I have proved the function being bijective, but I don't know how to inverse it. So if someone could please show the steps how to do it I would be very grateful. Below is the provided proofs for injective and surjective.

f: $\mathbb{Q} \rightarrow \mathbb{Q}$, f(x) = 2x+1
Let a,b $ \in \mathbb{Q}$, now assume f(a) = f(b)
Show that a=b:
Thus a=b function is injective.

Let a $\in \mathbb{Q}$, claim that $\frac{a-1}{2}$ maps to a. $\frac{a-1}{2} \in \mathbb{Q}$ since a $ \in \mathbb{Q}$
f($\frac{a-1}{2}$) = 2($\frac{a-1}{2}$)+1
Thus function is surjective.


We want to find the inverse to the function $f$, we denote this inverse as $f^{-1}=g$. Now by definition of the inverse we have $f(g(x))=x$. This just means that $g$ 'undoes' what $f$ did with $x$. So if you apply $g$ to $x$ and then $f$ we get out original $x$ back. So the composition of a function with its inverse is by definition the identity map.

Mathematically this translates into $$f(g(x)) = x.$$ We now just use the fact that we know what the function $f$ is to obtain the following expression $$2g(x)+1=x.$$ Here we just treated $g(x)$ as a variable and filled it into the expression for $f$. We can solve this now for $g$ to get $$g(x) = \frac{x-1}{2}.$$ We can now check to see that indeed $f(g(x)) = x$ and $g(f(x))=x$. Thus the $g$ that we found is indeed the inverse of $f$.

  • $\begingroup$ Is it possible if you could show it more thoroughly? I am having a hard time understanding how to go through the steps of it. Thanks! $\endgroup$ – J_SNSD Dec 1 '14 at 10:31
  • $\begingroup$ Sure, I will update my answer. $\endgroup$ – Slugger Dec 1 '14 at 10:32
  • $\begingroup$ I get it now! Thank you so much :) $\endgroup$ – J_SNSD Dec 1 '14 at 10:38
  • $\begingroup$ no worries, glad to help $\endgroup$ – Slugger Dec 1 '14 at 10:39
  • $\begingroup$ Nice. A more intuitive alternative to the usual practice. $\endgroup$ – Dan Christensen Dec 1 '14 at 17:52

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.