I Math Myth: The rationals are numbers

[martinbn]

It is wrong in the sense that they are representatives of equivalence classes and as such stand for entire sets, not numbers, because people associate a single element with the word number, not a set.

These sets are elements of another set. The set of equvalent classes. So each of them is a single element.

 

[fresh_42]

Whether you accept something isn't of any relevance. This point of view is one of my criticisms. "Because we always did so, it is right." Your "definition" isn't one. It is not even well-defined. I gave a definition in post #77, where we set $S=\mathbb{Z}^\times$ and $R=\mathbb{Z}.$ If you consider only quotients, how could you not distinguish $\dfrac{1}{1}$ from $\dfrac{12}{12}?$

 

[fresh_42]

These sets are elements of another set. The set of equvalent classes. So each of them is a single element.

All I want is to recognize this fact! We do not teach it to my best knowledge, @Infrared 's experience aside.

 

[dextercioby]

@fresh_42 This is a follow-up to my previous post here. No. 65 from page 3

IIRC, in Romania (I believe 5th grade, i.e. 11-12 y.o.) we were given a 3rd definition of „rational numbers”.

Definition 3

$$\mathbb Q_3 :=\left\{\frac{a}{b}\vert~ a\in\mathbb Z, b\in \mathbb Z\setminus\{0\}, \text{gcd}(a,b) =\{\pm 1\}\right\}$$

So the question for @fresh_42 is: are the elements of $\mathbb Q_3$ also equivalence classes? Cause teaching people about $\mathbb Q_1$ is utterly wrong, if $\mathbb Q_3$ is available.

 

[mathman]

It seems to me the issue is what does "=" mean. Mathematical definition means that $1=\frac{12}{12}$ is a valid statement. As images they are different.

 

[pbuk]

Whether you accept something isn't of any relevance.

Well yes, I can agree with that.

This point of view is one of my criticisms. "Because we always did so, it is right."

That was not what I was trying to say: I was trying to ask what was wrong with the way we always did it. I could equally well characterize your criticisim as 'Because we always did so, it is wrong'.

I gave a definition in post #77, where we set $S=\mathbb{Z}^\times$ and ##R=\mathbb{Z}.

But that is not a definition of the rationals, nor even a set that is bijective with the rationals. To get to the rationals from here I need to eliminate the duplicates with an equivalence relation - why is it better to travel in this direction?

If you consider only quotients, how could you not distinguish $\dfrac{1}{1}$ from $\dfrac{12}{12}?$

I am not sure what you mean here, but I can't distinguish any collections of symbols without context.

 

[fresh_42]

I think simple and go with Kronecker (almost):

"Die ganzen Zahlen hat der liebe Gott gemacht, alles andere ist Menschenwerk"
("God made the integers, all else is the work of man")

I believe that counting is the only valid base to start with. That gives us the semigroup $\mathbb{N}.$ I don't even consider $0$ a natural number. I think that naming something which isn't there is actually a human achievement, an Indian to be precise. Probably accounting in Babylon brought us the next extension to the additive group of integers $\mathbb{Z}$ which are naturally a ring. We were lucky that they are an integral domain, that allows the next extension to a quotient field $\mathbb{Q}$ in a very easy way. Next came the real numbers, which already require some topology to get there. No wonder that the ancient Greeks spoke of irrational numbers. And they had only the algebraic reals which arise from geometry. It took almost 2000 years to get a reasonable definition of all real numbers. The next and in some sense final step are the complex numbers, which are in my opinion again a set of equivalence classes $\mathbb{C}=\mathbb{R}[x]/(x^2+1).$

 

[Infrared]

$$\mathbb Q_3 :=\left\{\frac{a}{b}\vert~ a\in\mathbb Z, b\in \mathbb Z\setminus\{0\}, \text{gcd}(a,b) =\{\pm 1\}\right\}$$

It looks like $\frac{-2}{3}$ and $\frac{2}{-3}$ are distinct elements here. If so, you would definitely need to take equivalence classes. Or does $a/b$ already refer to the equivalence class, not just the pair $(a,b)$?

 

[jbriggs444]

It looks like $\frac{-2}{3}$ and $\frac{2}{-3}$ are distinct elements here. If so, you would definitely need to take equivalence classes.

One could pick out canonical exemplars from each class, for instance requiring the denominator to be strictly positive.

 

[Infrared]

Right, but I'm just reading what is written.

 

[fresh_42]

So the question for @fresh_42 is: are the elements of $\mathbb{Q}_3$ also equivalence classes?

