You should say "totally ordered". Well ordering is another property that, by the way, natural numbers also possess, namely that any subset has a least element, which you call "Archimedian principle".
Hej Michael! I think your videos are fantastic, even for us non-mathematicians who are just enamored by the beauty and logic of the subject. The best and most enjoyable I've seen so far on YT. But as it would be a shame to have any unclearness/confusion in such a strictly logical subject, I hope you read all the comments below from Angel Mendez-Rivera, also the ones commenting other people's comments. All the best from Sweden (where you might want to come and climb some day! ;-).
Hi, For fun: Poor 0, let go from the start...°° 9:22 : "so let's may be get rid of this, and we will move on to the next stuff", 9:41 : "so on and so forth", 14:36 : "so let's may be get rid of this and we'll look at a couple of more properties", 16:34 : "ok, great", 19:13 : "so let's get rid of this and keep going".
At 8:19 (and earlier) addition and multiplication are defined recursively. I think the pattern would become clearer if it was written as (e.g.): n + 1 = s(n) n + s(m) = s(n + m) n * 1 = n n * s(m) = nm + n In my presentation, the left operand is unchanged within each pair of cases, and the right operand is either 1 or s(m). [This is a case analysis on the structure of the natural numbers: every natural number is either min(N), here 1, or else s(k) for some k. If you've done functional programming you've probably done case analyses on the structure on recursively defined data (ADTs), especially lists. Notice the similarity. Natural numbers are isomorphic to the "list of unit" type.]
Also in Brazil Z+(more like Z^{+}) is the set of *non-negative* integers(a.k.a. positives and 0) Z+*(imagine + and * stacked vertically) is the positive integers(a.k.a. + excludes the negatives, * excludes 0) Sometimes you encounter a wild person who does things in the opposite way(like...N not including 0, N* including; Z+ being positive integers and Z+* including 0 back)
What you've called the Archimedean property is usually called well-ordering. The Archimedean property is something else (en.wikipedia.org/wiki/Archimedean_property).
I think we can also say that.. Because it states that for any a,b and b>0, and n belongs to natural no. Then nb>a so a is least animal.. So, i think it is true at a point .
One important thing you glossed over is proving that those definitions give a unique way to add and multiply any two numbers, both that it exists and that it's unique. I think that's a good first proof, since it's less obvious that these definitions work at all than that, if they do, they have the familiar properties of arithmetic.
Hi. If I am not mistaken the 4th axiom should be that 1 has no predecessor and not n is never successor of itself. Otherwise the set could be finite by looping towards 1.
Yes, the axiom that states there is no number whose successor is 1 is missing. Otherwise, the set {1, 2} with the addition table 1+1 = 2, 1 + 2 = 2 + 1 = 1, 2 + 2 = 2 satisfies the axioms.
Beyond the comments stating that the axioms are incomplete as we further need to state that there exists no element n in N such that S(n) = 1, wouldn't the resultant structure only guarantee that any set A obeying the inductive hypothesis has N as a subset? Consider A = {y_1, y_2, x_1, x_2, x_3, x_4,...} and define S(y_1) = y_2 and S(y_2) =y_1, for the rest of the elements define S(x_1) = x_2, S(x_2) = x_3, S(x_3) = x_4, and so on such that no element ever links back to x_1. The structure of A has N as a subset by the virtue of the x_i but it also has the loop consisting of the y_j which can be extended to be arbitrarily long.
That's not what Well-Ordered is, that's just the definition on an ordered set, that is trichotomy that is followed by real numbers and thus any of the subsets real numbers, Well-Ordered Set means what you have called the Archimedian Principle, that's generally called the well ordering principle
I'm still learning this stuff, so I might be wrong; however, with the Peano axioms as stated, would the set A={1,2} with the successor function s:A->A defined by s(1)=2 and s(2)=1 also satisfy the axioms? The first three axioms are clearly satisfied. It also seems that the fourth axiom is satisfied, since s(1) is 2 (not 1) and s(2) is 1 (not 2). Finally, the fifth axiom should also be true, since the only subset S of A satisfying 1 \in S and (n \in S implies s(n) \in S) is the set S=A.
There is also another peano's axiom that says that 1 is not a successor of any natural number, which your set doesn't satisfy since s(2)=1 by your definition. I don't think he mentioned this axiom in the video though.
But if we don't have an axiom that states that there's no natural number n that precedes 1 (in other words s(n)=1) we could have a looped set of natural numbers. For example, 1 and 2 are natural numbers such that s(1)=2 and s(2)=1. There's no way I can show this example contradicts any other axiom in the video or am I missing something?
