Math Myth: The rationals are numbers

[fresh_42]

$1=\frac{12}{12}$ is mathematics. 1 pizza not equal to 12 slices is not.

$1=\frac{12}{12}$ is sloppy mathematics, or convenient, if sloppy offends you. The pizza analogon is a rhetorical mean: figura per immutationem.

 

[Infrared]

The way I was taught is that $a/b$ is defined to be the equivalence class of $(a,b)$ with respect to the usual relation defining fractions. In that sense, say, $1/1=12/12$ is true because both sides are equivalence classes that coincide.

 

[fresh_42]

The way I was taught is that $a/b$ is defined to be the equivalence class of $(a,b)$ with respect to the usual relation defining fractions. In that sense, say, $1/1=12/12$ is true because both sides are equivalence classes that coincide.

Thank you. (cp. #35)

 

[Mark44]

I still don't get it. Why is it wrong to say that 1=12/12!?

It is as right or wrong as it is to say −2=5.

It is correct to say that $-2 \equiv 5 \mod 7$, but without providing the context that you're working with equivalence classes modulo 7, it is wrong to say that -2 = 5.

Further, it is wrong to write $1 \ne \frac {12}{12}$ unless you explicitly state that you're working with equivalence classes.

 

[fresh_42]

This is really astonishing with this debate: While I listed many reasons, why it can be reasonable to regard quotients as equivalence classes, all I received was: "We always did it so. Hence it must be right.", which is rather poor if you ask me.

 

[Mark44]

From post #1:

But $1\neq \dfrac{12}{12}.$ You can literally see that it is different: $5$ symbols instead of $1.$

Just because one expression takes 5 symbols to write while another takes only 1 does not mean that the two expressions are different in value. By this logic, one would conclude that the Pythagorean formula is incorrect; i.e., that $c^2 \neq a^2 + b^2$, with a and b being the sides of a right triangle, and c being the hypotenuse. Here we have two symbols on the left, and five on the right. Does that make the two expressions different?
What we are debating is your claim that the expressions 1 and $\frac {12}{12}$ are unequal, without explicitly stating that you are dealing with equivalence classes, and similarly saying that -2 and 5 are equal, without stating that the relation is actually equivalent, modulo 7.

 

[fresh_42]

From post #1:

Just because one expression takes 5 symbols to write while another takes only 1 does not mean that the two expressions are different in value. By this logic, one would conclude that the Pythagorean formula is incorrect; i.e., that $c^2 \neq a^2 + b^2$, with a and b being the sides of a right triangle, and c being the hypotenuse. Here we have two symbols on the left, and five on the right. Does that make the two expressions different?
What we are debating is your claim that the expressions 1 and $\frac {12}{12}$ are unequal, without explicitly stating that you are dealing with equivalence classes, and similarly saying that -2 and 5 are equal, without stating that the relation is actually equivalent, modulo 7.

This is simply not true.

Of course, we treat $1=\frac{12}{12}$ the same because we are interested in its value, however, they are only equal because $1\cdot 12 = 12 \cdot 1.$ It becomes clearer in its general form:

$$\dfrac{a}{b}\sim\dfrac{c}{d}\Longleftrightarrow a\cdot d= b\cdot c$$

'$\sim$' is strictly speaking an equivalence relation. It gathers really many quotients under one name

$$1=\left\{1,\dfrac{2}{2},\dfrac{-3}{-3},\dfrac{12}{12},\ldots\right\}$$

and the same is true for all other quotients. We take them as the same and write '$=$' instead of '$\sim$' because we are only interested in their values. But $1\neq \dfrac{12}{12}.$ You can literally see that it is different: $5$ symbols instead of $1.$

 

[Mark44]

This is simply not true.

Which part of what I wrote is not true? I quoted what you wrote directly from post #1.
From post #1:

But $1\neq \dfrac{12}{12}.$ You can literally see that it is different: $5$ symbols instead of $1.$

And my example of the Pythagorean formula is a counterexample to your assertion that having a different number of symbols makes a difference.

 

[fresh_42]

Which part of what I wrote is not true? I quoted what you wrote directly from post #1.
From post #1:

That I haven't mentioned the equivalence classes. The entire section is about it. You left out everything and extracted a single sentence. This is willfully misquoting. Is that our new standard?

