# Proving a relation is a mapping

In my abstract algebra book in the preliminary section there is a review of relations, mappings, etc. In the practice problems there are a few problems that check your understanding.

Example 1:

$$f(\frac{p}{q}) = \frac{3p}{3q}$$

This one is trivially a mapping because no matter what p, q you choose you will get a unique value from Q to Q.

Example 2:

$$f(\frac{p}{q}) = \frac{p + q}{q^2}$$

This one is not a mapping. A counter-example is $$f(1/2) \neq f(2/4)$$. These both map to different values so this cannot be a mapping.

But I am left kind of confused on the "proof" part. Showing a relation is not a mapping seems to be difficult because you have to search for a counter-example of a potentially hard problem. But proving something is a mapping seems extremely difficult. Is it possible to construct some kind of contradiction to prove it, or are you left sort of "guessing and checking"? I am self learning at the moment so I don't really have a professor to ask this to.

• Which abstract algebra book is this? I'd be interested to know, because the presentation of this is lacking in rigour and it may be instructive to see the original. Aug 14, 2022 at 22:00
• @PrimeMover Abstract Algebra: Theory and Applications. Available free. abstract.ups.edu/aata/sets-exercises.html here are the exercises. Aug 14, 2022 at 22:10
• Thanks. It's traditional, incidentally, when citing a work, to say who the author is as well, and its date of publication. There are many such books with similar titles. Suggest you might want to edit this information into the question itself. Aug 15, 2022 at 5:10

Proving that a mapping $$f$$ is well-defined (or equivalently that $$f$$ is a mapping) is something that comes up a lot when studying structures defined up to equivalence. The usual way to prove such a claim is to take two representatives $$a_1,a_2$$ of the same equivalence class (in you example you would take fractions $$a_1=\frac{p_1}{q_1}$$ and $$a_2=\frac{p_2}{q_2}$$ such that $$a_1=a_2 \Leftrightarrow p_1q_2=p_2q_1$$) and show that $$f(a_1)=f(a_2)$$. Sometimes this might present some difficulty but this is the standard method.

On the other hand, to prove that $$f$$ is not well defined requires you to find a specific counterexample which, again, can prove a bit tricky depending on the specific problem at hand.

That being said, I wouldn't stress too much about it, at least for now, since you will surely encounter many such proofs throughout your abstract algebra journey.

• I think you are underplaying the blatant abuse of notation that the OP is being subjected to. Aug 14, 2022 at 21:28
• @RobArthan Indeed but I guess this is what his reference book is using. Personally, I find the discussion about well-defined mappings to be unnecessary this early on since these examples can only further confuse the reader. Imo, this should be discussed either rigorously by defining equivalence classes or not at all. Aug 14, 2022 at 21:33
• I see, yes the book barely covered the definition of an equivalence class in the preliminaries but everything surrounding it was handwaved. I will wait for further chapters to learn more then. Thank you for clearing this up. I was very confused why proofs were so rigorously pursued and then these examples were given as proof. Aug 14, 2022 at 21:59

I've looked at the source work you're using: Thomas W. Judson's Abstract Algebra: Theory and Applications (which appears to be 2015).

It's a breathless avalanche of facts jumbled together with random examples and incidental asides, followed at the bottom of the slope by a morass of exercises. Comes across like a typical high-school textbook. Its fairly excusable that you have probably missed the crucial paragraph in the section "1.2 Sets and Equivalence Relations":

Consider the relation $$f: \mathbb Q \to \mathbb Z$$ given by $$f(p/q)=p$$. We know that $$1/2 = 2/4$$, but is $$f(1/2)$$ $$1$$ or $$2$$? This relation cannot be a mapping because it is not well-defined. A relation is well-defined if each element in the domain is assigned to a unique element in the range.

Appallingly muddled, particularly when later on down the page he uses exactly the same notation where $$p/q$$ is definitely specified that $$p/q$$ is "a rational number expressed in its lowest terms with a positive denominator."

I suggest that if you continue with this work you may want to bear in mind that it is less than optimal quality, and possibly back up your study with something a little more solid.

I continue to plug the useful An Introduction to Abstract Algebra by Thomas A. Whitelaw (Blackie, 1978) (if you can find it -- all exercises are with complete answers), or (tougher but rewarding) Allan Clark's Elements of Abstract Algebra (Dover, 1971) (it has no answers to exercises unfortunately, and the latter are not easy unless you are really on top of your material).

• Fantastic answer, really impressed by the fact that you are able to consider the context and provide suggestions. The minute I saw your comment I was about to ask you to write an answer, but it turns out you've already written one. Great stuff here. Aug 15, 2022 at 9:50
• I am glad you posted this. I was really feeling dumb with this textbook especially considering it is being used for the MIT OCW class. It certainly is difficult to deal with. I will look into your suggestions as alternatives. Thank you ! Aug 16, 2022 at 15:25