16:38 This step requires that we know _l_ = *1* · _l._ However, this is not given in the definition of multiplication, nor is it immediately obvious from any axiom. If we had commutativity, this would be trivial, but we don't, because we need the distributive theorem first. I think technically, this requires an entire lemma that _n_ = *1* · _n_ for all natural numbers _n,_ which you have to prove with induction. *Theorem* ∀ _n_ ∈ *N* ( *1* · _n_ = _n_ ). *Proof* Let U = { _n_ ∈ *N* | *1* · _n_ = _n_ }. Then *1* ∈ *N* because *1* · *1* = *1* (by definition). Given _k_ ∈ U , *1* · S( _k_ ) = *1* · _k_ + *1* (by definition), ... = _k_ + *1* (by _k_ 's membership in U), and ... = S( _k_ ) (by definition of addition). Thus for all _k_ ∈ U, S( _k_ ) ∈ U. So by the principle of induction, U = *N* , meaning ∀ _n_ ∈ *N* ( *1* · _n_ = _n_ ).∎
The proof of distributivity is a bit sloppy. The base case requires the use of the identity x*1=x, which was skipped, the inductive case exploited associativity of multiplication (easy to overlook this because parentheses were missing) which was not proven here (only assoc of addition) - this could be left as an exercise, but at least it should be noted.
Why is Peano axiom 4 (n != n+1, or if you like n != S(n)) necessary? I don't see how the other axioms allow you to prove n = n+1 in the absence of axiom 4.
Things to know about having pianos: You might not have a piano. You could always get a piano. If you got a piano, you wouldn't not have a piano. If two people would have just as many pianos as each other if they each got a piano, they have just as many now. If a club contains everyone who doesn't have a piano, and anyone in the club would stay in the club if they got a piano, then everyone is in the club, no matter how many pianos they have. Somehow, I find it easier to remember and accurately reproduce these facts about having pianos than the Peano Axioms written in cymbals.
I love to see those definitions of addition and multiplication. Years ago, I came up with those defintions myself in my head, and thought about how they could be used to prove various things. I am so happy to see that my definitions are correct AND accepted. I *kinda* wish I knew who originally wrote those formal definitions, but we all learn these things in a more *informal* way in elementary school. Thanks so much Michael!
Think I will follow along with this course. What text are you using please. Probably worth buying it. Haven't seriously explored number theory since I graduated with a bachelor of applied math 25 or so years ago... and that was mostly analysis and differential equations.
I big issue with that definition (and with his) is equality. By that definition, 5/2 and 10/4 are two different element of the set, and thus 5/2=10/4 is false. If we want this to be true, we need to define our rational numbers in a differen way. One approach is to define them from equivalence classes of ordered pairs of integers.
@@kilian8250 Really, you are just making a point of basically nothing. People looking at the given definition will know perfectly well that 5/2 and 10/4 are to be considered equal, especially in an introductory course in number theory. I think the need to justify the definition by means of equivalence classes really is unnecessary and goes beyond the scope (and by that I mean in the sense that it is just plainly too obvious to discuss in class. Sure, if somebody is to ask for a more rigorous definition, you could give them that, but it just doesn't fit in for the rest of the video to include such a specific detail which most students will not have problems with in this course)
@@joelganesh8920 yes of course, but I don’t know what level this course is at. If it’s a lower-lever course then I agree. I’m just saying that it’s not a formal definition.
with m + S(n) = S(m+n) where m=5 and n=4, is it valid to do S(m) + S(n) = S(S(m+n))? S(4) + S(3) = S(S(4+3)) = S(S(7)) = S(8) = 9 or does that require more steps?
I think that's the same thing is it not? from m + S(n) = S(m+n) you can have S(m) + S(n) = S(m + S(n)) = S(S(m + n)) by just applying the same property twice.
No, you cannot do it directly. You only have a rule, that allows you to move S() from the right summand onto the whole summation. However, you have to *prove* that it works for the left summand as well.
I just found this channel/playlist, I am a Computer Science student, will this playlist be of any use to me or just some videos? I want to overcome my fear of math, I want to understand it in order to become a great programmer.
one doesn't have subtraction in formal treatments of the natural numbers: one only has successor function S and addition and multiplication defined recursively in terms of S. a formal proof of commutativity of + would be by induction (say on the right summand), so you have to show that m+1 = 1+m for all natural numbers m as the base case, and for the inductive step you need to show that if m+n = n+m for all m then m+S(n) = S(n)+m for all m.
correct, there are many ways to put a partial order on the complex numbers (which as a set is just like pairs of real numbers) - for example we can define a+bi
I still don't really understand these proofs but maybe that's because i still need to finish my course which contains proof by induction. But very nice to already get some insight in number theory which i am gonna get as a course in my next semester
Hah, I just managed to figure out why I find the definition of the natual numbers as the positve integers so unnatural. That means that addition over N is not a group, because the additive identity element would not be in N. That, I think, means that the non-negative numbers are a more natural "natual numbers" than the positive ones.
yes, including zero gives you an additive identity, but note that you still don't get a group (what you do get - an associative operation with an identity element - is called a *monoid* or, less commonly, a semigroup with identity)
@@schweinmachtbree1013 You are correct, I managed to (somehow, don't ask me how) stuff it in a "modulo something". But, still, the lack of additive identity bothers me.
I assume you have to first define a set of all natural numbers n such that 1+n=n+1 and then show that this set is equal to the natural numbers using induction (1 obviously is in this set) and then you can define a new set of all natural numbers n such that for a fixed m, n+m=n+m and you've already proven in the first step that 1 is in this set and you can therefore use that as a base case to prove that addition is commutative for all natural numbers
since Peano's axioms implicitly define what natural numbers are, anything that satisfies Peano's axioms is automatically considered as the natural numbers.
