Wow, your teaching is extraordinary! Thank you so much for making this! I'm a mathematics undergrad in my first semester and this helped me out greatly! cheers!
Very interesting! I also wanted to let you know that I love your videos! I'm an upcoming freshman in college and I feel like these courses will help greatly with my discrete math course and many others as I go on. Thank you for all of these!
You're absolutely right that the Lebesgue measures of the sets [0,1] and [0,1) are the same. However, Lebesgue measure is one of many ways in which we can analyze sets. If you haven't already, take a look at the lecture on Topology, which goes more into detail about open and closed sets and can give you a better idea of just WHY these two intervals are fundamentally different. (By the way, consider the fact that the sets {1,2} and {1,2,3} have the same Lebesgue measure but are quite different!)
On "Constructing Z": [(3,1)] = 2 and [(1,3)] = -2. Multiply them together and you get -4 or [(6,10)] not [(10,6)]. You must have got the result of multiplication in the wrong order, unless I'm going mad. When I looked at the slide originally I figured that [(a,b)] meant -a+b but if it doesn't, then the slide is wrong.
Hello! Thanks for the very interesting series. I have one little correction: at 17:45 you said we had defined the set of real algebraic numbers by extending the rationals with all possible nth roots. There are real algebraic numbers that can't be expressed in terms of roots, however, as Abel (I guess) proved. For instance, the roots of x^5 - x + 1 are real and algebraic, but are not in the set you defined.
At 23:43, you say that real numbers can be put into a 1:1 correspondence with all possible Dedekind cuts of Q. So take that example of the Dedekind cut at the sqrt(2). This would produce a set A of all rational numbers less than sqrt(2) and a set B of all rational numbers greater than sqrt(2). Now suppose that we take a Dedekind cut at a new number x, but this results in the same set of A and B as before. Is it a necessity that x is equal to sqrt(2)? In other words, if the number "closest" to sqrt(2) is a rational number, then you are fine, but if there number closest to sqrt(2) is another irrational number, than at least 2 irrational numbers produce the same Dedekind cut sets. Can you prove there are not an infinite number of irrational numbers which result in the same Dedekind cut sets? It seems to me like the Dedekind cut proof is saying that only one irrational number exists between every two rationals... and it seems if this were true, the sets would have the same carnality, as I could define the one-to-one correspondence as in the video. I am sure I am missing something...
If the two number x and sqrt(2) are different but creates the same partition of A and B, then there would be no rationals between the two. That however cannot be the case, as the two numbers x and sqrt(2) differ by an amount larger than zero and any non zero size interval contains rational numbers. Sorry that I skipped the equivalence class gymnastics here.
Interesting thought - I'm not sure of the top of my head what number system would be created by looking at, say, Z x A, but I can assure you it would be countable because a Cartesian product of countable sets is countable. You're correct that we can't represent R as the Cartesian product of two countable sets. Also, we wouldn't use Aleph_1 for R (or R² or R³, etc.), but actually the next letter of the Hebrew alphabet, Beth_1. Look up "Beth numbers" on Wikipedia for more information.
Wow, good catch. [(a,b)] * [(c,d)] should in fact be [(ac+bd,ad+bc)]. It turns out my source on that construction, Interactive Real Analysis (tinyurl dot com slash m44l4t4), was actually incorrect ... I'll have to do a bit more fact-checking in the future. My apologies, and annotated accordingly.
Oh hey, perfect thing for me to do while waiting for the washroom! Okay, so usually "A is a subset of B" means that all elements of A are in B. So yes. The set of Naturals just constructed is not a subset of that set of integer constructed like so, by that definition. But we can still take the subset of the integers of the form [(a, b)] such that a ≥ b and get bijections (f([(a, b)]) = a - b. Can be well-defined since a ≥ b) between that subset and the naturals _which preserves addition, multiplication etc._ . That effectively means there is a set that basically behaves just like the natural numbers (with respect to arithmetics) inside of the integers. A pretty suitable compromise, if you ask me. Anyways, since we have this, to multiply an integer z with a natural n, you just multiply z with the corresponding equivalence class in the integers: [(n, 0)]. Similarly for addition. Alternatively, you can also define × : {(a, b) | a is in *Z* AND b is in *N*} -> *Z* with a × b = a * f(b), where f is the bijection between the natural numbers and the corresponding subset in the integers (same with addition)
22:20 . Be careful! The texts can be misleading! (Maybe he can add some annotations? It's been 5 years, can he even do that? eh) "... the lower set contains all the negative numbers, union * the solutions of the inequality x^2 ≤ 2 in the rational numbers *" This defines the interval {x | x is in *Q* AND x < sqrt(2)} without ever mentioning that irrational number. Also, clearly, this means its upper set, the complement, contains all rationals x ≥ sqrt(2). (and since sqrt(2) is irrational, it contains all rationals x > sqrt(2)). With this in mind, we define a real number as the union of sets of rational numbers, satisfying inequalities. Transcendental numbers like pi and e can be reached using (taylor) approximations and etc. Then, just define the corresponding operations on them. (There's a lot more information online. I'm not quite a professional, just a curious student --don't take my word, research more yourself :D)
Ah, but you see, it is! Suppose you DO make a set that allows so, and have the exact same constraints. Then, (3, -4) ~ (-3, 4). After all, (3 * 4) = (-3 * -4). Thus, they would be in the same equivalence class. So, allowing negative denominators does not give us any new element (nor does it cost us any). Really, these two sets are the same! (Well, isomorphic respecting arithmetics and etc.)