And my example of the Pythagorean formula is a counterexample to your assertion that having a different number of symbols makes a difference.

This depends on the context. It makes a difference indeed, e.g. in the theory of formal languages. However, Pythagoras hasn't anything to do with the subject. And, yes, I used rhetorical methods, because I wrote a pamphlet and not an article.

 

[stevendaryl]

This discussion is a little bit confusing. Saying that $a/b$ represents an equivalence class doesn't imply for example that $1/2 \neq 2/4$. It's a true equality because the denotation of the two expressions is the SAME equivalence class. In the equivalence class view of rationals, $1/2 =$ the set of all pairs $(n,m)$ such that $n \times 2 = m \times 1$. And that's the same set as the set of all pairs $(n,m)$ such that $n \times 4 = m \times 2$.

The problematic case is $1 = 12/12$. That can't literally be true under the interpretation of $12/12$ as an equivalence class. But it's also not true under that interpretation that $1 \approx 12/12$, because $12/12$ is an equivalence class of pairs. A pair can't be equivalent to a singleton.

What's really going on, which I mentioned earlier, is injections.

There is a unique way to interpret a natural as an integer, $n \mapsto$ the equivalence class of pairs $(n,0)$ where $(n,m) \approx (n',m')$ if $n+m' = n' + m$.

and there's a unique way to interpret an integer as a rational,
$n \mapsto$ the equivalence class of $(n,1)$ where $(n,m) \approx (n',m')$ if $n \times m' = n' \times m$

and there's a unique way to interpret a rational as a real, and there's a unique way to interpret a real as a complex number. So when someone writes, for example:

$1 = -(-1) = 1/1 = \pi/\pi = 1 + 0 i$

they're writing equal signs between completely different objects, which makes no sense. But if you first inject $1$ into the integers, then it makes sense that the result is equal to (not just equivalent to) $- (-1)$. If you then inject $-(-1)$ into the rationals, it makes sense to say that $-(-1) = 1/1$. Etc.

So in practice, when someone writes $1$, they mean either the natural, or the integer, or the rational, or the real, or the complex number, whichever one is needed for the problem at hand. And the ambiguity doesn't really cause any harm.

 

[Mark44]

That I haven't mentioned the equivalence classes. The entire section is about it. You left out everything and extracted a single sentence.

The title of the thread (now) is "Math Myth: the rationals are numbers".
I doubt that any mathematician would agree that "the rationals are number" is a myth.

Again, and for at least the 3rd time, writing $1 \neq \frac{12}{12}$ without including in that statement, that you're talking equivalence classes, is what I'm objecting to. IMO, failing to do that is akin to leaving off the qualifiers in a $\delta - \epsilon$ proof.

This depends on the context. It makes a difference indeed, e.g. in the theory of formal languages.

Which is not under discussion here, given the title of the thread.

 

[atyy]

I am only saying that $1\neq \dfrac{12}{12}.$

That's obviously true, they are in different positions! And the above quote is not the same as the original.

https://en.wikipedia.org/wiki/The_Treachery_of_Images

 

[martinbn]

The problematic case is $1 = 12/12$. That can't literally be true under the interpretation of $12/12$ as an equivalence class. But it's also not true under that interpretation that $1 \approx 12/12$, because $12/12$ is an equivalence class of pairs. A pair can't be equivalent to a singleton.

I disagree with this part. The singleton $n$ is just the usual notation for the equivalence class of the pair $(n,1)$, so the equality is perfectly fine. Even in school childern are thought that for fractions like $\frac51$ you don't have to write the denominator and write it simply as $5$.

 

[stevendaryl]

I disagree with this part. The singleton $n$ is just the usual notation for the equivalence class of the pair $(n,1)$,

That was my point. If naturals are singletons and rationals are equivalence classes of pairs, then they can never be equal, or even equivalent. But you can always "coerce" a natural to the corresponding rational.

 

[stevendaryl]

There's probably a way to introduce real numbers "all at once" without starting with naturals, proceeding to integers to rationals to reals.

