell if it is either, but I don't know how to write down the proof in a way that is applicable to any function if that's possible? If you could point me to good reference material, I'd be happy too. I have found more than enough, but I'm having a hard time applying it to my own problem, though.

$f:D_f \rightarrow [0, \infty[ \ with \ f(x)= \sqrt{x}$

Now I know that it it's injective and not surjective. How would I prove that?

  • 0
    Your function is surjective; Generally, for injectivity you just show that if $f(x)=f(y)$ then $x=y$. For surjectivity, it's usually harder, but principle is that given some $y$ in codomain, you are trying to find some $x$ in domain of $f$ such that $f(x)=y$ – 2017-02-02
  • 0
    @user160738 Oh right, I've forgot to include the interval in my thought process... Now I know what the definition for either of those are, but as far as I know, there are certain rules of having the right step sequence. I've learnt that if injectivity is true, I need to prove it with a general proof and if its false, with an example. Could you write down the proof with the right syntax for me? – 2017-02-02
  • 0
    Just apply the definition: if $\sqrt{x}=\sqrt{y}$ then is it possbily true that $x\neq y$? and for any $a\ge 0$, can you find some $x$ such that $\sqrt{x}=a$? – 2017-02-02

1 Answers 1

0

I guess that by $D_f$ you mean the domain of $f$, right?

The problem "is it injective or surjective or both"? Without the precise specification of the domain/codomain pair, is meaningless. Let me explain. A function $f: A \rightarrow B$ is defined to be the triple composed by: the set $A$, the set $B$, and the set-theoretic map $f$. It depends on the sets $A, B$ as well as the map $f$. Now, take your function:

$f:D_f \rightarrow [0, \infty[ \ with \ f(x)= \sqrt{x}$

If you see $D_f$ as a subinterval of the real numbers, then your functions is certainly both injective and surjective. But if $D_f$ you mean, for example a subset of the rational numbers, then the function is injective but not anymore surjective, which is perhaps the reason why you say that your function "is not surjective".

In other words, you must better specify the pair domain/codomain. If on the other hand $D_f$ is the set of nonnegative real numbers, then the function is also surjective, and this claim is equaivalent to the property (true in the field of real numbers) that every positive real number is the square root of another (positive) real number, which depends on the construction of the field $\mathbb R$, and it is by no means trivial.

Let's see injectivity. The claim that a given function is injective is equivalent to the following: $$ f(x) = f(y) \Rightarrow x = y $$

So let us suppose that for some $x, y$ we have $f(x)=f(y)$, i.e. $\sqrt x=\sqrt y$. Now square this equality and obtain $(\sqrt x)^2=(\sqrt y)^2$. As the square root of a (positive) number $a$, $\sqrt a$, is always defined to be a positive number, this amounts to say simply $x=y$, which proves that $f$ is indeed injective.

As I mentioned before, proof of surjectivity is definitely less trivial, and goes to the heart of the construction of the field of real numbers.

  • 0
    I'm not sure if I understood everything you said, but I guess in this case the domain and codomain are both $[0, \infty [$ So now thats a little clearer, but I still don't know how to write down the proof. – 2017-02-02
  • 0
    @Cheeseball I have edited and added the proof of injectivity, hope it is more clear now. – 2017-02-02