This is an observation. Not really anything for me to correct/explain. But, I already replied to all the others, so... Just one important thing, though. It does not make the assumption that these natural numbers are sets. But I believe that it also does not make the assumption that these numbers are _not_ sets. Just something interesting, I thought.
@CogitoErgoCogitoSum Great remark on the incompatibility of naturals as numbers and equivalence classes of pairs of naturals as set, though this is something that can be solved after we migrate from naturals to the equivalence classes. More specifically, in the new system a natural n would be the equivalence class of (n,0), and Z would be the set of all equivalence classes, so it would transpire that N is a subset of Z. As for the fractions with negative denominators, they can be defined afterwards by an easy clause such as a/-b df= -a /b, or we can embed it right into the original definition by allowing b to be a non-negative number as well. For the existence of sqrt(2), the way it's done is circular indeed, but there's a way to go about it without invoking sqrt(2): just define the lower set of Dedekind cut as {x in Q | x 2}, and once we prove that this is indeed a Dedekind cut, we've got the square root of 2! (easier said than done of course, as to prove that it's indeed a cut, we'd have to show that 1) A and B forms a partition of Q and 2) any element from A is smaller than any element from B and 3) A has no greatest element. As you can see, a level of sophistication with inequalities is needed here.) So in general, there's a bit of details to be filled in when it comes to these kinds of stuffs, but presentation-wise, I think the video summarizes the essence of construction of R from the ground up.
I like your series very much! Thanks. Right now I'm watching Math Foundations 116 by N. Wildberger. There Wildberger is saying: Dedecuts don't work. Can I ask Bill, what's your opinion about his arguments? Does it makes sense to you that you can't construct the real’s because there's not enough computing time, place in the universe, make an infinite amount of choices, etc?
jazzymarieke As much respect as I have for Dr. Wildberger, I have to disagree with his rather staunch stance on the real numbers, especially when he seems so willing to work with imaginary numbers and even infinity as an ideal point in projective geometry. To me, a big part of the beauty of mathematics is that it transcends the necessity to stay fixed in the real world --- if your mind can invent it, it exists. :) If you want to do something until infinity, you need not be constrained by the finiteness of the universe.
Since a Dedekind cut is written using a finite number of symbols picked from a finite alphabet, they can at best construct a countably infinite set. I.e. not the Real numbers. Or did I totally misunderstand how to set up bijections?
So, is the point of Dedekind cuts that we have already constructed the rational numbers and the algebraic numbers. And now we're using those to construct the transcendental numbers too?
What blows my mind is that the closed set [0,1] has greatest member 1. But if we remove the right end-point, [0,1), this set has no greatest member! But the two sets, considered geometrically, would seem to be congruent.
Also since a Dedekind cut requires a equation, isn't it in fact a construction of computable numbers, rather than real numbers? Surely it is not possible to construct a real with a fully random decimal expansion this way?
As we are constructing reals from dedekind cuts which can be represented by sets of rationals, can we say that there exist a bijection between reals and the power set of the rationals and thus proving reals has the same cardinality as the power set of rationals? (While the result that c = 2^N0 is indeed true, the proof in wikipedia is long and involves things like tangent functions, I am wondering is what I wrote above a valid proof for c=2^N0)
SInce the union of countable sets is countable, does that mean that the cartesian product between say the integers and the algebraic numbers must be countable. What might changing these cartesian products do? What kind of number system could be created? Also, since the real numbers are uncountable, does that mean it cannot be represented as a cartesian product between any other two countable sets? Is R^2 uncountable too? If so, what is its cardinality, aleph-two? Or aleph-one like the reals? TY
So if I understand correctly, a Dedekind cut at an irrational number is defined by two sets of rationals, one containing all smaller numbers, and the other containing all larger numbers. However there are supposed to be _many_ more irrational numbers (i.e. locations where we can cut) than rational numbers, so wouldn't there be an infinite amount of irrational numbers sandwiched in between _the same_ rational boundaries?
No, any non-zero size interval contains rational numbers. So any two different irrational numbers are separated by an infinite number of rational numbers !
@@1isten2me Doesn't that go counter the whole idea of Aleph Zero and Aleph One then? I think the whole problem here is that when you write down a Dedekind cut, you by necessity have to use a finite amount of ink. Or a finite amount of symbols picked from a finite set of symbols. But the cardinality of such a set is Aleph Zero, whereas the Reals demand to be Aleph One. So Dedekind cuts cannot possibly be a construction of *the reals*, at best they can be used to construct *some reals*.
