Very early difficulties II

Very early difficulties II

This is the second part of a series of posts about certain difficulties math students have in the very early stages of studying abstract math. The first post, Very early difficulties in studying abstract math, gives some background to the subject and discusses one particular difficulty: Some students do not know that it is worthwhile to try starting a proof by rewriting what is to be proved using the definitions of the terms involved.

Math StackExchange

The website Math StackExchange is open to any questions about math, even very easy ones. It is in contrast with Math OverFlow, which is aimed at professional mathematicians asking questions in their own field.

Math SE contains many examples of the early difficulties discussed in this series of posts, and I recommend to math ed people (not just RUME people, since some abstract math occurs in advanced high school courses) that they might consider reading through questions on Math SE for examples of misunderstanding students have.

There are two caveats:

  • Most questions on Math SE are at a high enough level that they don’t really concern these early difficulties.
  • Many of the questions are so confused that it is hard to pinpoint what is causing the difficulty that the questioner has.

Connotations of English words

The terms(s) defined in a definition are often given ordinary English words as names, and the beginner automatically associates the connotations of the meaning of the English word with the objects defined in the definition.

Infinite cardinals

If $A$ if a finite set, the cardinality of $A$ is simply a natural number (including $0$). If $A$ is a proper subset of another set $B$, then the cardinality of $A$ is strictly less than the cardinality of $B$.

In the nineteenth century, mathematicians extended the definition of cardinality for infinite sets, and for the most part cardinality has the same behavior as for finite sets. For example, the cardinal numbers are well-ordered. However, for infinite sets it is possible for a set and a proper subset of the set to have the same cardinality. For example, the cardinality of the set of natural numbers is the same as the cardinality of the set of rational numbers. This phenomenon causes major cognitive dissonance.

Question 1331680 on Math Stack Exchange shows an example of this confusion. I have also discussed the problem with cardinality in the section Cardinality.

Morphism in category theory

The concept of category is defined by saying there is a bunch of objects called objects (sorry bout that) and a bunch of objects called morphisms, subject to certain axioms. One requirement is that there are functions from morphisms to objects choosing a “domain” and a “codomain” of each morphism. This is spelled out in Category Theory in Wikibooks, and in any other book on category theory.

The concepts of morphism, domain and codomain in a category are therefore defined by abstract definitions, which means that any property of morphisms and their domains and codomains that is true in every category must follow from the axioms. However, the word “morphism” and the talk about domains and codomains naturally suggests to many students that a morphism must be a function, so they immediately and incorrectly expect to evaluate it at an element of its domain, or to treat it as a function in other ways.


If $\mathcal{C}$ is a category, its opposite category $\mathcal{C}^{op}$ is defined this way:

  • The objects of $\mathcal{C}^{op}$ are the objects of $\mathcal{C}$.
  • A morphism $f:X\to Y$ of $\mathcal{C}^{op}$ is a morphism from $Y$ to $X$ of $\mathcal{C}$ (swap the domain and codomain).

In Question 980933 on Math SE, the questioner is saying (among other things) that in $\text{Set}^{op}$, this would imply that there has to be a morphism from a nonempty set to the empty set. This of course is true, but the questioner is worried that you can’t have a function from a nonempty set to the empty set. That is also true, but what it implies is that in $\text{Set}^{op}$, the morphism from $\{1,2,3\}$ to the empty set is not a function from $\{1,2,3\}$ to the empty set. The morphism exists, but it is not a function. This does not any any sense make the definition of $\text{Set}^{op}$ incorrect.

Student confusion like this tends to make the teacher want to have a one foot by six foot billboard in his classroom saying


However, even that statement causes confusion. The questioner who asked Question 1594658 essentially responded to the statement in purple prose above by assuming a morphism that is “not a function” must have two distinct values at some input!

That questioner is still allowing the connotations of the word “morphism” to lead them to assume something that the definition of category does not give: that the morphism can evaluate elements of the domain to give elements of the codomain.

So we need a more elaborate poster in the classroom:

The definition of “category” makes no requirement
that an object has elements
or that morphisms evaluate elements.

As was remarked long long ago, category theory is pointless.

English words implementing logic

There are lots of questions about logic that show that students really do not think that the definition of some particular logical construction can possibly be correct. That is why in the chapter on definitions I inserted this purple prose:

A definition is a totalitarian dictator.

It is often the case that you can explain why the definition is worded the way it is, and of course when you can you should. But it is also true that the student has to grovel and obey the definition no matter how weird they think it is.

Formula and term

In logic you learn that a formula is a statement with variables in it, for example “$\exists x((x+5)^3\gt2)$”. The expression “$(x+5)^3$” is not a formula because it is not a statement; it is a “term”. But in English, $H_2O$ is a formula, the formula for water. As a result, some students have a remarkably difficult time understanding the difference between “term” and “formula”. I think that is because those students don’t really believe that the definition must be taken seriously.

Exclusive or

Question 804250 in MathSE says:

“Consider $P$ and $Q$. Let $P+Q$ denote exclusive or. Then if $P$ and $Q$ are both true or are both false then $P+Q$ is false. If one of them is true and one of them is false then $P+Q$ is true. By exclusive or I mean $P$ or $Q$ but not both. I have been trying to figure out why the truth table is the way it is. For example if $P$ is true and $Q$ is true then no matter what would it be true?”

I believe that the questioner is really confused by the plus sign: $P+Q$ ought to be true if $P$ and $Q$ are both true because that’s what the plus sign ought to mean.

Yes, I know this is about a symbol instead of an English word, but I think the difficulty has the same dynamics as the English-word examples I have given.

If I have understood this difficulty correctly, it is similar to the students who want to know why $1$ is not a prime number. In that case, there is a good explanation.

Only if

The phrase “only if” simply does not mean the same thing in math as it does in English. In Question 17562 in MathSE, a reader asks the question, why does “$P$ only if $Q$” mean the same as “if $P$ then $Q$” instead of “if $Q$ then $P$”?

Many answerers wasted a lot of time trying to convince us that “$P$ only if $Q$” mean the same as “if $P$ then $Q$” in ordinary English, when in fact it does not. That’s because in English, clauses involving “if” usually connote causation, which does not happen in math English.

Consider these two pairs of examples.

  1. “I take my umbrella only if it is raining.”
  2. “If I take my umbrella, then it is raining.”
  3. “I flip that switch only if a light comes on.”
  4. “If I flip that switch, a light comes on.”

The average non-mathematical English speaker will easily believe that (1) and (4) are true, but will balk and (2) and (3). To me, (3) means that the light coming on makes me flip the switch. (2) is more problematical, but it does (to me) have a feeling of causation going the wrong way. It is this difference that causes students to balk at the equivalence in math of “$P$ only if $Q$” and “If $P$, then $Q$”. In math, there is no such thing as causation, and the truth tables for implication force us to live with the fact that these two sentences mean the same thing.

Henning Makholm’ answer to Question 17562 begins this way: “I don’t think there’s really anything to understand here. One simply has to learn as a fact that in mathematics jargon the words ‘only if’ invariably encode that particular meaning. It is not really forced by the everyday meanings of ‘only’ and’ if’ in isolation; it’s just how it is.” That is the best way to answer the question. (Other answerers besides Makholm said something similar.)

I have also discussed this difficulty (and other difficulties with logic) in the abmath section on “only if“.


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