I think so. It is all hidden in the condition $gcd(a,b)=\pm 1$. How do you handle $\dfrac{12}{12}$ in such a case? It is not part of $\mathbb{Q}_3$ but part of $\mathbb{Q}$. So we must identify it with $\dfrac{1}{1}$ to get $\mathbb{Q}_3$ work. It always comes down to equivalences. E.g. in @pbuk's formula as quotients, the simple need to make it well-defined requires to answer what makes $\dfrac{2}{4}$ equal to $\dfrac{3}{6}.$ And the answer is always: because $2\cdot 6 = 3\cdot 4,$ which is exactly the formal definition as $S^{-1}R.$ The equality sign is a convenience, an abbreviation.

 

[dextercioby]

Yes, you are correct. I did intend integers, so indeed $\mathbb Q_3$ is also "not enough".
So there:

Definition 4

$$\mathbb Q_4 := \left\{\frac{a}{b}\vert a\in\mathbb Z, b\in\mathbb N^{*}, \text{gcd} (a,b) = \{\pm 1\}\right\}$$

This should do it.

 

[jbriggs444]

How do you handle $\dfrac{12}{12}$ in such a case?

That seems to be a question of notation, rather than something with much mathematical meat on its bones.

If we allow $\dfrac{1}{1}$ as a numeral that is literally the rational number 1 then we have removed the notation $\dfrac{1}{1}$ as possibly denoting an expression for the rational number 1 divided by the rational number 1 in the field of rational numbers.

Edit to add:

In practice, the expression evaluates to the same thing as the numeral, so it is a distinction without much of a difference.

So I guess I've argued my way around to your position. We have this equivalence class of expressions which we customarily refer to as all being equal (or equivalent) to one another.

 

[pbuk]

E.g. in @pbuk's formula as quotients, the simple need to make it well-defined requires to answer what makes $\dfrac{2}{4}$ equal to $\dfrac{3}{6}.$ And the answer is always: because $2\cdot 6 = 3\cdot 4,$ which is exactly the formal definition as $S^{-1}R.$ The equality sign is a convenience, an abbreviation.

Are you saying that I must accept a category theory foundation before I can do any maths?

 

[fresh_42]

That seems to be a question of notation, rather than something with much mathematical meat on its bones.

If we allow $\dfrac{1}{1}$ as a numeral that is literally the rational number 1 then we have removed the notation $\dfrac{1}{1}$ as possibly denoting an expression for the rational number 1 divided by the rational number 1 in the field of rational numbers.

Of course, there is always the view of the ancient Greeks, i.e. a rational (sic!) number is the ratio of two lengths. But even this leads to equivalence classes since ratios can be equal even if the lengths are not.

Whichever I look at it, I see these classes. However, I don't want to revolutionize teaching, I only want it to be mentioned. Closer to real mathematics and away from algorithms. I find it embarrassing if people in quiz shows are asked to calculate e.g. $3^3$ and then say, that they were always bad at mathematics. Heck, this ain't mathematics. Make it two classes, calculation and mathematics, but do not pretend it was the same.

 

[fresh_42]

Are you saying that I must accept a category theory foundation before I can do any maths?

The principle of equivalence relations can hardly be called category theory. Relations are important, equivalence relations even more. It does no harm to teach it. 1 hour at most. The bargain, however, is much more if it comes to other examples.

 

[pbuk]

The principle of equivalence relations can hardly be called category theory.

But surely if you elevate equivalence to a higher level than equality then that is where you will end up?

 

[atyy]

Following Bourbaki's discussionm it is likely that even $1 \neq 1$ due to physics, even if we allow equivalence classes of symbols at different positions.

 

[trees and plants]

Following Bourbaki's discussionm it is likely that even $1 \neq 1$ due to physics, even if we allow equivalence classes of symbols at different positions.

Can you show me how $1\neq1$ can hold? As i know equality is an equivalence relation, so the reflexive property holds https://en.wikipedia.org/wiki/Equivalence_relation .

 

[trees and plants]

It is wrong in the sense that they are representatives of equivalence classes and as such stand for entire sets, not numbers, because people associate a single element with the word number, not a set.

Could you provide an example or more when saying that people associate a single element with the word number, not a set?

 

[fresh_42]

Could you provide an example or more when saying that people associate a single element with the word number, not a set?