This actually turns out to be a very interesting question: it depends whether you define the natural numbers in so-called first order logic or second-order logic (I should note - as others have pointed out - that the peano axioms were incorrectly stated here, so see Wikipedia, for example). If you do things in first-order logic then there is only one model of the axioms up to isomorphism ( although there are different implementations, which are necessarily isomorphic - for example, von Neumann implemented the natural numbers within set theory as 0 = {}, 1 = {0} = {{}}, 2 = {0, 1} = {{}, {{}}}, 3 = {0, 1, 2} = {{}, {{}}, {{}, {{}}}}, etc. (i.e. given by the rule 0 = {} and n+1 = {0, ..., n}), whereas Zermelo implemented them as 0 = {}, 1 = {0} = {{}}, 2 = {1} = {{{}}}, 3 = {2} = {{{{}}}}, etc. (i.e. given by the rule 0 = {} and n+1 = {n}) - Zermelo's construction is much cleaner, but von Neumann's construction plays nicer with the ordering on the natural numbers, since n is less than m precisely when n is an element of m). *However* , if one does things in second-order logic, then there is not just one model of the (second-order) axioms up to isomorphism! - the models of the axioms which are non-isomorphic to the "proper natural numbers" are called *non-standard models* - see for example the section "Nonstandard models" in the wikipedia article for the Peano axioms.
@@angelmendez-rivera351 "the natural numbers without 0 cannot be constructively extended to the natural numbers including 0": Yes they can; there is a semigroup-to-monoid construction (mentioned in most semigroup theory texts) - If the semigroup-without-identity is S, then it is common to denote the resulting monoid by S^0 (if we are using additive notation). The construction is thus: adjoin an element 0 and extend the existing operation of (S, +) to S \union {0} by defining 0+0 := 0 and x+0 = 0+x := x for all x in S. One can check that the associative law is preserved, so we really do get a monoid. It is also very important note that the identity element in a monoid is unique! - and therefore if our semigroup S was already a monoid, having an identity 0', then it turns out that the element we are formally adjoining, 0, must actually equal 0', since 0 + 0' := 0' and 0 + 0' = 0, and so we aren't actually adjoining anything at all - that is, S^0 = S.
I don't understand his insistence on not including 0 in N. Since he talks about + in N, doesn't he need an additive identity in N? In other words, an element x (i.e. 0) such that x+n = n, for all n, in N.
an identity is nice to have but I'm sure one can do without it (plus, leaving 0 out of the natural numbers when defining them axiomatically prompts a lot of debate in the comments section which is good for Michael's channel :P)
lol, does your formulation of the Peano axiom in this vid prevent loops? Don't you need to include that 1 isn't the successor of any nat number? Edit: nvm you're assuming well ordering which should banish loops
you were right before your edit - there are several mistakes in this video. his axiom "there does not exists n such that n+1=n" should be "there does not exist n such that n+1=1", i.e. "1 is not a successor" (because Michael is choosing to omit 0 from N in his definition). And by the way, Michael says "well-ordering" for the property "for all m=/=n, mn" but he should have said total ordering (a.k.a. linear ordering); a well-ordering is something else.
@@bertblankenstein3738 You can only do that after you define subtraction. That complicates things because the natural numbers are not closed under subtraction.
Hey Michael, this is awesome! but something caught my attention. I'm new the community so I'm not sure, but what happened to your arm? Regardless, it's wonderful to see this course, and I am very interested in number theory combined with real analysis. If that's something you could cover.
Excluding 0 from the Natural Numbers here got me thinking about some interesting ways the definitions change when you include 0. For one thing, if you include 0 then you also presumably are defining it as the additive identity (i.e. n + 0 = n by definition). So when you define addition using the successor function you get something like n + 0 = n s(0) = 1 m + s(n) = s(m + n) Likewise, with multiplication, I believe you get n*0 = 0 n*1 = n m * s(n) = mn + m (In a ring where you have additive inverses and distributivity of multiplication over addition, n*0 = 0 falls out as a consequence. But not yet having distributivity or an inverse to addition I think you'd probably have to define n*0 = 0 in the base case.) As far as whether or not to include or exclude 0, either works as long as it's clear up front like it is in the video which way you're going with it. Personally I like defining the Natural numbers as the cardinalities of finite sets, and since 0 is the cardinality of the empty set that means it gets included. The downside being simply that occasionally you want to work in N+ which is the set of strictly positive integers 1,2,3,... simply because every now and then a theorem doesn't hold for 0 that does for the other numbers. That's just my preference though. 🤷♂️
I found it a bit unusual that Michael used the language of *set theory* to expound Peano Arithmetic. That's unusual for several reasons: • When you're starting with nothing, you cannot assume that a logically sound set theory exists! 🙄 • If you want to use the language of set theory, you have to define such a theory first-ZFC or any other-before getting into Peano. • And if you did start with an axiomatic set theory-ZFC or any other-you can derive the arithmetic of natural numbers from it, without having to consider Peano axioms at all. Most other _'Foundations of Mathematics'_ courses that I've seen start with logic and symbols even before attacking the axioms of any one system. Then, coming to first-order logic systems, they do teach both Peano and ZFC. But they don't mix up the notations; they are taught as separate but equal. They also mention, either casually or for comparison, other foundational systems like Presburger, Robinson, NBG, Category etc. Then, after the arithmetic of natural numbers is established (independently in both Peano and ZFC), Peano is put to rest. Further development of arithmetic (rational, real, complex, quaternion, whatever) as well as abstract algebra (fields, rings etc.) is done using set theory entirely. For working with number theory-which is Michael's goal-it really doesn't matter which foundational system you start with. Either Peano or ZFC is independently sufficient. There's nothing actually wrong with mixing up the two, since they're equivalent for this purpose. That's why I said 'unusual'.