@David Rutten While the rational numbers are countable, the set of Dedekind cuts is not. I think I see your concern with a set definition, e.g. the set of dedekind cuts, without a way to systematically produce or describe each elements, even with an infinite amount of ink. I got interested in these questions from watching NJ Wilberger’s videos recently, he has a strong opinion on any matter related to the real numbers :-)
@@1isten2me I briefly met Wildberger at the ISGG in '14 in Innsbruck. It was indeed the first time I've heard about a serious modern mathematician rejecting the Real numbers (in their current form). I've since become fairly convinced by his arguments, but I realise that I'm in no way an expert on this, so it would be easy for someone like him to fool someone like me. Sadly, nobody seems to be putting out any proper counter-arguments to his objections. So, given that every Real number is constructed via a unique Dedekind cut, how does one perform basic arithmetic on two of those cuts if they represent numbers which cannot be written down using a finite amount of symbols in their set-builder notation?
@David Rutten I am not sure about that either. Maybe the answer is that you don’t. And in that respect dr Wildberger might be right that most of the real numbers are hidden in eternal darkness. At the same time I like the simplicity of having a set that includes all the computable and describable numbers, where all the nice operations just work. I have not seen a single “real” issue exposed yet. Have you?
So then, since Q is constructed from the Cartesian product of Z and N, and Z is constructed from N x N, then Q is technically constructed from an ordered triple of N, correct? (N x N) x N?
At 5:12, what about + shouldn't you define that operation after you have defined the Natural numbers? + is in fact a function NxN->N, so basing the natural numbers axiomatically on an operation like + may result in circular reasoning if you then use N to define the operation +. Or is there a way out of this?
When we're constructing N using Peano's axioms, we're only really using "+ 1" as notation for what the successor function does because we know what addition is already. We haven't defined TRUE addition yet, and we can do it in terms of the successor function. If I wanted, I could have pulled a bit of notation from computer science and written s(a) = a++. Either way, the successor operator is unary, not binary. We'd be building the binary operation of addition out of it.
Josh Kaiser suggests below that the gaps in Q may contain more than one real number. I have been trying to understand this for a couple of days now and was thinking the same. Q is countable, R is not. If Q is countable, the gaps in Q must be countable. So the gaps contain an uncountable number of reals divided by a countable number of gaps, resulting in an infinite number of real numbers in every gap. Just intuitive reasoning, not proof, I know. How about this reasoning: the lower set has no least upper bound and the upper set has no greatest lower bound as subsets of Q can't have these. Again intuitively, it feels like the unique real number in the gap could be the supremum/infimum which if included in the lower and upper sets would result in their closure. I am probably committing errors here. Can a set of numbers in Q be closed by adding a real number? Even if this is correct it does not answer my first question. Actually, one more point: Real numbers with a finite number of decimal places are rational numbers, so they must be countable. I guess real numbers are uncountable because they can have an infinite number of decimal places. Are rational numbers allowed to have infinite numbers in the numerator or denominator? Both together would be undefined of course, but otherwise allowing this would seem to allow the rationals to have the same "granularity" as the reals, and hence why is one countable and the other not. Sorry if my engineering viewpoint is making the pure mathematicians cringe!
No - left-unique means injective. It may sounds a little backwards at first, but think of it this way - left-uniqueness means an element on the right (codomain) can be pointed to by ONLY ONE UNIQUE element on the left (domain). No two left elements point to the same right element.
At 21:12 the number line is claimed to contain only rational numbers, yet at 22:18 we want to make a cut on this number line at an irrational number. I don't understand how could we find an irrational number from a number line that contains only the rational numbers?
What we're doing is using the solution set to the inequality "x < 0 or x² ≤ 2", in the *rational* numbers, to "cut" or "split" the rational number line into two parts. What ends up happening is that the left part doesn't end up having a maximum value (that's rational, at least), so there's a "hole" between the left part and the right part. That's where √2 lives -- you're right, it wasn't part of the original *rational* number line, so we include it in the *real* number line, along with all other places where we can "cut" and not find a rational number. Think of it this way as an analogy -- imagine you had only the integers, and you looked at the inequality "2x ≤ 5". Notice that you could find a solution that satisfies the "less than" part -- x=2 (or anything lower) -- but no solution that satisfies the "equal to" part. That would lead you to wonder if perhaps there's another number, one that isn't an integer, that could satisfy it -- in this case 2.5.
One of the typos I had fixed in annotations before TH-cam got rid of them all. :/ Eventually I'll go back through these and add errata. The last one should be [(6,10)].
I can follow the logic but I have more of a philosophical question. Why is it necessary to "construct" the Natural Numbers if they are already there? In this construction you use natural numbers so the argument seems cyclical. Many thanks.