That was a guess. What do you associate if you hear the word number? Something like $7$ or something like $\left\{\dfrac{a}{b}=7\,|\,a\in \mathbb{Z},b\in \mathbb{Z}^\times\right\}?$

 

[trees and plants]

That was a guess. What do you associate if you hear the word number? Something like $7$ or something like $\left\{\dfrac{a}{b}=7\,|\,a\in \mathbb{Z},b\in \mathbb{Z}^\times\right\}?$

You mean the use of the number in non scientific language? In math i think rationals are numbers. But i do not think a definition of a number mathematically is easy to make. There are of course examples of sets with numbers.

Somewhere i read i think that things can be defined by giving examples not necessarily definitions.

Excuse me if i make any mistake.

 

[Dale]

That was a guess. What do you associate if you hear the word number? Something like $7$ or something like $\left\{\dfrac{a}{b}=7\,|\,a\in \mathbb{Z},b\in \mathbb{Z}^\times\right\}?$

I think that people do associate the rationals with the entire equivalence class. They don’t know the term and they wouldn’t be able to recognize it written as you did, but the concept is there. People easily recognize 12/12 of a pizza and 8/8 of a pizza as being distinct from each other but both equivalent to 1 pizza.

 

[stevendaryl]

I think that there is something not quite about right saying "we have to be talking about equivalence classes in order to say $\frac{12}{2} = \frac{1}{1}$. If you think of $\frac{x}{y}$ as an expression denoting the value of a binary function on the arguments $x,y$, then there is no implication that there are any equivalence classes involved. If I say that $sin(\pi) = sin(2\pi)$, that doesn't mean I'm dealing with equivalence classes. It just means that the function $(x,y) \rightarrow \frac{x}{y}$ is not one-to-one.

There is often an ambiguity when we talk about an expression such as $\frac{x}{y}$ whether we're talking about the expression or the value denoted by the expression. When asking whether $\frac{12}{12} = \frac{1}{1}$, we're talking about the denotation. When asking "What is the denominator of $\frac{2}{3}$?" we're talking about the expression.

 

[fresh_42]

I think that people do associate the rationals with the entire equivalence class. They don’t know the term and they wouldn’t be able to recognize it written as you did, but the concept is there. People easily recognize 12/12 of a pizza and 8/8 of a pizza as being distinct from each other but both equivalent to 1 pizza.

People will consider it as equal value (sic!) of pizza, but they will clearly see the difference.

The concept of being equivalent isn't too hard to teach students. I simply do not see, why we insist to introduce it as equality. Teach it right and add: "From now one we will write $1=\dfrac{12}{12}$ because we are only interested in values, not in representations. That's why we write $=$ and not $\equiv,$ since nobody wants to count lines.

Such an approach is only marginally more work than what is taught anyway, but it would be closer to math and farther from calculations.

Edit: Nobody performs the double slit experiment and avoids the word interference.

 

[stevendaryl]

The concept of being equivalent isn't too hard to teach students. I simply do not see, why we insist to introduce it as equality. Teach it right and add: "From now one we will write $1=\dfrac{12}{12}$ because we are only interested in values, not in representations.

Well, if we used $=$ only for representations, and not values, then we would have very little occasion to use it. You could write 1=1, 2=2, 3=3, 1/2 = 1/2, but that's very boring mathematics.

 

[atyy]

Can you show me how $1\neq1$ can hold? As i know equality is an equivalence relation, so the reflexive property holds https://en.wikipedia.org/wiki/Equivalence_relation .

Generically one would expect $a \neq b$ since $a$ and $b$ are different. $1\neq1$ is simply the same principle. The two $1$s which appear identical to you are with high probability not identical, since they are not made of exactly the same number of atoms. Or if you are reading on a computer screen, there are also very likely non-uniformities in the display. Thus microscopically $1\neq1$.

Basically, symbols that may appear identical to you are not physically identical.

 

[dextercioby]

I think it's all getting muddy now.

[...]Teach it right and add: "From now one we will write $1=\dfrac{12}{12}$ because we are only interested in values, not in representations. That's why we write $=$ and not $\equiv,$ since nobody wants to count lines.
[...]

Let me add my comment on this part. I do not know if I am right, from the perspective of the most advanced mathematics, but there is a clear distinction between $=$ and $\equiv$.

$$ 3+4 = 7 \tag{1} $$ simply means that the result of the internal operation on $\mathbb N$ for certain chosen elements is identical (completely replaceable) with another particular element of $\mathbb N$ (this would be your "we are only interested in values"),

while

$$ \sin (x+y) \equiv \sin x \cos y + \sin y \cos x \tag{2} $$

means that for whatever elements of $\mathbb R$, the exact element (the value) in $\mathbb R$ in the LHS is identical (we would shift to using the same sign as in the previous paragraph) to the element in the RHS. For $(1)$ we reserve the terminology "equality", while for the second, the terminology "identity".

