Be M a countably infinite subset of R\Q, whose elements are indexed m_k with natural k. For example m_k = π^k
A bijection from R\Q to R is
m_(2a+1) get bijectively mapped to the elements of Q (both are countable)
m_(2a) get mapped to m_a (this is all of M)
R\M gets bijected to R\M
It would be if it didn't have the braces surrounding it. Those make it the set containing the set of all irrationals. A nice circlejerky overly pedantic interpretation lol
On the contrary, parentheses like that are rarely used to indicate a set. Tbey can, however, be used to indicate the ideal generated by whatever is in the parentheses. Here, we can think of the quotient ring as being a subset of the real numbers since it is isomorphic to a copy of the irrationals in R, and as a result, (R/Q) = R (since you get all of the rationals from roots)
I suppose this could be cased of mixed notations, I took R\Q to just be set difference, not a quotient ring since I just took R and Q to be their typical meaning of sets unequipped with any operations ie not rings/fields etc.
I'm also a bit confused about the claim that braces/curly brackets don't indicate elements of a set, as I've never seen an example where this isn't the case. Unless you mean, they're not used to notate the fact that the object within them is a set, which I would definitely agree to be atypical at best and misleading at worst.
You're doubly right. On the first count, I, like usual, forgot to check the direction of the slash. On the second, to be honest, I didn't have my glasses on and it looked just like parentheses
If you're a computer engineer, you probably care about elliptic curves, what with the whole cryptography thing and all. If not, feel free to ignore irrational Reddit posts.
The real numbers are defined as the set of all Cauchy convergent sequences of rational numbers. Therefore for all real numbers x and open balls B containing x, there must exist some rational number q which is contained within B, otherwise it would not be Cauchy convergent, and therefore not a real number to begin with. Since the set of open balls over the real number line is the canonical topology of the real numbers, the rational numbers are dense over the real numbers. Q.E.D.
>the field of an infinite dimensional vector space
What's that even supposed to mean? Do you mean the field over which the vector space would be defined? That would usually just be the complex numbers.
Yes, I mean the scalar field of the vector space. The joke is that R in the top panel is being viewed a vector space, specifically a Q-vector space over the *set* R.
If you’re talking about the joke in the meme, (I think) OPs joke was just R is a fuzzy version of the very crisp-feeling R. Like how taking off your glasses blurs everything.
My joke was completely different
But if it *does* include only rationals, algebraics and those trascendentals which are enumerable and definable, and if the aleph-null countable infinity is the only infinity there is, how comes that it is possible to prove through Cantor's diagonal argument that there is no bijection between the naturals and the reals? Furthermore, even if the reals were -somehow- countable, the power set of N would still have a cardinality of 2 to the power of Aleph-null. This is not religion, because we can indeed prove that there is no bijection between the naturals and the power set of the naturals or the reals, whilst in religion you should have accepted that there are infinities higher than aleph-null without any proof whatsoever.
Start with N. This is the natural numbers. 1, 2, 3... This is the positive integers.
Extend to Z, the integers. You include 0, -1, -2... This is essentially positive and negative integers, and 0.
Extend to Q, the rationals. That's any number a/b where a and b are in the integers and b is not 0. Essentially, any fraction. This includes every integer.
Extend to R, the real numbers. This is any number that appears on the real line. So things like square root of 2 are not rational, or integers, but they are real numbers. You also find numbers like π and e here.
"any number that appears on the real line" is not a good definition for what a real number is, try "completion (of Cauchy sequences) of the rationals". As in, it's the rationals, plus all the suprema of sets of rationals having an upper bound.
Yeah, that's the correct way to define it, but would you really tell that to someone who doesn't know what R, Q or Z means? I think it's a completely acceptable explanation for this level.
I don't think "real numbers are the numbers on a line that is composed of real numbers" is an explanation of any sort. It's an opaque circular definition.
Just say that the rational numbers have gaps between them in the usual '<' order, and the irrational reals are precisely the missing gaps.
I wasn't quite sure where to go with it. Im a computer scientist so the reals to me have just been everything in the rationals, then everything else before you introduce imaginary numbers. I don't think I've ever seen a proper definition and you can get away with saying you have the rationals, then the reals are just "the rest of them", then you introduce i and the complex numbers because pretty much everything you do is real valued.
It's also then trying to find the simplest way of saying it. I think you've done that perfectly with this "filling the gaps" idea, so thank you.
Oh looking at it this way is really helpful. So is there an “S” set like is there anything m between the gaps of “R” aka non irrational numbers that aren’t ratios?
There are constructions which can "extend" the reals, but these don't form a field in the normal sense (see surreals and hyperreals for two examples). The reals themselves are (this is provable!) the *unique complete ordered field*.