I'm not really using the natural numbers in the construction. Rather, they're being defined axiomatically. What the construction does is start off by assuming one number exists - namely, 1 (or zero if you'd rather start there - I've been moving in that direction myself) and then show how to "generate" all the rest of the natural numbers from there. What may interest you as well is how Zermelo-Fraenkel set theory constructs the natural numbers ... almost from nothing! We start with 0 represented by ∅, the null set ... then once we have a number a, to get the next number a+1 (the "successor" of a in Peano's terms), we form the set a ∪ {a}. So, to start off: 0 = ∅ 1 = ∅ ∪ {∅} = {∅} 2 = {∅} ∪ {{∅}} = {∅,{∅}} 3 = {∅,{∅}} ∪ {{∅,{∅}}} = {∅,{∅},{∅,{∅}}} ... And so on. Crazy, isn't it? Effectively what we're doing is making each natural number correspond with the set containing all natural numbers less than it. So: 0 = ∅ 1 = {0} 2 = {0,1} 3 = {0,1,2} ... Does this help more?
I think it's easier for us to imagine "outward" infinity, 1,2,3,..., because we can extrapolate from our finite experience of moving around in space. But moving in "inward" space is also infinite, e.g., magnifying a Mandelbrot set, but there's no analogous experience in life, and I can't comprehend it. I'm a mysterian and think that the human mind has fundamental limitations. We can prove certain things but not apprehend them, e.g., different levels of infinity.
I would like to address your comment at time: 2:33, where you say "We are going to start at square 1, literally" I understand you actually are going to start with the square of one in a non metaphorical sense but because the concept of sqrt(1) is an abstract comment I feel like as a propositional statement that comment would evaluate to false. PS. I think my girlfriend hates that you taught us about formal argumentation.
+Dylan Jorgensen The "literally" part really applied only to "one", not so much for "square" --- no squaring or square rooting was meant to be implied :P (Although nowadays, if I were to do this series again, I would have instead started at zero --- I rather like the modern construction of the natural numbers starting with the empty set!)
+Bill Shillito Well, even if you would like to change some of the details about this series the unique thing is that it gives you a sense of a "math narrative". I think humans are wired for meaning over facts and this series gave me sense for "game" (quarrels, winners, losers, victories, etc..) for how math has evolved, making the facts the pieces that move the story forward was the genius part.
I don't understand why Peano's axioms 3, 4, and 5 are required. Surely 3 is already the default implication. Only if axiom 2 is specifically said to work both ways do you need a specific rule saying that one is the lowest natural number. I especially do not understand how equality is defined using successors. _If_ you can compare successors for equality, why can't you compare the original numbers for equality? And if this turns into a recursion where you have to compare the successors of the successors, etc., you've broken maths. Axiom 5 seems to be unnecessary for the construction of the Naturals, it sounds very much like an ad hoc axiom needed by Set theory. If we're only just now constructing naturals, aren't we getting a bit ahead of ourselves already juggling infinite sets?
Axiom 3, is needed because it's the default choice. You need somewhere to start, it's known as the axiom of choice. If you don't have it you could start anywhere, I could say Axiom 3 starts at 5 and discard the previous ones. Axiom 4 is needed otherwise the Natural Numbers are not one to one, this goes back to relation and sets theory if you have 3 the S(3) = 4 or can it equal S(3) = 5 so then you have ambiguity and everything collapses Axiom 5, it's more of a conclusive statement and you don't know what exactly "a" is, it could be 1 or 100002325234322342342323443242243324 but you know the next number following "a" where if I give you a number you would be able to give me a number using the function and thus we have all of the Natural numbers, infinity is hard to juggle with and is one of the biggest hurdles to overcome.
There s no where in this video that you actually CONSTRUCT the irrational numbers. You just assume they are uncountable without any justification. If we ignore the difficulties of constructing Dedekind cuts from Q for the transcendentals, you still need to convert an enumerable set of Dedekind cuts to an unenumerable set of the Reals. How do you do that?
I really appreciate that I do not have to go to a prestigious college to get such an amazing quality lecture. Thanks Bill for doing this.
Wow, your teaching is extraordinary! Thank you so much for making this! I'm a mathematics undergrad in my first semester and this helped me out greatly!
cheers!
Very interesting! I also wanted to let you know that I love your videos! I'm an upcoming freshman in college and I feel like these courses will help greatly with my discrete math course and many others as I go on. Thank you for all of these!
You're absolutely right that the Lebesgue measures of the sets [0,1] and [0,1) are the same. However, Lebesgue measure is one of many ways in which we can analyze sets. If you haven't already, take a look at the lecture on Topology, which goes more into detail about open and closed sets and can give you a better idea of just WHY these two intervals are fundamentally different. (By the way, consider the fact that the sets {1,2} and {1,2,3} have the same Lebesgue measure but are quite different!)
Great video Mr Shillito! It solved many of my questions. Thank you very much!
Years ago I struggled with Dedekind cut, "Why we need them". Now it seems obvious. The best series ever.