 

[Infrared]

@PeroK I don't think I really see the distinction that you're making.

In (1), you have a sum of two natural numbers is equal to a third. In (2), you have that a sum of two functions (of two variables) is equal to a third. Besides the addition taking place in two different monoids ($\mathbb{N}$ versus $\mathbb{R}^{\mathbb{R}\times\mathbb{R}}$), what is the key distinction?

 

[atyy]

Myth: Natural numbers are numbers. They are NOT numbers. They are sets!

 

[jbriggs444]

Myth: Natural numbers are numbers. They are NOT numbers. They are sets!

No, no. It's both a floor wax and a dessert topping.

[And it's less filling and still tastes great].

 

[PeroK]

It's certainly true that $1 == \frac{12}{12}$, as we can see from this Python code:

if 1 == 12/12:
print("Fresh 42 is wrong!")
else:
print("Fresh 42 is right!")

 

[WWGD]

We may have to leave it up to Bill Clinton: " It depends on what 'is' is."

 

[mathman]

This thread is absurd!

 

[ShayanJ]

The way I see it, every little detail in mathematics is decided based on the question "so what?", "what does it lead to?". And I don't just mean something that is useful in engineering, physics, etc. I'm also including those aspects that are just in there so that mathematicians can explore more abstract and sophisticated ideas.
So this is how I judge this question. Yeah, that maybe true technically, but what does it lead to? How does it change anything? Why should we care?
I think this is how it should be decided. If someone can come up with a situation in which this actually makes a difference in some theorem or mathematical procedure, then I'm ready to accept it.
Otherwise it might be the first case of "too pedantic even for a pure mathematician" that I have ever seen!

 

[Office_Shredder]

This isn't too pedantic for a pure mathematician, defining the field of fractions for an integral domain is pretty standard undergrad/intro graduate course stuff.

 

[mathman]

The original statement is absurd. "numbers" is a word to defined. In all mathematical discourse rationals fit the definition.

 

[weirdoguy]

The original statement is absurd.

Most of these myths are absurd in one way or another, but I guess it was intended to be a little bit controversial.

 

[S.G. Janssens]

I am afraid the subtleties that the original statement tries to address are already obvious to mathematicians, while in its current form it probably mostly confuses readers that have not seen such subtleties before. Either way, the point is missed.

 

[pinball1970]

Whether you accept something isn't of any relevance. This point of view is one of my criticisms. "Because we always did so, it is right." Your "definition" isn't one. It is not even well-defined. I gave a definition in post #77, where we set $S=\mathbb{Z}^\times$ and $R=\mathbb{Z}.$ If you consider only quotients, how could you not distinguish $\dfrac{1}{1}$ from $\dfrac{12}{12}?$

I have read some of the thread but not every single post so this may have cropped up.

12/12 is not equal to one has been discussed

How about the identically equal sign? one extra bar? Can that not distinguish?

So 12/12 =1 but 12/12 is identically equal to 12/12?

12/12≡12/12?

 

[trees and plants]

How can you prove that $\frac{12}{12}\neq 1$? Show me a proof if you know.What is your definition of equality?

 

[jbriggs444]

The sets are not equal

In mathematics, a set is a collection of distinct elements.[1][2][3] The elements that make up a set can be any kind of things: people, letters of the alphabet, numbers, points in space, lines, other geometrical shapes, variables, or even other sets.[4] Two sets are equal if and only if they have precisely the same elements.[5]

12/12 has 12 elements 1 has 1, could you use that?

I thought that 12/12 had infinitely many elements, each element being an ordered pair. Each ordered pair would naturally be equivalent to every other ordered pair in the set under our standard equivalence relation for the rationals. An ordered pair, following the Kuratowski definition would be a two-element set {a,{a,b}}. In the case of this particular equivalence class, each ordered pair would collapse to {a,{a}}.

So the infinite set would be something like { {1,{1}}, {2,{2}}, {3,{3}}, ... {12,{12}}, ... }

We would likely adopt the notation where 0 = {}, 1 = {0}, 2 = {0,1}, etc. So, for instance, 12 would be a twelve element set.

Of course, this is assuming that "12/12" is to be interpreted as a ratio of naturals. If it is to be a ratio of signed integers, there are additional layers to the construction.

Meanwhile, we all understand that none of this folderol is relevant to what someone means when they write 12/12=1.

 

[Mark44]

The thread seems to have run its course, and is now closed.