What is continuum?

[idle marlin]

Is it an infinite set of infinities? The book I'm reading doesn't really clarify the difference between infinity and a continuum, but I'm guessing there is some logical difference. The book speaks about continuum in the context of introducing the concept of different sizes of infinities, it has (tried to) explain to me what uncountable infinities are using Cantor's diagonal method. I'm a complete noob to this.

How does continuum relate to infinity?

[rapid arch]

I'm not an expert, but this is a brief description I can offer you.

There are different classifications of infinity. The first you'll come to be acquainted with are countable and uncountable. A set that is countably infinite means that every element in that set maps one-to-one with a natural number. Naturals, integers, and even the rational numbers are all countably infinite.

What is not countably infinite is the set of all real numbers. There exist real numbers, like sqrt(2) and pi, which are not rational. This is important in the idea of bounded sequences. Consider the set of all the rational numbers x such that x²<2. This set is pretty interesting. It's bounded above by 2. That is, 2>x for every x in this set. But does the set have a least upper bound? That is, an upper bound smaller than any other upper bound? No. It does not. The least upper bound would be sqrt(2), but that's not a rational number. And any upper bound you could think of that is a rational number, there will always be a rational number between it and sqrt(2).

But what does this all mean? It means that there's a gap/break in rational numbers (I'm getting kinda hand-wavy now, but you can look into Dedekind cuts if you want the full story). This is the key difference between the rationals and the reals. The rationals will have this break at sqrt(2), but the set of real numbers contains sqrt(2), so there are no breaks there, or anywhere else for that matter. Hence, continuum.

I probably didn't get that 100% correct, so other people can feel free to weigh in.

[idle marlin]

1. My book says real ("irrational") numbers are numbers which can't be represented as an ordinary fraction m/n where m is a whole number and n is a natural number. The book also points out that fractions with periods in them (like 0.(3) in decimal) are not irrational numbers because their seeming "infinite" nature comes from the number system's representation, so with another number system those numbers would not have a period. But I don't get it. Why would a number not be representable as m/n? I mean, if we take pi, I've heard it has actually a very big period, like it was found out finally (maybe I heard it wrong?). But suppose we're in the past and we don't know that yet, so we think pi does not have period and still seems infinite. Why are we sure it's not representable as m/n if we don't know the whole number? Maybe it is, once the number is calculated fully. Maybe later humanity will find out that it has period. And if it has period, shouldn't it be representable using some other number system?

I don't get sqrt(2) either. I mean, I know what it means - square root from 2 - but I don't understand the implications of the number and how it's calculated (if at all).

[idle marlin]

2. Why would the least upper bound be sqrt(2)? (I think) I get the part about there always being a rational number lesser than picked previously, but why sqrt(2) is the one that's the least?

... is it because, when you actually try to do sqrt(2), you'll end up with number that tries to be as close to some value as possible but can't quite reach that point, ever? Like, as in infinitely close to N? Is this what real number is? Like, if you try to represent it in another number system, the property of this particular number is such that it will try to reach the N value in any number system, while some other number would stop behave in an "infinite" manner once represented in a particular number system? Like, real numbers are numbers that behave "infinitely" because that's the number's property, and non-real numbers behave "infinitely" (and, at first glance, seem to have the same property because of that) because of how the numbers are viewed, but not of what the numbers actually are, not because of their own properties.

3. ... and so, if we have all numbers that can be represented as m/n above (rational numbers), it doesn't cover the cases where numbers aren't... static? And continuum is called continuum because it includes numbers that continue to "change" their values in order to reach some desired N, so these numbers are more accurate (granular) than rational numbers?

and if so, can a real number be "infinitely close to" another real number?

sorry if I'm asking stupid questions, this all is new to me but very curious

... a continuum is an infinity of uncountable infinities? or both countable and uncountable? or, if it's an infinity of uncountable infinities, that'd automatically mean there'll be no breaks, so countable infinities aren't needed for continuum to be a continuum?

