1
$\begingroup$

My question is: Given sets $A$ = {$a_1, a_2$} and B = {$b_1$}, let the function $f$ from $A$ to $B$ be given by the following set of ordered pairs, $f$ = { ($a_1, b_1$), ($a_2, b_1$) }. If $f$ has an inverse function, call it $g$, and write $g$ as a set of ordered pairs. If $f$ does not have an inverse function, explain why it doesn’t.

I was thinking that there cannot be an inverse function because the domain of the inverse function, $g$ = {($b_1, a_1$), ($b_1, a_2$)}, is $b_1$, which is the same and is pointing to different ranges. For a function to exist the domain has to be unique correct?

$\endgroup$
  • $\begingroup$ Your reasoning is correct but your wording is strained. "The domain of the inverse function g"=> g is not the inverse function and g is not a function at all.. The domain of the relation g is b1. "which is the same" The same as what? "Pointing to different ranges". It's a single value pointing to two values so so the range is {a1,a2} is a single range. Not two different ranges. "the domain has to be unique". {b1} is unique. $\endgroup$ – fleablood Dec 11 '15 at 17:43
  • $\begingroup$ Better wording might be. a1 and a2 are different values in the domain of f yet f(a1) and f(a2) are the same value b1. (Such a function by definition is not injective.) Such a function can not have an inverse, g, as g(b1) would be defined to be the value x such that f(x) = b1. As there are two such values, a1 and a2 the definition is ambiguous. (And if we specified one over the other, say g(b1) = a1. Then g(f(a2)) = a1 and not a2 failing the condition that an inverse does invert all values.) $\endgroup$ – fleablood Dec 11 '15 at 17:53
0
$\begingroup$

Your reasoning is basically correct (you need to assume $a_1\neq a_2$ though!)

Your wording is a little bit off though, I believe. The range is the set of all possible values a function has. A single value (like $a_1$ or $a_2$) is just called, well, a value. For example, the range of $f$ is $B$. The domain of a function is again a set: the set of all possible "inputs". In your case, the domain of $f$ is $A$. Calling $g$ an inverse function is a bit troublesome, since it is of course not a function. You can call $g$ the inverse relation of $f$. The convention "Function = Graph" makes every function surjective (onto) by default. So eventually, you want to define functions in the following way: $$f = (A, \{ (a_1,b_1),(a_2,b_1) \}, B)$$ or similarly. Here $f$ is surjective so it does not actually make a difference, but if you had $B = \{b_1,b_2\}$ with $b_1\neq b_2$ it would matter.


Here is a more common way to prove your statement:

If $f$ has an inverse, then $f$ is bijective, hence injective (one-to-one), but $f(a_1)=f(a_2)$ even though $a_1\neq a_2$ (Contradiction).

$\endgroup$

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.