@@angelmendez-rivera351 oh, thanks mate! But it feels a bit strange to me tho, do the axioms of both perspectives imply each other? Otherwise there would exist some disagreement on both parts, wouldn't there?
for those interested, we can go one "why?" further and ask "but how do we know *that*": the fact that every number times 1 is itself - i.e. that 1 is an identity element for multiplication - is an axiom, while the fact that every number times 0 is equal to 0 - i.e. that 0 is an absorbing element for multiplication - is a theorem -- the proof it as follows: for any n we have n*0 + 0 = n*0 = n*(0+0) = n*0 + n*0, and hence cancelling an n*0 from the extreme left and right hand sides we get 0 = n*0 (in this proof I have used various properties, namely that any number plus 0 is itself - i.e. that 0 is an identity element for addition; that multiplication distributes over addition; and that natural numbers can be cancelled from both sides of an equation - k+n = k+m => n=m. therefore the proof holds not only in the natural numbers, but in any structure in which these properties hold - for example, in any ring)
A course in number theory? This would be a great contribution to the math RUclips. Great job, Michael!
Yes I really need that!
@Rick Does Math as in whole of theory?
@Rick Does Math could you share the playlist link I am unable to see it on the channel
@Rick Does Math Yes, also Richard Borcherds has a whole lecture playlist on number theory:
ruclips.net/p/PL8yHsr3EFj52Qf7lc3HHvHRdIysxEcj1H
@@adityaekbote8498 Here:
ruclips.net/p/PL22w63XsKjqwAgBzVFVqZNMcVKpOOAA7c
Really excited to follow along with this, love the way definitions of concepts are really clearly spelled out and considered!
I do hope this carries on.
Thank you, professor.
I literally just read this stuff in the very beginning of Jacobson's Lectures in Abs. Alg. Very good book, I recommend.
You should say "totally ordered". Well ordering is another property that, by the way, natural numbers also possess, namely that any subset has a least element, which you call "Archimedian principle".
Hej Michael! I think your videos are fantastic, even for us non-mathematicians who are just enamored by the beauty and logic of the subject. The best and most enjoyable I've seen so far on YT. But as it would be a shame to have any unclearness/confusion in such a strictly logical subject, I hope you read all the comments below from Angel Mendez-Rivera, also the ones commenting other people's comments. All the best from Sweden (where you might want to come and climb some day! ;-).
5:52 seems like you are saying that we only know how to add 1 (which we call the successor operation) and we build our notion of addition out of this.
Hi,
For fun:
Poor 0, let go from the start...°°
9:22 : "so let's may be get rid of this, and we will move on to the next stuff",
9:41 : "so on and so forth",
14:36 : "so let's may be get rid of this and we'll look at a couple of more properties",
16:34 : "ok, great",
19:13 : "so let's get rid of this and keep going".
At 8:19 (and earlier) addition and multiplication are defined recursively.
I think the pattern would become clearer if it was written as (e.g.):
n + 1 = s(n)
n + s(m) = s(n + m)
n * 1 = n
n * s(m) = nm + n
In my presentation, the left operand is unchanged within each pair of cases, and the right operand is either 1 or s(m).
[This is a case analysis on the structure of the natural numbers: every natural number is either min(N), here 1, or else s(k) for some k. If you've done functional programming you've probably done case analyses on the structure on recursively defined data (ADTs), especially lists. Notice the similarity. Natural numbers are isomorphic to the "list of unit" type.]
Question:. Should Peano's 4th axiom end with n+1 does not equal 1, i.e. 1 is not a successor of any number?
@@angelmendez-rivera351 what about the 5th axiom (as it’s presented) is wrong?
Hello, In French, the number zero is considered a natural number. We use the notation N* to represent the set of natural numbers without zero.
Also in Brazil
Z+(more like Z^{+}) is the set of *non-negative* integers(a.k.a. positives and 0)
Z+*(imagine + and * stacked vertically) is the positive integers(a.k.a. + excludes the negatives, * excludes 0)
Sometimes you encounter a wild person who does things in the opposite way(like...N not including 0, N* including; Z+ being positive integers and Z+* including 0 back)
Much appreciated that you lay these foundations too often overlooked by math teachers.
Yeah, they don't bother to teach these axioms.
@@HaiNguyen-qx3db the same way they don't teach real analysis. It's just not necessary.
This whole series is wonderful. Thank you very much for the efforts.
Bring more such contents on Number Theory
What you've called the Archimedean property is usually called well-ordering. The Archimedean property is something else (en.wikipedia.org/wiki/Archimedean_property).
I think we can also say that..
Because it states that for any a,b and b>0, and n belongs to natural no. Then nb>a so a is least animal..
So, i think it is true at a point .
@@amanpants275 I don't grab what your point is.