[rapid arch]

Why would a number not be representable as m/n
Let's prove that sqrt(2) is irrational

Assume sqrt(2) is rational

Then it can be represented as m/n for some integer m and natural n.

Next, assume m and n share no common factors. That is, it's a fully simplified fraction (e.g. if it were 2/4, you would write it as 1/2)

If m/n = sqrt(2), then m²/n²=2, which you can rewrite as m²=2n²

This means that m² is even. However, the only way that m² is even is if m itself is even. Thus, m must have a factor of 2 in it. If m has a factor of 2, then there must be some integer k such that m=2k.

Replacing m with 2k in m²=2n² gives us 4k²=2n², which simplifies to n²=2k²

This means that n² has a factor of 2 in it, which means n has a factor of 2 in it.

And this is where the problem arises, we've found that m and n both have a factor of 2, but we required that m and n have no common factors.

This leads to a contradiction. This contradiction arose from assuming that sqrt(2) was rational. Therefore, it must be that sqrt(2) is not rational.

[rapid arch]

if we take pi, I've heard it has actually a very big period, like it was found out finally (maybe I heard it wrong?).
This is indeed incorrect. Pi was proven irrational, so it has no period. The proof for its irrationality is quite a bit harder, and not something I personally know, but it is indeed irrational.

[rapid arch]

Why are we sure it's not representable as m/n if we don't know the whole number?
There's more to knowing a number than just its digits. Just like how I proved sqrt(2) was irrational, I didn't need to know a single digit of sqrt(2).

[rapid arch]

Seeing as how it's proven that pi (and e) are irrational, this will not be the case.
Proving that certain numbers are irrational is indeed a difficult thing to do though, make no mistake there. Mathematicians still don't know if certain combinations like e^pi (or was it pi^e?) are irrational.

[rapid arch]

Square roots are an interesting classification of numbers. Mathematicians define square root of a number n to be, "what non-negative number, x, when I square it, gives me n?"

Defining and calculating square roots are two different problems though.

[rapid arch]

Why would the least upper bound be sqrt(2)? (I think) I get the part about there always being a rational number lesser than picked previously, but why sqrt(2) is the one that's the least?
Remember, our set is all rational numbers x such that x²<2. We first show that sqrt(2) is indeed an upper bound. sqrt(2)²=2, so x²<sqrt(2)². However, this implies that x<sqrt(2). This shows that sqrt(2) is indeed an upper bound. To show that there is no other upper bound less than sqrt(2) is actually a bit tricky. But consider taking some really small positive number a>0. Assume sqrt(2)-a was an upper bound of the set, then that must mean x<sqrt(2)-a for every x. But we can rewrite this as x+a<sqrt(2), or x²+2ax+a²<2. However, since x is part of our set, it must be that x²<2, so we can substitute this into our inequality to get 2+2ax+a²<2, which simplifies to 2ax+a²<0. We know a>0, so we can divide both sides of that inequality by a to get 2x+a<0.

However, this must be true for every x in that satisfies x²<2. Notice that 0²<2, so 0 is in the set. Plugging in x=0 to the inequality gives us a<0. However, we assumed a>0, so we're at a contradiction.

This contradiction arose from assuming there was some a>0 such that sqrt(2)-a was an upper bound of the set, so it must be that this is not true.

[rapid arch]

... is it because, when you actually try to do sqrt(2)
This kind of thinking is flawed. Because sqrt(2) is irrational, it has no repeating period, so it cannot be represented as a fraction, or as a finite or repeating decimal number (in any number system). You can approximate sqrt(2) using fractions or decimals, but none will ever be exact. But yeah, you get closer and closer the more you approximate. This is a concept of limits, which you can learn rigorously in real analysis, a college course that discusses theorems used in calculus.