Couldn't we just say:

• 0 is a real
• 1 is a real
• if $x$ and $y$ are reals, then so are $x+y$ and $x \times y$ and $x - y$.
• If $x$ and $y$ are reals, and $y \neq 0$, then $x/y$ is a real.
• Then basic facts about $+$, $\times$ and $-$ and $/$.

This way, $/$ is not an operation for forming rationals from integers, it's just a binary operation on reals.

 

[martinbn]

That was my point. If naturals are singletons and rationals are equivalence classes of pairs, then they can never be equal, or even equivalent. But you can always "coerce" a natural to the corresponding rational.

My point was that only that you can, but you do. And there is no coercion. It is part of the definition.

https://en.wikipedia.org/wiki/Field_of_fractions

..
The embedding of $R$ in $Frac(R)$ maps each $n\in R$ to the fraction $\frac{en}e$ for any nonzero $e\in R$ (the equivalence class is independent of the choice $e$). This is modeled on the identity $\frac n1 = n$.
..

 

[jbriggs444]

There's probably a way to introduce real numbers "all at once" without starting with naturals, proceeding to integers to rationals to reals.

Couldn't we just say:

• 0 is a real
• 1 is a real
• if $x$ and $y$ are reals, then so are $x+y$ and $x \times y$ and $x - y$.
• If $x$ and $y$ are reals, and $y \neq 0$, then $x/y$ is a real.
• Then basic facts about $+$, $\times$ and $-$ and $/$.

This way, $/$ is not an operation for forming rationals from integers, it's just a binary operation on reals.

Without some additional flesh behind the "basic facts", the two element field GF(2) fits the above definition.

If you want the reals, you could ask for a "complete, ordered, archimedean field". If you want the rationals, you could skip the "complete" and, perhaps, require that it be "countable" instead.

The utility of constructions such as equivalence classes is, to me, that they provide some assurance that the asked-for structure exists. [Or "has a model" or whatever the appropriate term would be].

 

[stevendaryl]

My point was that only that you can, but you do. And there is no coercion. It is part of the definition.

https://en.wikipedia.org/wiki/Field_of_fractions

The definition of what? If you define fractions to be equivalence classes of ordered pairs of naturals, then a natural is not a fraction. Every natural can be associated with a fraction (the article uses the term "embedding"; if you are embedding one set into another, that does not mean that the first set is a subset of the second set).

Actually, in category theory, maybe it does? In category theory, there is no such thing as "the" naturals or "the" rationals. There are just objects that work like the naturals, or the rationals, etc. So it's pointless to ask whether the naturals is a subset of the rationals.

 

[stevendaryl]

Without some additional flesh behind the "basic facts", the two element field GF(2) fits the above definition.

Well, yes, a structure can't be pinned down without axioms.

 

[stevendaryl]

The utility of constructions such as equivalence classes is, to me, that they provide some assurance that the asked-for structure exists. [Or "has a model" or whatever the appropriate term would be].

That's exactly right.

 

[pbuk]

That was my point. If naturals are singletons and rationals are equivalence classes of pairs, then they can never be equal, or even equivalent. But you can always "coerce" a natural to the corresponding rational.

Yes, this. When we use the = sign it is implicit that the the thing on the LHS is a member of the same set as the thing on the RHS. To say that $1 = \dfrac{12}{12}$ is not a theorem because we must interpret the LHS as an integer and the RHS as a quotient pair is nonsense IMHO.

Similarly I do not accept that if we admit $1 = \dfrac{12}{12}$ then we must admit $-2 = 5$: clearly these are not the same element of the set of integers and if we interpret 5 as an element of the set of integers modulo 7 then -2 is not even in the set.

I recognise that @fresh_42's Insight article was intended to be provacative, but of all the contentious propositions this one seems to me to be particularly unattractive.

 

[stevendaryl]

Similarly I do not accept that if we admit $1 = \dfrac{12}{12}$ then we must admit $-2 = 5$: clearly these are not the same element of the set of integers and if we interpret 5 as an element of the set of integers modulo 7 then -2 is not even in the set.

There's always an ambiguity as to whether something like $-2$ is supposed to be a canonical name for an element, or whether it is to be interpreted as the unary minus operator applied to the number 2. With the latter interpretation, $-2$ is an element of the integers modulo 7.