Legendary Bill is back ❤️❤️❤️
Well order , supremum infimum law of trichotomy and etc everything revised Thanks man🙏
On "Constructing Z": [(3,1)] = 2 and [(1,3)] = -2. Multiply them together and you get -4 or [(6,10)] not [(10,6)]. You must have got the result of multiplication in the wrong order, unless I'm going mad. When I looked at the slide originally I figured that [(a,b)] meant -a+b but if it doesn't, then the slide is wrong.
Hello! Thanks for the very interesting series. I have one little correction: at 17:45 you said we had defined the set of real algebraic numbers by extending the rationals with all possible nth roots. There are real algebraic numbers that can't be expressed in terms of roots, however, as Abel (I guess) proved. For instance, the roots of x^5 - x + 1 are real and algebraic, but are not in the set you defined.
At 23:43, you say that real numbers can be put into a 1:1 correspondence with all possible Dedekind cuts of Q. So take that example of the Dedekind cut at the sqrt(2). This would produce a set A of all rational numbers less than sqrt(2) and a set B of all rational numbers greater than sqrt(2). Now suppose that we take a Dedekind cut at a new number x, but this results in the same set of A and B as before. Is it a necessity that x is equal to sqrt(2)?
In other words, if the number "closest" to sqrt(2) is a rational number, then you are fine, but if there number closest to sqrt(2) is another irrational number, than at least 2 irrational numbers produce the same Dedekind cut sets. Can you prove there are not an infinite number of irrational numbers which result in the same Dedekind cut sets? It seems to me like the Dedekind cut proof is saying that only one irrational number exists between every two rationals... and it seems if this were true, the sets would have the same carnality, as I could define the one-to-one correspondence as in the video. I am sure I am missing something...
If the two number x and sqrt(2) are different but creates the same partition of A and B, then there would be no rationals between the two. That however cannot be the case, as the two numbers x and sqrt(2) differ by an amount larger than zero and any non zero size interval contains rational numbers. Sorry that I skipped the equivalence class gymnastics here.
Interesting thought - I'm not sure of the top of my head what number system would be created by looking at, say, Z x A, but I can assure you it would be countable because a Cartesian product of countable sets is countable.
You're correct that we can't represent R as the Cartesian product of two countable sets. Also, we wouldn't use Aleph_1 for R (or R² or R³, etc.), but actually the next letter of the Hebrew alphabet, Beth_1. Look up "Beth numbers" on Wikipedia for more information.
Wow, good catch. [(a,b)] * [(c,d)] should in fact be [(ac+bd,ad+bc)]. It turns out my source on that construction, Interactive Real Analysis (tinyurl dot com slash m44l4t4), was actually incorrect ... I'll have to do a bit more fact-checking in the future. My apologies, and annotated accordingly.
Oh hey, perfect thing for me to do while waiting for the washroom!
Okay, so usually "A is a subset of B" means that all elements of A are in B. So yes. The set of Naturals just constructed is not a subset of that set of integer constructed like so, by that definition.
But we can still take the subset of the integers of the form [(a, b)] such that a ≥ b and get bijections (f([(a, b)]) = a - b. Can be well-defined since a ≥ b) between that subset and the naturals _which preserves addition, multiplication etc._ . That effectively means there is a set that basically behaves just like the natural numbers (with respect to arithmetics) inside of the integers. A pretty suitable compromise, if you ask me.
Anyways, since we have this, to multiply an integer z with a natural n, you just multiply z with the corresponding equivalence class in the integers: [(n, 0)]. Similarly for addition. Alternatively, you can also define × : {(a, b) | a is in *Z* AND b is in *N*} -> *Z* with a × b = a * f(b), where f is the bijection between the natural numbers and the corresponding subset in the integers (same with addition)
22:20 . Be careful! The texts can be misleading! (Maybe he can add some annotations? It's been 5 years, can he even do that? eh) "... the lower set contains all the negative numbers, union * the solutions of the inequality x^2 ≤ 2 in the rational numbers *" This defines the interval {x | x is in *Q* AND x < sqrt(2)} without ever mentioning that irrational number.
Also, clearly, this means its upper set, the complement, contains all rationals x ≥ sqrt(2). (and since sqrt(2) is irrational, it contains all rationals x > sqrt(2)).
With this in mind, we define a real number as the union of sets of rational numbers, satisfying inequalities. Transcendental numbers like pi and e can be reached using (taylor) approximations and etc. Then, just define the corresponding operations on them. (There's a lot more information online. I'm not quite a professional, just a curious student --don't take my word, research more yourself :D)
Ah, but you see, it is! Suppose you DO make a set that allows so, and have the exact same constraints. Then, (3, -4) ~ (-3, 4). After all, (3 * 4) = (-3 * -4). Thus, they would be in the same equivalence class. So, allowing negative denominators does not give us any new element (nor does it cost us any). Really, these two sets are the same! (Well, isomorphic respecting arithmetics and etc.)
This is an observation. Not really anything for me to correct/explain. But, I already replied to all the others, so... Just one important thing, though. It does not make the assumption that these natural numbers are sets. But I believe that it also does not make the assumption that these numbers are _not_ sets. Just something interesting, I thought.