... and so, if we have all numbers that can be represented as m/n above (rational numbers), it doesn't cover the cases where numbers aren't... static? And continuum is called continuum because it includes numbers that continue to "change" their values in order to reach some desired N, so these numbers are more accurate (granular) than rational numbers?
I kinda get what you're saying here, but the way you're looking at is a little strange to me. In short, a rational is just a number that can be written in fraction form. And an irrational is a real number that is not rational. That's really all there is to it.

[idle marlin]

... is it OK if I think about it for a day or two and then come back to this thread with new questions? I need to think of what you've said and come up with proper questions

[idle marlin]

1. why are we substituting 2 into x² in x²+2ax+a²<2 if we defined x² < 2, so it can't be 2?

[idle marlin]

2. the way you talk of irrational numbers - it's like you're defining ther behavior by some by-product of this behavior. Irrational numbers can't be written in fraction form. But why? When we tried to write sqrt(2) as a fraction, it turned out that this number has some property X that prevents it from being written in fraction form, while rational numbers don't have this property. I mean, m/n is kind of an end quantity. Once you've found the correct fraction to represent the number, you know exactly how much that number is, but you can't do that with irrational numbers. So that must mean the quantity they represent is... not exact? And the fact that we can't write irrational numbers in fraction form is just the consequence of that, a symptom of the property X but not full definition? Irrational numbers aren't rational numbers. But what are irrational numbers? What is property X that makes the set of irrational numbers completely non-overlapping with rational numbers? How do matematicians imagine this property in their head? Like, you could prove that some left hand expression is equivalent to some right hand expression, but proof isn't always the same as explanation, what is the dynamics of irrational numbers, of this property X? When I think of 2^n * 2^m, I imagine a row of n 2s all multiplied with each other and a row of m 2s in same manner, so when I multiply those two rows I join the rows into one and now there are n+m 2s in the row of factors, so I get 2^(n + m) and that's how I know how multiplication of powers of the same number works, I have this dynamic abstraction in my head that is the real explanation, not just a proof. What could be an abstraction for irrational numbers? Because I'm only familiar with rational numbers, I can dig those, but I don't have the slightest how to wrap my head around irrational numbers, it's like I have a blind spot here without an abstraction.

[inner temple]

This is a really cool discussion and I hope y'all don't mind if I jump in.

"Continuum" just means "the real number line".
And sometimes it also means "the size of infinity of the real number line", but that use is less common and can be expressed by the phrase "continuumly many".

So that must mean the quantity they represent is... not exact?
This is not right, but I can see why you'd think it. All of your experience of numbers has been either decimals in base 10 or fractions, so you're trying to impose that. Let's back up for a moment.

Suppose you only know about whole numbers. You can answer 7+4 and you can answer 7-4. But what happens when you try 4-7?

Your number system doesn't have an answer to that. So you increase the scope of what you consider to be a number. You if you want to answer to any subtraction, you know have the integers: ..., -2, -1, 0, 1, 2, ...

You can now solve 5+7 and 6-9 and 12x4 and 12÷4, but what about 11÷4?

You need rational numbers.

I think the difficulty now is this: what question are you trying to solve whose answer is going to force you to include all of the real numbers?

There are actually 3 possibilities here (that I know of; there are probably more), and the simplest of this is least upper bounds.

[rapid arch]

I'm using a few properties of inequalities.
If A<B and C<D, then A+C<B+D

And if A<B and B<C, then A<C

[inner temple]

Least upper bounds are not really a process that gives you any sort of intuition for what it spits out, so I would recommend we now turn our focus to real numbers, which are often defined in high school textbooks as "any number you can draw on the real number line". It's a bit circuitous, but I think it's still helpful.

A real number is what it is. When we talk about pi and we say that pi=3.14 or that pi=3.14159265358979..., we're not defining pi. pi already exists and it's perfectly fine as it is. You could similarly say that 4 is a rational number because 4=8/2, but that doesn't really help the number 4.