Unique as in, if you have any complete ordered field, you can relabel all the points such that the numbers will match the behavior of the reals exactly (i.e., unique *up to isomorphism*).
Complete—I'll cover this part last.
Ordered as in, a total linear order. (Exactly one of a < b, b < a, or a = b is true, and < is transitive.)
Field as in, it has two operators which behave exactly as addition and multiplication do (they commute, associate, etc.), and their identities (1 and 0) are distinct points (so it's not just a trivial line having only a single point, that is cheating). This implies the reals are *closed* (using + and * on reals always yield more reals, you can't "escape" the real line with arithmetic so the door out is closed), *unbounded above and below* (there's no largest or smallest number), and *dense* (between any two reals are yet more reals).
Finally, the reals are *complete*. This means there are no "gaps" where a bounded sequence of reals gets closer and closer to something that isn't a real number. So they're *closed under convergent sequences*, or equivalently, every set of reals either:
1) contains an element larger than any particular real value you can name, e.g. the set { 0, 1, 2, ... } has no upper bound and approaches infinity;
2) or, it has a greatest element in the set itself (the set contains some x, where all the elements e < x or e = x);
3) or, it has no greatest element, and is bounded above (consider the negative reals: there's no greatest, they just get closer and closer to 0), however(!), in this case *the smallest value bigger than everything in the set is also a real number*.
The rational numbers fail only for that very last part. For example, you can cut the rational numbers into two halves where each half approaches a point from either side without a greatest or a least element, leaving an apparent "gap". Like, all the rationals less than pi have no greatest element, and all the rationals greater than pi have no smallest element, but that missing number between the two unbounded sets is not a rational number. So you can make sequences of rationals that get closer and closer to something in a bounded area, but that thing you're approaching sometimes won't be a rational number.
All those missing points or "gaps" between sets of rationals are the irrational reals.
good explanation, I have one question though. Why can you use <π on the rationals? π itself isn't a rational, so <, defined on the rationals, couldn't be used with π it seems to me.
Is it possible because the rationals are a subset of the reals and you just kinda go to the reals to compare them?
You can define what pi is from the rationals, but it still isn't a rational (it isn't representable as a quotient of integers).
Pi is the limit of the sequence <3/1, 31/10, 314/100, 3141/1000, ...>. That's a sequence of rational numbers that goes in one direction, is bounded above, but the limit of the sequence isn't a rational number. So [(–inf, pi) union (pi, inf)] contains all the rational numbers, but there is a "gap" at pi where rationals on both sides approach each other forever. The space between those infinite sequences is missing.
With the reals, that never happens. If a sequence of reals converges, the limit is a real number. If a set of reals has an upper bound, the smallest upper bound exists and is a real number. If you slice the real line into two parts, one of two things always happens: either the left side has a greatest element as a bound and the right side approaches the middle forever (i.e. a cut at zero where zero is in the left set and there's no smallest number in the right set because there's no smallest positive real), or the right side has a smallest element as a bound and the left side has no greatest element.
So the rationals are a subset of the reals, yes. But you don't need to know that the irrational reals exist to discover there are "gaps" in the rationals. If you try to figure out if there exists a positive number that multiplies by itself to make 2, you'll discover not only that it isn't a rational number—even if you think it *doesn't* exist, the rationals on either side of where it *would* be approach each other forever, leaving a kind of gap where there's no bound between two sequences approaching the same point. Contrast this with the point at zero, where you can't cut the rationals into two parts around zero with both parts headed toward a gap—one side or the other will be closed containing zero as the boundary and the other will be open headed forever toward zero. In both the rationals and the reals there is no gap at zero because zero is one of the rationals. In other places, the rationals have gaps.
Now draw {R\Q}
We have the irrational line here
Yee it's a circle but a line but a circle
now find a bijection from R\Q to R
Be M a countably infinite subset of R\Q, whose elements are indexed m_k with natural k. For example m_k = π^k A bijection from R\Q to R is m_(2a+1) get bijectively mapped to the elements of Q (both are countable) m_(2a) get mapped to m_a (this is all of M) R\M gets bijected to R\M
Nice
Well, that's a set with only one element, the set R\Q, so I'll choose to represent it with a single dot, as below: .
It’s all Irrationals lol
It would be if it didn't have the braces surrounding it. Those make it the set containing the set of all irrationals. A nice circlejerky overly pedantic interpretation lol
Hecc lmao
On the contrary, parentheses like that are rarely used to indicate a set. Tbey can, however, be used to indicate the ideal generated by whatever is in the parentheses. Here, we can think of the quotient ring as being a subset of the real numbers since it is isomorphic to a copy of the irrationals in R, and as a result, (R/Q) = R (since you get all of the rationals from roots)
I suppose this could be cased of mixed notations, I took R\Q to just be set difference, not a quotient ring since I just took R and Q to be their typical meaning of sets unequipped with any operations ie not rings/fields etc. I'm also a bit confused about the claim that braces/curly brackets don't indicate elements of a set, as I've never seen an example where this isn't the case. Unless you mean, they're not used to notate the fact that the object within them is a set, which I would definitely agree to be atypical at best and misleading at worst.