Especially, positive rational numbers admit Archimedean property while not being well ordered
That is great. And when you finish the course could you please make a full video of this course?
I already watched most of your old Number Theory Playlist
One important thing you glossed over is proving that those definitions give a unique way to add and multiply any two numbers, both that it exists and that it's unique. I think that's a good first proof, since it's less obvious that these definitions work at all than that, if they do, they have the familiar properties of arithmetic.
Hi. If I am not mistaken the 4th axiom should be that 1 has no predecessor and not n is never successor of itself. Otherwise the set could be finite by looping towards 1.
Yes, the axiom that states there is no number whose successor is 1 is missing. Otherwise, the set {1, 2} with the addition table 1+1 = 2, 1 + 2 = 2 + 1 = 1, 2 + 2 = 2 satisfies the axioms.
Can you solve the following equation in natural numbers: x^2 + y^2 = 5xy +7 ?? Thanks
Has no natural solutions mate
There's
1) (0,√7)
2) (0,-√7)
3) (√7,0)
4) (-√7,0)
Not sure if there are more
Pretty sure that's all
x^2 + y^2 = 5xy + 7
1) If (x,y) = (u,v) is a solution, then (x,y) = (v,u) is also a solution.
2) If (x,y) = (u,v) is a solution, then (x,y) = (-u, -v) is also a solution.
3) If (x,y) = (u,v) is a solution, then (x,y) = (v, 5v-u) is also a solution. (Proof left to the reader.)
Integer solutions (x,y) =
(-1, 1) , (1, -1)
(1, 6) , (6, 1) , (-1, -6) , (-6, -1)
(6, 29) , (29, 6) , (-6, -29) , (-29, -6)
(29, 139) , (139, 29) , (-29, -139) , (-139, -29)
(139, 666) , (666, 139) , (-139, -666) , (-666, -139)
(666, 3191) , (3191, 666) , (-666, -3191) , (-3191, -666)
(3191, 15289) , (15289, 3191) , (-3191, -15289) , (-15289, -3191)
etcetera.
(I'm not sure if this generates _all_ possible integer solutions though.)
I'm not necessarily proud of my skills in number theory, so im really excited for this series!
Professor Penn, thank for a solid analysis and foundation of Number Theory One.
Please correct the errors (i.e. well/total ordering, 4th axiom, Archimedean property, etc.)
Beyond the comments stating that the axioms are incomplete as we further need to state that there exists no element n in N such that S(n) = 1, wouldn't the resultant structure only guarantee that any set A obeying the inductive hypothesis has N as a subset?
Consider A = {y_1, y_2, x_1, x_2, x_3, x_4,...} and define S(y_1) = y_2 and S(y_2) =y_1, for the rest of the elements define S(x_1) = x_2, S(x_2) = x_3, S(x_3) = x_4, and so on such that no element ever links back to x_1.
The structure of A has N as a subset by the virtue of the x_i but it also has the loop consisting of the y_j which can be extended to be arbitrarily long.
It will be an amazing Video Series also you can read the Number Theory Book of Burton(Beautiful Book) to make some excercises ...
That's not what Well-Ordered is, that's just the definition on an ordered set, that is trichotomy that is followed by real numbers and thus any of the subsets real numbers, Well-Ordered Set means what you have called the Archimedian Principle, that's generally called the well ordering principle
How to proof that multiplication is conmutative if we dont know that l*1=1*l??? (Base case for induction)
Thank you so much for making this course.
wow i watched the prev playlist and understood a little, covered basic set theory, coming back its all clear now
I think it should be the well-ordering principle, not the archmiedean principle?
Gödel, Escher, Bach prepared me for this
I'm still learning this stuff, so I might be wrong; however, with the Peano axioms as stated, would the set A={1,2} with the successor function s:A->A defined by s(1)=2 and s(2)=1 also satisfy the axioms?
The first three axioms are clearly satisfied. It also seems that the fourth axiom is satisfied, since s(1) is 2 (not 1) and s(2) is 1 (not 2). Finally, the fifth axiom should also be true, since the only subset S of A satisfying 1 \in S and (n \in S implies s(n) \in S) is the set S=A.
There is also another peano's axiom that says that 1 is not a successor of any natural number, which your set doesn't satisfy since s(2)=1 by your definition. I don't think he mentioned this axiom in the video though.
19:26 Homework that you need to do before watching the second video 👀
19:50 Good Place To Stop
But if we don't have an axiom that states that there's no natural number n that precedes 1 (in other words s(n)=1) we could have a looped set of natural numbers. For example, 1 and 2 are natural numbers such that s(1)=2 and s(2)=1. There's no way I can show this example contradicts any other axiom in the video or am I missing something?
16:38 This step requires that we know _l_ = *1* · _l._ However, this is not given in the definition of multiplication, nor is it immediately obvious from any axiom. If we had commutativity, this would be trivial, but we don't, because we need the distributive theorem first. I think technically, this requires an entire lemma that _n_ = *1* · _n_ for all natural numbers _n,_ which you have to prove with induction.
*Theorem* ∀ _n_ ∈ *N* ( *1* · _n_ = _n_ ).
*Proof* Let U = { _n_ ∈ *N* | *1* · _n_ = _n_ }.