In the case of pi, it's actually worse, because pi isn't 3.14, and calling it 3.14159... often leads to misunderstandings.

The infinite decimal representation 3.14159... isn't moving or changing. It's entirely static. The whole thing appears all at once. And it just is a single number in the same way that 7 is a single number.

(Side note: I hate when people say that "pi is infinite" because that's misleading. They should say "the decimal representation of pi is non-terminating".)

You might ask what we can do with real numbers. Everything we can do with rational numbers we can do with real numbers. Add, subtract, multiply, divide. What's 4÷pi? It's 4/pi.

4/pi is a real number, and there's no simpler way to express it.

That's something that many of my students struggle to get comfortable with.

[rapid arch]

I may have written this backwards actually. I'll correct the proof when I have some time

[idle marlin]

so, basically, 4 and sqrt(2) aren't some different types of quantities (types which would have differept type properties, like they would behave differently in what quantity they mean), they are exactly the same type - definite quantity, static, finite.

and the thing with irrational numbers is that the mathematical language used to expressed them (fractions) can't actually express them. like you can't express 1/3 in decimal, so you have to use periodic fractions to clarify what you mean, and you can't express irrationals using fractions or non-secondary language construct (as in, not through an operation like sqrt()), so you have to use limits and functions to clarify the meaning. periodic fractions and irrational numbers aren't "infinite" in the same way, but the concept of their inability to be expressed finitely is the same, it's just the problem that makes them "inexpressable" lies in different expression types. number system for periodic fractions, fractions for irrational numbers.

so "irrationality" of numbers isn't actually a property of numbers, it's a property of the language used to express them, and that's why everyone everywhere defines irrationals as "something that can't be expressed using X". because it's like saying "there are numbers just like other numbers, with the exception that you can't write these using fractions, so enjoy your math". like, "you don't see them in this language, but they are there, here's the proof: sqrt(2)"

... and if we come up with some other mathematical language or an extension to the existing one that would be able to express irrational numbers concisely like it can ordinary fractions, then we wouldn't even need the distinction between them, since it was purely practical: see these numbers, don't do this to them, it doesn't work in the language we've chosen to operate with numbers, and you'll be fine

... right?

like in that physics theory about 5th dimension - if you don't see it, it still works, you just don't understand why some stuff that uses that dimension behaves the way it does in your 4D understanding of the world

and that's when the contradictions come up in seemingly correct mathematical expressions when using irrationals like you would rationals - those are the symptoms of the flaw in our way of looking at things, those aren't the properties of the things themselves

[inner temple]

I really appreciate that you repeat what you understand. It's a great way to be sure you've understood.

You have made some good realizations, but some misunderstandings.

Irrationality is still a property of the number, just the same as rationality is a property of the number, and the same as primality for example. 5 is a prime number regardless of, say, the base.

The math actually gets a bit messier. There are too many real numbers for us to be able to define all of them. We simply can't describe the vast majority of real numbers!

The problem is that we have countably many symbols, and a description can only be finite in length, so we can only describe countably many numbers, but there are uncountably many real numbers.

I'm not sure if this is what you wanted to know, but at this point I feel like we've diverged a little into general math discussion.

[idle marlin]

it seems closely related to my original question, yes

in order to understand what the book is saying about irrational numbers in the context of continuum, i need to understand the properties of irrational numbers. countable and uncountable infinities were also discussed, and Cantor's diagonal method is still unclear to me

... but I need to think on new info again

[inner temple]

I recommend that you look around for a few different explanation of Cantor's diagonal argument. 3blue1brown probably has one, but there are lots out there regardless. When you've thought on it for a while, feel free to keep asking here or ping me or message me. I'm curious to see your realizations. 🙂

[idle marlin]

... but I have one question I kinf of forgot to ask

if you, idk, subtract irrational number from a rational, you'll always get irrational number?