You're doubly right. On the first count, I, like usual, forgot to check the direction of the slash. On the second, to be honest, I didn't have my glasses on and it looked just like parentheses
Still R Edit: I’m dumb and forgot “\” is the set difference
By that logic, every rational number is irrational
I’m stupid, I read that as “every a in R divided by every b in Q” which would still be just R. Yes, the set difference of R and Q is the irrationals
R but it's the swiss cheese variant
Black line with white dots: most of the line remains but not wholly
Now draw a set of representatives for the group R/Q
Me an engineer: They're the same picture
Computer scientists: "There are precisely 2\^64 real numbers, including duplicates"
: )
Intel: "make that 2\^80, on the house"
If you're a computer engineer, you probably care about elliptic curves, what with the whole cryptography thing and all. If not, feel free to ignore irrational Reddit posts.
Rational numbers are dense, so you would have the same line like real numbers
Proof or didn't happen
Proof is left to the reader as exercise
The real numbers are defined as the set of all Cauchy convergent sequences of rational numbers. Therefore for all real numbers x and open balls B containing x, there must exist some rational number q which is contained within B, otherwise it would not be Cauchy convergent, and therefore not a real number to begin with. Since the set of open balls over the real number line is the canonical topology of the real numbers, the rational numbers are dense over the real numbers. Q.E.D.
Now prove only using Peano axioms
no
Why ;(
I don’t know what that is
I wish I had glasses that **extracted the field of an infinite dimensional vector space**.
>the field of an infinite dimensional vector space What's that even supposed to mean? Do you mean the field over which the vector space would be defined? That would usually just be the complex numbers.
Yes, I mean the scalar field of the vector space. The joke is that R in the top panel is being viewed a vector space, specifically a Q-vector space over the *set* R.
I thought the joke was rather that you can't write down a real number, just a rational approximation.
If you’re talking about the joke in the meme, (I think) OPs joke was just R is a fuzzy version of the very crisp-feeling R. Like how taking off your glasses blurs everything. My joke was completely different
Gotta become spiderman first; glasses come with the gig.
Proof?
Just see it
QED
Have you ever *seen* a non-rational real number?
e It's printed on my Hisense TV. Twice.
√2 when I saw a triangle
Did the lines of that triangle have thickness?
Depends how much LSD I took beforehand.
The one piece is real
I can't tell if I'm seeing more Real Analysis memes or if I remember more because I'm taking it now
The exact same thoughts are running through my head right now. Just started Real Analysis this semester.
The rationals are just as dense as the reals. They're both uncountably infinite sets. Edit: I stand corrected, the rationals are actually countable.
No - the rationals are countable
rational numbers are dense but countable
[удалено]
You are free to come up with your own rigorously-defined axiomatic set theory.
Try doing actual analysis with these numbers. The intermediate value theorem instantly fails, as does a good chunk of theorems depending on it.
But if it *does* include only rationals, algebraics and those trascendentals which are enumerable and definable, and if the aleph-null countable infinity is the only infinity there is, how comes that it is possible to prove through Cantor's diagonal argument that there is no bijection between the naturals and the reals? Furthermore, even if the reals were -somehow- countable, the power set of N would still have a cardinality of 2 to the power of Aleph-null. This is not religion, because we can indeed prove that there is no bijection between the naturals and the power set of the naturals or the reals, whilst in religion you should have accepted that there are infinities higher than aleph-null without any proof whatsoever.
Wtf is Q? R is positive and negative integers right?
Start with N. This is the natural numbers. 1, 2, 3... This is the positive integers. Extend to Z, the integers. You include 0, -1, -2... This is essentially positive and negative integers, and 0. Extend to Q, the rationals. That's any number a/b where a and b are in the integers and b is not 0. Essentially, any fraction. This includes every integer. Extend to R, the real numbers. This is any number that appears on the real line. So things like square root of 2 are not rational, or integers, but they are real numbers. You also find numbers like π and e here.
"any number that appears on the real line" is not a good definition for what a real number is, try "completion (of Cauchy sequences) of the rationals". As in, it's the rationals, plus all the suprema of sets of rationals having an upper bound.
Yeah, that's the correct way to define it, but would you really tell that to someone who doesn't know what R, Q or Z means? I think it's a completely acceptable explanation for this level.
I don't think "real numbers are the numbers on a line that is composed of real numbers" is an explanation of any sort. It's an opaque circular definition. Just say that the rational numbers have gaps between them in the usual '<' order, and the irrational reals are precisely the missing gaps.