There are different philosophies about the foundations of mathematics. In some approaches, it's assumed that for anything more complicated than the naturals, you're always dealing with an equivalence relation, rather than equality (or rather, a congruence relation, because all the operations have to respect the equivalence relation).

 

[pbuk]

There's always an ambiguity as to whether something like $-2$ is supposed to be a canonical name for an element, or whether it is to be interpreted as the unary minus operator applied to the number 2. With the latter interpretation, $-2$ is an element of the integers modulo 7.

Hmm good point. But integers modulo 7 are an equivalence class not a set and it is therefore correct to write $(-2) \equiv 5$, or alternativey $0 - 2 \equiv 5$ but still not $(-2) = 5$ nor $0 - 2 = 5$.

There are different philosophies about the foundations of mathematics. In some approaches, it's assumed that for anything more complicated than the naturals, you're always dealing with an equivalence relation, rather than equality (or rather, a congruence relation, because all the operations have to respect the equivalence relation).

That seems irrational, or at least unnecessarily complex. Surely in the real world a thing can be equal to itself, not just equivalent?

 

[fresh_42]

We could endlessly debate the words rational number, representative of an equivalence class, quotient, or whatever. Probably without ever coming to a conclusion. It is hard to think about new ideas for what has been taught for centuries. The usage of wrong in the context was a rhetorical mean, less a mathematical statement. I see $\mathbb{Q}$ as $\left(\mathbb{Z}-\{0\}\right)^{-1}\mathbb{Z}$ (Atiyah, MacDonald: Introduction to Commutative Algebra, Chp. 3). Yes, it is an algebraic approach, because I think that the curriculum at school follows the embeddings semigroup - ring - field: $\mathbb{N} \subseteq \mathbb{Z} \subseteq \mathbb{Q}$, such that it is natural to follow this algebraic way once it has been entered.

Anyway. I remember names like endoplasmic reticulum from biology at school. Mainly because of the teacher's accent rather than what it means. I remember that I had to learn those horrible names like 2,6,8-trihydroxypurin in chemistry. And we learned about the double-slit experiment and the weak decay in physics.

Only in mathematics, we refuse to teach anything other than triangles, and calculations. It is (normally) not mathematics, it is counting. What I wanted to initiate with this article was a discussion, why we don't teach mathematics more scientifically as we do in other STEM areas? Mathematics seems to prefer to be the big surprise at university. A habit that is wrong in my opinion, and that fails to inspire kids. Instead, we are hunting them down in algorithms. One after the other.

 

[stevendaryl]

That seems irrational, or at least unnecessarily complex. Surely in the real world a thing can be equal to itself, not just equivalent?

I don't remember the details of the construction, but I will try to give a little bit of the flavor.

Start with a term language, which is defined by a collection of constant symbols and function symbols of various numbers of arguments. A term is anything you can form starting with constant symbols and applying function symbols. Then a "type" is such a term language together with a partial equivalence relation on terms. "Partial" because some terms don't denote any element of the type (for example, $0/0$ doesn't denote any element of the rationals). A term $t$ is in the type $T$ is $t \approx t$ according to the partial equivalence relation of $T$.

Then if $A$ are types and $B$ are types, and you have a function $f$ that takes terms of $A$ and returns terms of $B$, then we say $f$ is of type $A \rightarrow B$ if whenever $t_1 \approx t_2$ is true according to the equivalence relation of $A$, then $f(t_1) \approx f(t_2)$ according to the equivalence relation of $B$. The equivalence relation for $A \rightarrow B$ is extensionality:

$f \approx g$ if and only if for all $t$, $f(t) \approx g(t)$

Etc.

This is different from the construction in terms of equivalence classes because you are always dealing with concrete objects, terms, rather than equivalence classes. And as long as objects are always identified with exactly one type, there is no confusion in using $=$ for $\approx$ in that type.

 

[Dale]

Is there a widely accepted official definition of "number". If so then the question of whether or not the rationals are numbers has nothing to do with whether or not they are also an equivalence class. The question of whether or not they are numbers is determined only by whether or not they fit the definition of number.