@CogitoErgoCogitoSum Great remark on the incompatibility of naturals as numbers and equivalence classes of pairs of naturals as set, though this is something that can be solved after we migrate from naturals to the equivalence classes.
More specifically, in the new system a natural n would be the equivalence class of (n,0), and Z would be the set of all equivalence classes, so it would transpire that N is a subset of Z.
As for the fractions with negative denominators, they can be defined afterwards by an easy clause such as a/-b df= -a /b, or we can embed it right into the original definition by allowing b to be a non-negative number as well.
For the existence of sqrt(2), the way it's done is circular indeed, but there's a way to go about it without invoking sqrt(2): just define the lower set of Dedekind cut as {x in Q | x 2}, and once we prove that this is indeed a Dedekind cut, we've got the square root of 2!
(easier said than done of course, as to prove that it's indeed a cut, we'd have to show that 1) A and B forms a partition of Q and 2) any element from A is smaller than any element from B and 3) A has no greatest element. As you can see, a level of sophistication with inequalities is needed here.)
So in general, there's a bit of details to be filled in when it comes to these kinds of stuffs, but presentation-wise, I think the video summarizes the essence of construction of R from the ground up.
I like your series very much! Thanks.
Right now I'm watching Math Foundations 116 by N. Wildberger. There Wildberger is saying: Dedecuts don't work.
Can I ask Bill, what's your opinion about his arguments?
Does it makes sense to you that you can't construct the real’s because there's not enough computing time, place in the universe, make an infinite amount of choices, etc?
jazzymarieke As much respect as I have for Dr. Wildberger, I have to disagree with his rather staunch stance on the real numbers, especially when he seems so willing to work with imaginary numbers and even infinity as an ideal point in projective geometry. To me, a big part of the beauty of mathematics is that it transcends the necessity to stay fixed in the real world --- if your mind can invent it, it exists. :) If you want to do something until infinity, you need not be constrained by the finiteness of the universe.
I am also puzzled by dr Wildberger’s attempt to differentiate between good and bad axioms. What is that all about?
Since a Dedekind cut is written using a finite number of symbols picked from a finite alphabet, they can at best construct a countably infinite set. I.e. not the Real numbers. Or did I totally misunderstand how to set up bijections?
You are the best!
So, is the point of Dedekind cuts that we have already constructed the rational numbers and the algebraic numbers. And now we're using those to construct the transcendental numbers too?
What blows my mind is that the closed set [0,1] has greatest member 1. But if we remove the right end-point, [0,1), this set has no greatest member! But the two sets, considered geometrically, would seem to be congruent.
Bill, you are da real mvp
Also since a Dedekind cut requires a equation, isn't it in fact a construction of computable numbers, rather than real numbers? Surely it is not possible to construct a real with a fully random decimal expansion this way?
As we are constructing reals from dedekind cuts which can be represented by sets of rationals, can we say that there exist a bijection between reals and the power set of the rationals and thus proving reals has the same cardinality as the power set of rationals? (While the result that c = 2^N0 is indeed true, the proof in wikipedia is long and involves things like tangent functions, I am wondering is what I wrote above a valid proof for c=2^N0)
SInce the union of countable sets is countable, does that mean that the cartesian product between say the integers and the algebraic numbers must be countable. What might changing these cartesian products do? What kind of number system could be created?
Also, since the real numbers are uncountable, does that mean it cannot be represented as a cartesian product between any other two countable sets? Is R^2 uncountable too? If so, what is its cardinality, aleph-two? Or aleph-one like the reals? TY
Up
So if I understand correctly, a Dedekind cut at an irrational number is defined by two sets of rationals, one containing all smaller numbers, and the other containing all larger numbers. However there are supposed to be _many_ more irrational numbers (i.e. locations where we can cut) than rational numbers, so wouldn't there be an infinite amount of irrational numbers sandwiched in between _the same_ rational boundaries?
No, any non-zero size interval contains rational numbers. So any two different irrational numbers are separated by an infinite number of rational numbers !
@@1isten2me Doesn't that go counter the whole idea of Aleph Zero and Aleph One then? I think the whole problem here is that when you write down a Dedekind cut, you by necessity have to use a finite amount of ink. Or a finite amount of symbols picked from a finite set of symbols. But the cardinality of such a set is Aleph Zero, whereas the Reals demand to be Aleph One. So Dedekind cuts cannot possibly be a construction of *the reals*, at best they can be used to construct *some reals*.
@David Rutten While the rational numbers are countable, the set of Dedekind cuts is not. I think I see your concern with a set definition, e.g. the set of dedekind cuts, without a way to systematically produce or describe each elements, even with an infinite amount of ink. I got interested in these questions from watching NJ Wilberger’s videos recently, he has a strong opinion on any matter related to the real numbers :-)
@@1isten2me I briefly met Wildberger at the ISGG in '14 in Innsbruck. It was indeed the first time I've heard about a serious modern mathematician rejecting the Real numbers (in their current form). I've since become fairly convinced by his arguments, but I realise that I'm in no way an expert on this, so it would be easy for someone like him to fool someone like me. Sadly, nobody seems to be putting out any proper counter-arguments to his objections.