Then *1* ∈ *N* because *1* · *1* = *1* (by definition).
Given _k_ ∈ U , *1* · S( _k_ ) = *1* · _k_ + *1* (by definition),
... = _k_ + *1* (by _k_ 's membership in U), and
... = S( _k_ ) (by definition of addition).
Thus for all _k_ ∈ U, S( _k_ ) ∈ U. So by the principle of induction, U = *N* , meaning ∀ _n_ ∈ *N* ( *1* · _n_ = _n_ ).∎
Not all heroes wears capes! Thanks for making this content available.
Some wear funny math t shirts
@@mathlegendno12 which I don't get
The proof of distributivity is a bit sloppy. The base case requires the use of the identity x*1=x, which was skipped, the inductive case exploited associativity of multiplication (easy to overlook this because parentheses were missing) which was not proven here (only assoc of addition) - this could be left as an exercise, but at least it should be noted.
Why is Peano axiom 4 (n != n+1, or if you like n != S(n)) necessary? I don't see how the other axioms allow you to prove n = n+1 in the absence of axiom 4.
Things to know about having pianos:
You might not have a piano.
You could always get a piano.
If you got a piano, you wouldn't not have a piano.
If two people would have just as many pianos as each other if they each got a piano, they have just as many now.
If a club contains everyone who doesn't have a piano, and anyone in the club would stay in the club if they got a piano, then everyone is in the club, no matter how many pianos they have.
Somehow, I find it easier to remember and accurately reproduce these facts about having pianos than the Peano Axioms written in cymbals.
Lmao that was nice!
"It's a Piano Postulate! There are five of them, and I've arranged them all for flute."
Looking forward to the course. Treat us gently, please. (I'm glad that your pronunciation of Peano's name improved after the first attempt.)
I love to see those definitions of addition and multiplication. Years ago, I came up with those defintions myself in my head, and thought about how they could be used to prove various things. I am so happy to see that my definitions are correct AND accepted. I *kinda* wish I knew who originally wrote those formal definitions, but we all learn these things in a more *informal* way in elementary school.
Thanks so much Michael!
Think I will follow along with this course. What text are you using please. Probably worth buying it. Haven't seriously explored number theory since I graduated with a bachelor of applied math 25 or so years ago... and that was mostly analysis and differential equations.
Math is really great, this is verry important and funny at the same time
if we dont know what is 5+4 then how could we know 5+3=8?
I think because we started with 5 + 1, then 5 + 1 + 1, then 5 + 1 + 1 + 1
Michael, what textbook will you be using for teaching this course?
😂😂😂😂😂😂
The thumbnail is killing me lol!
Me too 😃
Brilliant teaching, thank you.
Very Helpful Thank You!!! Please make more of these soon
I didn't finish the video but already know that its amazing
this is gold.
What textbook will you be using Michael?
Why not define ℚ = { p / q | p∈ℤ, q∈ℕ }?
I suppose your variant is more common(I've personlly been taught like that), but I don't think it's a big problem to define as Michael did.
That's an equivalent definition. Another possibility would be to add another restriction: gcd(p, q)=1. But that still defines the same set of numbers.
I big issue with that definition (and with his) is equality. By that definition, 5/2 and 10/4 are two different element of the set, and thus 5/2=10/4 is false. If we want this to be true, we need to define our rational numbers in a differen way. One approach is to define them from equivalence classes of ordered pairs of integers.
@@kilian8250 Really, you are just making a point of basically nothing. People looking at the given definition will know perfectly well that 5/2 and 10/4 are to be considered equal, especially in an introductory course in number theory. I think the need to justify the definition by means of equivalence classes really is unnecessary and goes beyond the scope (and by that I mean in the sense that it is just plainly too obvious to discuss in class. Sure, if somebody is to ask for a more rigorous definition, you could give them that, but it just doesn't fit in for the rest of the video to include such a specific detail which most students will not have problems with in this course)
@@joelganesh8920 yes of course, but I don’t know what level this course is at. If it’s a lower-lever course then I agree.
I’m just saying that it’s not a formal definition.
Thank you SOOOO MUCH
Ok, great! Thanks, Prof. Penn.
The last video will be about Abraxas conjecture proven by Gaal Dornick!
Thanks Micheal I need this because I am Beggineer
with m + S(n) = S(m+n) where m=5 and n=4,
is it valid to do S(m) + S(n) = S(S(m+n))?
S(4) + S(3) = S(S(4+3)) = S(S(7)) = S(8) = 9
or does that require more steps?
I think that's the same thing is it not?
from m + S(n) = S(m+n)
you can have S(m) + S(n) = S(m + S(n)) = S(S(m + n)) by just applying the same property twice.
No, you cannot do it directly. You only have a rule, that allows you to move S() from the right summand onto the whole summation. However, you have to *prove* that it works for the left summand as well.
Why the sound is too weak?
I just found this channel/playlist, I am a Computer Science student, will this playlist be of any use to me or just some videos?
I want to overcome my fear of math, I want to understand it in order to become a great programmer.
was it useful
Am i allowed to assume S(1+(m-1))=S(m) when proving commutativity holds over addition?
one doesn't have subtraction in formal treatments of the natural numbers: one only has successor function S and addition and multiplication defined recursively in terms of S. a formal proof of commutativity of + would be by induction (say on the right summand), so you have to show that m+1 = 1+m for all natural numbers m as the base case, and for the inductive step you need to show that if m+n = n+m for all m then m+S(n) = S(n)+m for all m.