That said, I don't know the official definition of number. Does anyone else?

 

[mathman]

Is "one" different from "1" or "I"(Roman)?

 

[fresh_42]

Throughout mathematics there are various entities referred to as “numbers”; in modern mathematics it would be more accurate to refer to anyone of various types called “number systems”, and simply define a number to be a term of such a type.

A numeral on the other hand is a syntactic representation of a number, part of a system of numeration.

It is interesting to try to describe the general conditions under which something comes to be designated as a “number”. After all, such number systems tend to form commutative rings or at least commutative rigs, but not all commutative rigs are considered to consist of “numbers”, or at least that is not how the language is used in practice.

The root notion, known to the great ancient civilizations and particularly ancient Hellenic civilization, is that of (the system of) natural numbers, and in some way or other each of the various number systems are extensions of that one.

https://ncatlab.org/nlab/show/number

 

[fresh_42]

Is "one" different from "1" or "I"(Roman)?

Yes. One can see it. In order to be the same, you have to provide context.

 

[dextercioby]

How do you define rational numbers? As a set, of course, but a set of what?

1st definition: A number $p$ is called a rational number iff is an element of this set:

$$\mathbb Q_1 :=\left\{\frac{a}{b}\vert a\in\mathbb Z, b\in \mathbb Z\setminus\{0\}\right\}$$

According to this definition: $\left(\frac{1}{1}=:\right) 1_{\mathbb Q_1} \neq \frac{12}{12}_{\mathbb Q_1}$

2nd definition: Let us endow $\mathbb Q_1$ with an equivalence relation, because there are solid reasons to believe $\mathbb Q_1$ has too many elements for the purpose of "clean" mathematics. For this we exploit the fact that $\mathbb Z$ is closed under multiplication.

$$\forall p,q\in\mathbb Q_1, p\equiv \frac{a}{b}, q\equiv \frac{c}{d}$$

$$ \frac{a}{b}\sim \frac{c}{d} \Leftrightarrow a\cdot d =_{\mathbb Z} b\cdot d$$

From here we simply define:

$$ \mathbb Q_2 := \mathbb Q_1 / \sim $$

So we obviously have that $\left[\frac{1}{1}\right]_{\mathbb Q_2} = \left[\frac{12}{12}\right]_{\mathbb Q_2}$.

So to say simply that $1=12/12$ is true only by ignoring the very definition of $\mathbb Q_2$ by scrambling the definition of $\mathbb Q_1$. What @fresh_42 is saying is that the standard mathematics education system (from elementary 1st grade to the end of high-school) simply uses improperly defined mathematical objects and perhaps even the highest level of mathematics education attained (college/university, before specializing to a PhD) does not properly define "rational numbers".

 

[Dale]

How do you define rational numbers?

First, how do you define numbers?

 

[dextercioby]

As elements of particular sets, i.e. sets with particular properties of their elements.

 

[Office_Shredder]

Why can't we just say the rational numbers are the smallest field of characteristic 0. We know the integers embed into it, and by axioms of a field 1=12/12.

You might ask how we know this field exists (I.e. there exists one that embeds into all others). It's pretty easy to prove your $\mathbb{Q}_2$ is it, so we're done. But I haven't defined the field as an equivalence class, it just exists, contains integers, and let's you divide by integers as well. Then saying $1\neq 12/12$ is like saying $9\neq 3^2$ because they're only equal after you apply evaluation of functions which you haven't done yet.

The construction of the rationals as equivalence classes is a convenient way to prove everything is on the up and up, but most people don't spend any time thinking about Peano's construction of the integers and it doesn't seem to be an issue either. This is like arguing that we need to teach everyone that 2 is the successor of 1, not just, you know, 2.

 

[Dale]

As elements of particular sets, i.e. sets with particular properties of their elements.

By that definition the rationals certainly are numbers.

 

[dextercioby]

By that definition the rationals certainly are numbers.

Sure, that is the definition of $\mathbb Q_1$ in my post. But because of it's void of the equivalence relation, it has "redundant elements", i.e. it has both $\frac{1}{1}$ and $\frac{14}{14}$ as distinct elements.