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Let $f: \mathbb R^* \to \mathbb R$ with $f(x) = \frac{x+1}x,$ where $\mathbb R^*$ is the set of all real numbers different from zero.
Determine whether or not $f$ is an onto function.

I know that this is not onto. But how do I go about disproving it in a formal way. I know the range does not contain the element $1$ whilst the codomain does. And since the codomain is not equal to the range the function is not onto. But like I said how do I formally disprove it??

Stephen Donovan
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user1138115
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  • Does this answer your question:https://math.stackexchange.com/questions/235548/proving-functions-are-surjective – Bumblebee Jan 09 '23 at 04:33
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    If you've proven $1$ is not mapped to by any real number, that's all you have to do. An onto function maps to every element of the codomain, so if you construct an element of the codomain which isn't mapped to (which you have done correctly) then the function cannot be onto. If you're confused about showing that $1$ isn't mapped to in a more formal sense, I would suggest supposing that there was some $x$ such that $\frac{x+1}x = 1$ and showing how that would lead to a contradiction, so there cannot exist any such $x.$ – Stephen Donovan Jan 09 '23 at 04:35
  • @StephenDonovan I was thinking find the inverse of the function x+1/x as
    follows y= x+1/x

    y= x+1/x xy=x+1

    xy-x=1

    x(y-1)=1

    x=1/y-1

    Next we’ll interchange x and y i.e., replace x with y and y with x. So we’ll get:

    y=1/x-1 . I was thinking that since the domain of the inverse of a function (f) is equal to the range of the actual f() maybe use this method??? Is this actually correct?

     – user1138115 Jan 09 '23 at 04:43
  • That idea works, yes. You can use that to find which elements of the codomain are mapped to, and if you find any element of the codomain which are absent from that set, then you can use that to conclude as I indicated. It's really just the same argument disguised with more algebra, but it could be very useful for finding your counterexample if you don't already have one. – Stephen Donovan Jan 09 '23 at 04:48

2 Answers2

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Hint: Put $f(x)=y.$ Then $(x+1)/x=y.$ From this one obtains $x=\frac{-1}{1-y}.$

Can you see what happens when $y=1?$

Nik
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  • x would be undefined. So that shows that when y=1 does not have a pre-image hence it is not surjective right? But f(x) = y this part how do you formally write this? – user1138115 Jan 09 '23 at 05:10
  • @user1138115 letting $f(x)=y$ is the formal way to begin with and yes you're idea is correct. – Nik Jan 09 '23 at 05:33
  • If a function is onto then for all y ∈ R there exists x∈R* such that y=f(x). Then we can say lets test this condition for y=(x+1)/x right? – user1138115 Jan 09 '23 at 05:38
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Let's recall the following definiton.

A function $f \colon X \to Y$ is called onto (i.e. surjective) if for each $y \in Y$, there exists a $x \in X$ such that $f(x) = y$, which means that the range is exactly the codomain $f(X) = \{f(x) \mid x \in X\} = Y$.

Therefore, a function $f \colon X \to Y$ is not onto if and only if there exists a $y \in Y$ such that $f(x) \neq y$ for all $x \in X$.

For the given function $f \colon \mathbb{R}^{*} \ni x \mapsto 1 + \frac{1}{x} \in \mathbb{R}$, we know that there exists a $y = 1 \in \mathbb{R}$ and $f(x) = 1 + \frac{1}{x} \neq 1 = y$ for all $x \in \mathbb{R}^{*}$, which completes the proof.

Your approach works as well and gives a finer consequence. In fact, you have shown that the restriction function $g \colon \mathbb{R}^{*} \to \mathbb{R} \setminus \{1\} = \{y \in \mathbb{R} \mid y \neq 1\}$ is a one-to-one and onto function (i.e. bijection).

However, in order to give a formal proof, it is better to avoid using the imprecise phrase "the inverse of the function" if you haven't proved that the function is one-to-one and onto. Because a function is invertible if and only if it is one-to-one and onto.

CYK
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