Is the inverse of this function unique

Question

Let $f$ be a function from any set(Say $K$) to any set (say $P$)

Now: $f(x)=2x+1$

My question:Is it necessary that the inverse of the function is $\frac{x-1}{2}$?

This is a problem given in my brother's book along with the answer but somehow I believe that some other may exist as its inverse though I could find none.

Answer 1

It is unique. Solving $x=2y+1$ for $y$ shows $y=(x-1)/2$ is the only solution.

Answer 2

$$\begin{equation}\begin{split}y=2x+1&\Leftrightarrow y-1=2x,\\&\Leftrightarrow x=\dfrac{y-1}{2}.\end{split}\end{equation}$$

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So the inverse of $f(x)=2x+1$ is $f^{-1}(x)=\dfrac{x-1}{2}$ which is unique.

Answer 3

Take a look at this page: http://en.m.wikipedia.org/wiki/Bijection,_injection_and_surjection It will give you a good hindsight about the topic. (You can find the answer to your specific question at bijection section)

In your case, yes, the two functions are bijective. One way to check this is by calculating ${fof^{-1}(x)} $ and the other way around and you should get x. But you also have to consider the domain and co-domain of your function. For example, ${\sqrt x}$ is the inverse of ${x^2}$, but if you take a look at the domains, they are not bijective.

Answer 4

Maybe a more algebraic point of view may help. I do not know what is your background but I hope you get something out of my answer. Let $X$ be a set and consider \begin{equation*} X^X= \{f \mid f:X\to X \} \end{equation*} i.e. the sets of functions from $X$ to $X$. To this set, we associate the operation "$\circ$" of composition of functions. We notice that the pair $(X^X,\circ)$ is a monoid (you need to be able to verify this). Thus, we can look at $(X^X)^\times$, the set of invertible elements of $X^X$ relative to the composition $\circ$ which forms a group. The elements of $X^X$ are bijective functions from $X$ to $X$.

But, inverses in groups are unique. In fact, suppose we have two elements $g_1,g_2\in X^X$ such that for $f\in X^X$, we have $g_1\circ f=f\circ g_1=\mathrm{id}_X$ and $g_2\circ f=f\circ g_2=\mathrm{id}_X$. We then have \begin{equation*} g_1=g_1\circ\mathrm{id}_X =g_1\circ ( f\circ g_2)=(g_1\circ f)\circ g_2=\mathrm{id}_X\circ g_2=g_2 \end{equation*} which shows $g_1=g_2$. And so, the inverse is unique.

Relative to your example, your set is $X=\mathbb{R}$ and it is easy to verify that the linear function you gave is bijective. This shows that it's inverse is unique.

As it may look, this answer to your question is probably "too much". My idea was to give you an insight into the usefulness of algebraic methods. I hope you learned a bit from this!