Complex numbers would also be an example of a partially ordered set, yes?
I suppose you could privilege an axis to order the entire set.
correct, there are many ways to put a partial order on the complex numbers (which as a set is just like pairs of real numbers) - for example we can define a+bi
I still don't really understand these proofs but maybe that's because i still need to finish my course which contains proof by induction. But very nice to already get some insight in number theory which i am gonna get as a course in my next semester
Can anybody help me in knowing what does defined recursively or a recursive function?
Hah, I just managed to figure out why I find the definition of the natual numbers as the positve integers so unnatural. That means that addition over N is not a group, because the additive identity element would not be in N. That, I think, means that the non-negative numbers are a more natural "natual numbers" than the positive ones.
yes, including zero gives you an additive identity, but note that you still don't get a group (what you do get - an associative operation with an identity element - is called a *monoid* or, less commonly, a semigroup with identity)
@@schweinmachtbree1013 You are correct, I managed to (somehow, don't ask me how) stuff it in a "modulo something". But, still, the lack of additive identity bothers me.
Thanks
I'm a little late here i know .
But what book are you covering ?
How do we prove those successor function formulas?
Brilliant!
Hi Michael, how do you prove the commutative property of addition without the number zero?
I assume you have to first define a set of all natural numbers n such that 1+n=n+1 and then show that this set is equal to the natural numbers using induction (1 obviously is in this set) and then you can define a new set of all natural numbers n such that for a fixed m, n+m=n+m and you've already proven in the first step that 1 is in this set and you can therefore use that as a base case to prove that addition is commutative for all natural numbers
In there anything which satisfies the Peano axioms, but is not equal to the natural numbers?
since Peano's axioms implicitly define what natural numbers are, anything that satisfies Peano's axioms is automatically considered as the natural numbers.
@@angelmendez-rivera351 yeah, he missed a lot.
This actually turns out to be a very interesting question: it depends whether you define the natural numbers in so-called first order logic or second-order logic (I should note - as others have pointed out - that the peano axioms were incorrectly stated here, so see Wikipedia, for example). If you do things in first-order logic then there is only one model of the axioms up to isomorphism ( although there are different implementations, which are necessarily isomorphic - for example, von Neumann implemented the natural numbers within set theory as 0 = {}, 1 = {0} = {{}}, 2 = {0, 1} = {{}, {{}}}, 3 = {0, 1, 2} = {{}, {{}}, {{}, {{}}}}, etc. (i.e. given by the rule 0 = {} and n+1 = {0, ..., n}), whereas Zermelo implemented them as 0 = {}, 1 = {0} = {{}}, 2 = {1} = {{{}}}, 3 = {2} = {{{{}}}}, etc. (i.e. given by the rule 0 = {} and n+1 = {n}) - Zermelo's construction is much cleaner, but von Neumann's construction plays nicer with the ordering on the natural numbers, since n is less than m precisely when n is an element of m). *However* , if one does things in second-order logic, then there is not just one model of the (second-order) axioms up to isomorphism! - the models of the axioms which are non-isomorphic to the "proper natural numbers" are called *non-standard models* - see for example the section "Nonstandard models" in the wikipedia article for the Peano axioms.
What about concrete objects?
I think by well-ordered you mean totally ordered, yes? But well-ordering is also true.
I never understood why you'd include the identity element of multiplication in ℕ - but not the identity element of addition.
@@angelmendez-rivera351 "the natural numbers without 0 cannot be constructively extended to the natural numbers including 0": Yes they can; there is a semigroup-to-monoid construction (mentioned in most semigroup theory texts) - If the semigroup-without-identity is S, then it is common to denote the resulting monoid by S^0 (if we are using additive notation). The construction is thus: adjoin an element 0 and extend the existing operation of (S, +) to S \union {0} by defining 0+0 := 0 and x+0 = 0+x := x for all x in S. One can check that the associative law is preserved, so we really do get a monoid. It is also very important note that the identity element in a monoid is unique! - and therefore if our semigroup S was already a monoid, having an identity 0', then it turns out that the element we are formally adjoining, 0, must actually equal 0', since 0 + 0' := 0' and 0 + 0' = 0, and so we aren't actually adjoining anything at all - that is, S^0 = S.
-54/99000 is a well known rational? ;-)
It’s funny how peano axioms sounds like piano axioms, and people say musicians are also good at maths
I don't understand his insistence on not including 0 in N. Since he talks about + in N, doesn't he need an additive identity in N? In other words, an element x (i.e. 0) such that x+n = n, for all n, in N.
an identity is nice to have but I'm sure one can do without it (plus, leaving 0 out of the natural numbers when defining them axiomatically prompts a lot of debate in the comments section which is good for Michael's channel :P)
lol, does your formulation of the Peano axiom in this vid prevent loops? Don't you need to include that 1 isn't the successor of any nat number?