So, given that every Real number is constructed via a unique Dedekind cut, how does one perform basic arithmetic on two of those cuts if they represent numbers which cannot be written down using a finite amount of symbols in their set-builder notation?
@David Rutten I am not sure about that either. Maybe the answer is that you don’t. And in that respect dr Wildberger might be right that most of the real numbers are hidden in eternal darkness. At the same time I like the simplicity of having a set that includes all the computable and describable numbers, where all the nice operations just work. I have not seen a single “real” issue exposed yet. Have you?
So then, since Q is constructed from the Cartesian product of Z and N, and Z is constructed from N x N, then Q is technically constructed from an ordered triple of N, correct? (N x N) x N?
At 10:12 I think your rule for multiplication is swapped because (a-b)(c-d)=(ac+bd)-(ad+bc)
You are correct, his sample at 10:48 is 2 * -2 = 4, which is incorrect, the rule is swapped.
I have a question : this work with any type of nubmer or just decimals?
At 5:12, what about + shouldn't you define that operation after you have defined the Natural numbers? + is in fact a function NxN->N, so basing the natural numbers axiomatically on an operation like + may result in circular reasoning if you then use N to define the operation +. Or is there a way out of this?
When we're constructing N using Peano's axioms, we're only really using "+ 1" as notation for what the successor function does because we know what addition is already. We haven't defined TRUE addition yet, and we can do it in terms of the successor function. If I wanted, I could have pulled a bit of notation from computer science and written s(a) = a++. Either way, the successor operator is unary, not binary. We'd be building the binary operation of addition out of it.
How do we order a set S? Eg. π>e?
Josh Kaiser suggests below that the gaps in Q may contain more than one real number. I have been trying to understand this for a couple of days now and was thinking the same. Q is countable, R is not. If Q is countable, the gaps in Q must be countable. So the gaps contain an uncountable number of reals divided by a countable number of gaps, resulting in an infinite number of real numbers in every gap. Just intuitive reasoning, not proof, I know.
How about this reasoning: the lower set has no least upper bound and the upper set has no greatest lower bound as subsets of Q can't have these. Again intuitively, it feels like the unique real number in the gap could be the supremum/infimum which if included in the lower and upper sets would result in their closure. I am probably committing errors here. Can a set of numbers in Q be closed by adding a real number? Even if this is correct it does not answer my first question.
Actually, one more point: Real numbers with a finite number of decimal places are rational numbers, so they must be countable. I guess real numbers are uncountable because they can have an infinite number of decimal places. Are rational numbers allowed to have infinite numbers in the numerator or denominator? Both together would be undefined of course, but otherwise allowing this would seem to allow the rationals to have the same "granularity" as the reals, and hence why is one countable and the other not.
Sorry if my engineering viewpoint is making the pure mathematicians cringe!
Why must it follow from the fact that Q is countable that the gaps in Q are countable?
At 04:25 you actually mean to say right unique, not left unique for them to be injective right? Kudos to the awesome work though!!
No - left-unique means injective. It may sounds a little backwards at first, but think of it this way - left-uniqueness means an element on the right (codomain) can be pointed to by ONLY ONE UNIQUE element on the left (domain). No two left elements point to the same right element.
Yeah my bad, watching two and a half straight hours of your videos took its toll on my brain :p
At 21:12 the number line is claimed to contain only rational numbers, yet at 22:18 we want to make a cut on this number line at an irrational number. I don't understand how could we find an irrational number from a number line that contains only the rational numbers?
What we're doing is using the solution set to the inequality "x < 0 or x² ≤ 2", in the *rational* numbers, to "cut" or "split" the rational number line into two parts. What ends up happening is that the left part doesn't end up having a maximum value (that's rational, at least), so there's a "hole" between the left part and the right part. That's where √2 lives -- you're right, it wasn't part of the original *rational* number line, so we include it in the *real* number line, along with all other places where we can "cut" and not find a rational number.
Think of it this way as an analogy -- imagine you had only the integers, and you looked at the inequality "2x ≤ 5". Notice that you could find a solution that satisfies the "less than" part -- x=2 (or anything lower) -- but no solution that satisfies the "equal to" part. That would lead you to wonder if perhaps there's another number, one that isn't an integer, that could satisfy it -- in this case 2.5.
@@BillShillito This is helpful, thank you! I admire your efforts bringing us up here.
Pls how can I get ur lectures from chapter 1
I think there's an issue in your construction of integers. You have the [(3,1)]x[(1,3)]=[(10,6)]. Doesn't that equate to 2 x -2 = 4?
One of the typos I had fixed in annotations before TH-cam got rid of them all. :/ Eventually I'll go back through these and add errata.
The last one should be [(6,10)].