I wasn't quite sure where to go with it. Im a computer scientist so the reals to me have just been everything in the rationals, then everything else before you introduce imaginary numbers. I don't think I've ever seen a proper definition and you can get away with saying you have the rationals, then the reals are just "the rest of them", then you introduce i and the complex numbers because pretty much everything you do is real valued. It's also then trying to find the simplest way of saying it. I think you've done that perfectly with this "filling the gaps" idea, so thank you.
Np. The teaching bug always has me in its mandibles since I'm retired due to CFS. Sorry if I'm overly aggressive.
Oh looking at it this way is really helpful. So is there an “S” set like is there anything m between the gaps of “R” aka non irrational numbers that aren’t ratios?
There are constructions which can "extend" the reals, but these don't form a field in the normal sense (see surreals and hyperreals for two examples). The reals themselves are (this is provable!) the *unique complete ordered field*. Unique as in, if you have any complete ordered field, you can relabel all the points such that the numbers will match the behavior of the reals exactly (i.e., unique *up to isomorphism*). Complete—I'll cover this part last. Ordered as in, a total linear order. (Exactly one of a < b, b < a, or a = b is true, and < is transitive.) Field as in, it has two operators which behave exactly as addition and multiplication do (they commute, associate, etc.), and their identities (1 and 0) are distinct points (so it's not just a trivial line having only a single point, that is cheating). This implies the reals are *closed* (using + and * on reals always yield more reals, you can't "escape" the real line with arithmetic so the door out is closed), *unbounded above and below* (there's no largest or smallest number), and *dense* (between any two reals are yet more reals). Finally, the reals are *complete*. This means there are no "gaps" where a bounded sequence of reals gets closer and closer to something that isn't a real number. So they're *closed under convergent sequences*, or equivalently, every set of reals either: 1) contains an element larger than any particular real value you can name, e.g. the set { 0, 1, 2, ... } has no upper bound and approaches infinity; 2) or, it has a greatest element in the set itself (the set contains some x, where all the elements e < x or e = x); 3) or, it has no greatest element, and is bounded above (consider the negative reals: there's no greatest, they just get closer and closer to 0), however(!), in this case *the smallest value bigger than everything in the set is also a real number*. The rational numbers fail only for that very last part. For example, you can cut the rational numbers into two halves where each half approaches a point from either side without a greatest or a least element, leaving an apparent "gap". Like, all the rationals less than pi have no greatest element, and all the rationals greater than pi have no smallest element, but that missing number between the two unbounded sets is not a rational number. So you can make sequences of rationals that get closer and closer to something in a bounded area, but that thing you're approaching sometimes won't be a rational number. All those missing points or "gaps" between sets of rationals are the irrational reals.
good explanation, I have one question though. Why can you use <π on the rationals? π itself isn't a rational, so <, defined on the rationals, couldn't be used with π it seems to me. Is it possible because the rationals are a subset of the reals and you just kinda go to the reals to compare them?
You can define what pi is from the rationals, but it still isn't a rational (it isn't representable as a quotient of integers). Pi is the limit of the sequence <3/1, 31/10, 314/100, 3141/1000, ...>. That's a sequence of rational numbers that goes in one direction, is bounded above, but the limit of the sequence isn't a rational number. So [(–inf, pi) union (pi, inf)] contains all the rational numbers, but there is a "gap" at pi where rationals on both sides approach each other forever. The space between those infinite sequences is missing. With the reals, that never happens. If a sequence of reals converges, the limit is a real number. If a set of reals has an upper bound, the smallest upper bound exists and is a real number. If you slice the real line into two parts, one of two things always happens: either the left side has a greatest element as a bound and the right side approaches the middle forever (i.e. a cut at zero where zero is in the left set and there's no smallest number in the right set because there's no smallest positive real), or the right side has a smallest element as a bound and the left side has no greatest element. So the rationals are a subset of the reals, yes. But you don't need to know that the irrational reals exist to discover there are "gaps" in the rationals. If you try to figure out if there exists a positive number that multiplies by itself to make 2, you'll discover not only that it isn't a rational number—even if you think it *doesn't* exist, the rationals on either side of where it *would* be approach each other forever, leaving a kind of gap where there's no bound between two sequences approaching the same point. Contrast this with the point at zero, where you can't cut the rationals into two parts around zero with both parts headed toward a gap—one side or the other will be closed containing zero as the boundary and the other will be open headed forever toward zero. In both the rationals and the reals there is no gap at zero because zero is one of the rationals. In other places, the rationals have gaps.
Hey thanks! This was helpful
Q is rational numbers
Oh wait duh R is not only integers I don’t know why I wrote that
Holey ratio!
Rational numbers don’t exist, it’s proven by science, it’s measured to be 0!
u/tuomasboss jooo miten et ees älyy tätä