Edit: nvm you're assuming well ordering which should banish loops
you were right before your edit - there are several mistakes in this video. his axiom "there does not exists n such that n+1=n" should be "there does not exist n such that n+1=1", i.e. "1 is not a successor" (because Michael is choosing to omit 0 from N in his definition). And by the way, Michael says "well-ordering" for the property "for all m=/=n, mn" but he should have said total ordering (a.k.a. linear ordering); a well-ordering is something else.
Is this the actual topic ?
lmao piano axioms :
1. piano has 88 distinct keys
2. any number of keys n can be played at the same time
Definition:
a piece is a sequence of keys
Can anyone please post the answers to the h.w questions, thanks.
I did not get the joke on your shirt
I think it is to do with vertex operator algebras (Michael's area of research)
0 is a natural number for generating odd numbers : 2k + 1 k€ N
Could also use 2k-1 to get those odd numbers.
@@bertblankenstein3738 You can only do that after you define subtraction. That complicates things because the natural numbers are not closed under subtraction.
Hey Michael, this is awesome! but something caught my attention. I'm new the community so I'm not sure, but what happened to your arm? Regardless, it's wonderful to see this course, and I am very interested in number theory combined with real analysis. If that's something you could cover.
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Excluding 0 from the Natural Numbers here got me thinking about some interesting ways the definitions change when you include 0.
For one thing, if you include 0 then you also presumably are defining it as the additive identity (i.e. n + 0 = n by definition). So when you define addition using the successor function you get something like
n + 0 = n
s(0) = 1
m + s(n) = s(m + n)
Likewise, with multiplication, I believe you get
n*0 = 0
n*1 = n
m * s(n) = mn + m
(In a ring where you have additive inverses and distributivity of multiplication over addition, n*0 = 0 falls out as a consequence. But not yet having distributivity or an inverse to addition I think you'd probably have to define n*0 = 0 in the base case.)
As far as whether or not to include or exclude 0, either works as long as it's clear up front like it is in the video which way you're going with it. Personally I like defining the Natural numbers as the cardinalities of finite sets, and since 0 is the cardinality of the empty set that means it gets included. The downside being simply that occasionally you want to work in N+ which is the set of strictly positive integers 1,2,3,... simply because every now and then a theorem doesn't hold for 0 that does for the other numbers. That's just my preference though. 🤷♂️
3:26 -99/54000 = -11/6000.
(Yeah, I know they're equivalent under a quotient group.)
I found it a bit unusual that Michael used the language of *set theory* to expound Peano Arithmetic. That's unusual for several reasons:
• When you're starting with nothing, you cannot assume that a logically sound set theory exists! 🙄
• If you want to use the language of set theory, you have to define such a theory first-ZFC or any other-before getting into Peano.
• And if you did start with an axiomatic set theory-ZFC or any other-you can derive the arithmetic of natural numbers from it, without having to consider Peano axioms at all.
Most other _'Foundations of Mathematics'_ courses that I've seen start with logic and symbols even before attacking the axioms of any one system. Then, coming to first-order logic systems, they do teach both Peano and ZFC. But they don't mix up the notations; they are taught as separate but equal. They also mention, either casually or for comparison, other foundational systems like Presburger, Robinson, NBG, Category etc.
Then, after the arithmetic of natural numbers is established (independently in both Peano and ZFC), Peano is put to rest. Further development of arithmetic (rational, real, complex, quaternion, whatever) as well as abstract algebra (fields, rings etc.) is done using set theory entirely.
For working with number theory-which is Michael's goal-it really doesn't matter which foundational system you start with. Either Peano or ZFC is independently sufficient. There's nothing actually wrong with mixing up the two, since they're equivalent for this purpose. That's why I said 'unusual'.
You are right ruclips.net/p/PL54Pt_mZzBqibWHgesgEICeQHnwHom8xz
Number theory course please
so you telling me that the distributive property of multiplication and the associative of sum aren't axioms?? omg
@@angelmendez-rivera351 oh, thanks mate! But it feels a bit strange to me tho, do the axioms of both perspectives imply each other? Otherwise there would exist some disagreement on both parts, wouldn't there?
Please imc 2021 p2
thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
We all know that 0.1 = 0 or 1.0 = 0
But hơw we know that? Because every number times 1 is itself and every number times 0 is equal to 0
What are you trying to say?
for those interested, we can go one "why?" further and ask "but how do we know *that*": the fact that every number times 1 is itself - i.e. that 1 is an identity element for multiplication - is an axiom, while the fact that every number times 0 is equal to 0 - i.e. that 0 is an absorbing element for multiplication - is a theorem -- the proof it as follows: for any n we have n*0 + 0 = n*0 = n*(0+0) = n*0 + n*0, and hence cancelling an n*0 from the extreme left and right hand sides we get 0 = n*0 (in this proof I have used various properties, namely that any number plus 0 is itself - i.e. that 0 is an identity element for addition; that multiplication distributes over addition; and that natural numbers can be cancelled from both sides of an equation - k+n = k+m => n=m. therefore the proof holds not only in the natural numbers, but in any structure in which these properties hold - for example, in any ring)
I wonder what happens when you show this stuff to people with dyscalculia.
I wonder.
Anyone has done the homework？
How dare you not consider 0 a natural number!
That is a course I like to follow. I hatelove numbertheory! :-)
beautiful
Epic
Second
First