@@BillShillito wow I never thought I'd get a reply from the creator on such an old video. You made a great video, props to you
Isn't the successor function typically denoted with a capital S?
Different people use different notation. The idea is the important thing.
@@BillShillito alright. Just checking because I thought I could've been mixing stuff up.
Bill se que hablas espanol....he visto tus videos todos tus videos, Gracias!!!!!
Bill, I know that you speak Spanish.... I've watched your videos, all of your videos, Thank you!!!!!
I can follow the logic but I have more of a philosophical question. Why is it necessary to "construct" the Natural Numbers if they are already there? In this construction you use natural numbers so the argument seems cyclical. Many thanks.
I'm not really using the natural numbers in the construction. Rather, they're being defined axiomatically. What the construction does is start off by assuming one number exists - namely, 1 (or zero if you'd rather start there - I've been moving in that direction myself) and then show how to "generate" all the rest of the natural numbers from there.
What may interest you as well is how Zermelo-Fraenkel set theory constructs the natural numbers ... almost from nothing! We start with 0 represented by ∅, the null set ... then once we have a number a, to get the next number a+1 (the "successor" of a in Peano's terms), we form the set a ∪ {a}. So, to start off:
0 = ∅
1 = ∅ ∪ {∅} = {∅}
2 = {∅} ∪ {{∅}} = {∅,{∅}}
3 = {∅,{∅}} ∪ {{∅,{∅}}} = {∅,{∅},{∅,{∅}}}
...
And so on. Crazy, isn't it? Effectively what we're doing is making each natural number correspond with the set containing all natural numbers less than it. So:
0 = ∅
1 = {0}
2 = {0,1}
3 = {0,1,2}
...
Does this help more?
Mind boggling
Bill I didn't understand whythe square root of 2 is a rational number
5:37 nice reference!
I think it's easier for us to imagine "outward" infinity, 1,2,3,..., because we can extrapolate from our finite experience of moving around in space. But moving in "inward" space is also infinite, e.g., magnifying a Mandelbrot set, but there's no analogous experience in life, and I can't comprehend it. I'm a mysterian and think that the human mind has fundamental limitations. We can prove certain things but not apprehend them, e.g., different levels of infinity.
I would like to address your comment at time: 2:33, where you say "We are going to start at square 1, literally" I understand you actually are going to start with the square of one in a non metaphorical sense but because the concept of sqrt(1) is an abstract comment I feel like as a propositional statement that comment would evaluate to false. PS. I think my girlfriend hates that you taught us about formal argumentation.
+Dylan Jorgensen The "literally" part really applied only to "one", not so much for "square" --- no squaring or square rooting was meant to be implied :P
(Although nowadays, if I were to do this series again, I would have instead started at zero --- I rather like the modern construction of the natural numbers starting with the empty set!)
+Bill Shillito Well, even if you would like to change some of the details about this series the unique thing is that it gives you a sense of a "math narrative". I think humans are wired for meaning over facts and this series gave me sense for "game" (quarrels, winners, losers, victories, etc..) for how math has evolved, making the facts the pieces that move the story forward was the genius part.
min. 15:39 >>> x=14/3=8/3 😀
16:00
actualy im like the meme 11:25
That damn mouse pointer....
I don't understand why Peano's axioms 3, 4, and 5 are required. Surely 3 is already the default implication. Only if axiom 2 is specifically said to work both ways do you need a specific rule saying that one is the lowest natural number.
I especially do not understand how equality is defined using successors. _If_ you can compare successors for equality, why can't you compare the original numbers for equality? And if this turns into a recursion where you have to compare the successors of the successors, etc., you've broken maths.
Axiom 5 seems to be unnecessary for the construction of the Naturals, it sounds very much like an ad hoc axiom needed by Set theory. If we're only just now constructing naturals, aren't we getting a bit ahead of ourselves already juggling infinite sets?
Axiom 3, is needed because it's the default choice. You need somewhere to start, it's known as the axiom of choice. If you don't have it you could start anywhere, I could say Axiom 3 starts at 5 and discard the previous ones.
Axiom 4 is needed otherwise the Natural Numbers are not one to one, this goes back to relation and sets theory if you have 3 the S(3) = 4 or can it equal S(3) = 5 so then you have ambiguity and everything collapses
Axiom 5, it's more of a conclusive statement and you don't know what exactly "a" is, it could be 1 or 100002325234322342342323443242243324 but you know the next number following "a" where if I give you a number you would be able to give me a number using the function and thus we have all of the Natural numbers, infinity is hard to juggle with and is one of the biggest hurdles to overcome.
There s no where in this video that you actually CONSTRUCT the irrational numbers. You just assume they are uncountable without any justification. If we ignore the difficulties of constructing Dedekind cuts from Q for the transcendentals, you still need to convert an enumerable set of Dedekind cuts to an unenumerable set of the Reals. How do you do that?
You can't construct the Irrationals at all, you cannot even count them. All you can say is R/Q is I and that's it
like Tom Lehrer?