[inner temple]

That's a good question, and you're right. irrational - rational will always give irrational

You might try to prove that as an exercise.

[inner temple]

Err, I read the order backwards, but the order doesn't matter and the + or - doesn't matter because multiplying by -1 doesn't change whether a number is rational or irrational.

[rapid arch]

Adding, subtracting, multiplying or dividing (not zero) a rational and irrational will always give you an irrational

[idle marlin]

wow, I think I got the Cantor's diagonal proof, maybe

if we say set A includes all entities of type T, and if we can find an entity of type T that is guaranteed to be different from all entities in set A, then this entity is not in the set, but the set was defined to include ALL such entities

the thing with Nth digit of the Nth entity in the set is just a clever algo to find an entity that is different from all other entities in the set

and the reason it succeeds is because we have "infinite" entities in a set that encompasses all of those entities

that's why it works both for a set of all infinite binary sequences and for a set of all irrational numbers (which are "infinite" in how they are written)

... and if we have an uncountable set I (irrationals) and a countable set R (rationals), then when we define set E (real numbers) which includes both rationals and irrationals, the uncountable nature of irrationals makes the whole parent set E uncountable too, even though it includes rationals from a countable set

and that's why a set of all real numbers is uncountable. if at least one thing in it isn't countable, the whole set is uncountable

[idle marlin]

i think that's what you've been trying to tell me there

[idle marlin]

i'm still unclear on how real numbers sequence works

like, there's always an irrational number between any two rationals

why is that? because between a and b rationals you can have an irrational number that is "infinitely close to a", which means it's less than b but not quite equal to a? also, since you can always find more "accurate" rational which is not more or not less than other rational, that means that you can have as many "in-betweens" for two rational numbers as you want, and in each of those "in-betweens" you'll always have at least one irrational number

like, you have 1/4 and 2/4, and there must be an irrational between them, because there's always an irrational between two rationals

but you can find infinitely many rationals between 1/4 and 2/4, like you can find 2/8 + 1/8 (which is more than 1/4 (=2/8) and less than 2/4 (=4/8)), and now you have at least one irrational between 1/4 and 3/8 and then one more between 3/8 and 2/4, so you can have infinitely many irrationals between any two rationals

... and that's why it's continuous - irrationals bind the rationals together, fill the gaps

[idle marlin]

... my book also says that a set of all irrational numbers is a continuum ("has power/magnitude of continuum" - in my poor translation)

if a set of all real numbers is continuum
and if a set of all irrational numbers is continuum

... then this means both sets are "uninterrupted", which means some irrational numbers must be equivalent to rational numbers... ?

i mean, we took the set of all real numbers, and it's continuum. and then we removed all rationals from it to get only irrationals and it's still continuum. that must mean those rationals we removed are somehow represented in the set of all irrationals (for it to be continuous). and the rationals were actually duplicating some of the quantities represented by the sequence of irrationals in the real numbers set. but that must mean one quantity that a rational represents couldn't have been bound 1-1 to a quantity that an irrational represents, otherwise we would have been able to represent irrationals the same way as rationals. so... quantity represented by a rational number corresponds to >1 quantities represented by irrational numbers

and that's why the least upper bound is an irrational number

[rapid arch]

For any two rational numbers, $a$, and $b$, you can create the irrational number $a+\frac{b-a}{\sqrt{2}}$ and it is between $a$ and $b$.

[jolly rover]

bonimy

[inner temple]

You've made some good progress, but also done misunderstandings. The arithmetic of infinities does not work how you think it works.

Indeed, irrationals and reals have the same cardinality (which you can think of as size), but one is contained in the other.

Consider the set of natural numbers and the set of natural numbers starting at 7.

[rapid arch]

One is a proper subset of the other

[inner temple]

I can match the elements of one set to the other set by pairing off x with x+7.

This matches the two sets perfectly (called a bijection), which means the two sets are the same size.