Peano Arithmetic -- Number Theory 1

[@Bodyknock]

1:23 Small point but "if for any two elements m,n either m

[@schweinmachtbree1013]

This is a reupload of a video from the main channel so Michael/Justin could have at least taken a few minutes to look at the top comments and fix the errata that were pointed out with on-screen annotations...

[@mathmajor]

I apologize for the errors, we actually recorded an alternate version that did not have the errors, but I mistakenly edited the wrong version due to a miscommunication. Thanks for letting us know!

[@MGSchmahl]

The assertion that m!=n implies either m>n or m, n.

[@chamelius]

3:25 "minus ninety-nine over fifty-four thousand" Is it bad that my brain immediately went: "That's not reduced... that's negative eleven over six thousand..."

[@stefanalecu9532]

In this particular case it isn't hard to reduce it, but you wouldn't be able to as quickly do -99/53982

[@AndersBjornTH]

Unless zero in included in the list of natural numbers, the recursion at 05:59 will not work for 1 + 1 = 2. For example, 1 + S(0) = S(1 + 0) = 2.

[@xoppa09]

Note that there are nonstandard models of the peano axioms which may include nonstandard elements. We can call any set that satisfies the peano axioms a 'peano structure' or a peano axiom (PA) model. Any set N (not necessarily ℕ = {1,2,3... } (or ℕ = {0, 1,2,3... } depending on convention) ) that satisfies the Peano axioms is a model of the Peano axioms. Such sets might not look like the standard natural numbers but still satisfy all the axioms. Non-standard models, which include "extra elements," exist in certain logical frameworks but are not what we call "the natural numbers."

[@matrixmash9907]

Love the video for number theory; so excited to see more since I discovered this channel. I did my senior capstone on Godels theorem and had to use PA extensively. While the material is not new for myself, it's wonderful how well you present and explain everything. Also "this is kind of abstract and sticky" is probably the best comment I've heard in a long time regarding pure maths. Keep up the amazing work!

[@chessematics]

8:31 Thats how I used to multiply as a child because i hated memorising tables.

[@yevengyklaus7066]

Perhaps I am confused, at 5:36 you state that we don't know how to add (yet), however we know of the successor function, this implies that the resulant of S(n) is a natural number meaning all natural numbers of the output of a function?

[@MrRyanroberson1]

after working with the peano axioms for long enough, i thought about a binary version, where instead of working with a successor function, you instead work with True, False, the "pair" function, and partial function evaluation. With this, you can also define arithmetic. The "pair" function takes a first, a second, and a function, then it evaluates the function with those two items. True takes a first and a second, returns the first. False takes a first and a second, returns the second. Binary numbers can be encoded as True/False sequences in a recursive pair tree - pair(True, pair(False, True)) ~ 1 + 4 = 5.

[@PeperazziTube]

The successor function in the example should really be used as 5 + 4 = 5 + s(3) = s(5+3) = s(5+s(2)) = s(s(5+2)) + s(s(5+s(1)) = s(s(s(5+1))) = s(s(s(s(5))) = s(s(s(6)) = s(s(7)) = s(8) = 9

[@pawemarsza9515]

He said "we do it recursively, so we assume we know how to add smaller integers"

[@thefourthbrotherkaramazov245]

Sorry but you are wrong. Watch again, his statement about it being done recursively allows him to skip explicitly doing it as you did each time. Which is trivial and excessively tedious.

[@atreidesson]

Oh well, by your axioms N = {1; 2} can be naturals if 2 + 1 -> 1... Maybe, the 4th one should be "n + 1 != 1" instead of "n + 1 != n", because the Induction axiom takes care of any cycles without 1 (with the help of injectivity)

[@Spacexioms]

What times we live in, I’m completing my math job online while watching these High quality lectures

[@MrRyanroberson1]

an interesting thing i just realized: the peano axioms WITHOUT the uniqueness condition give rise to modular arithmetic. let s(n) = e^(2ipi/5) * n, where the initial "0" element is the number 1. This gives rise to mod 5 arithmetic

[@abrahammekonnen]

5:35 I think a better way of saying this is that we don't know how to add arbitrary natural numbers m,n we only know how to add 1 to a natural number n that we already know.

[@Carmenifold]

you've got me doing homework for fun

[@maxpercer7119]

he was about to say, he was laying down the bricks of number theory

[@SuperTommox]

This is a beautiful topic but it seems so hard to take it all in.

[@agrajyadav2951]

Thank you sooooo much professor!!!

[@xoppa09]

why is induction on n sufficient to show any three numbers satisfy addition associatively. associativity involves three variables, not one variable.

[@matematicacommarcospaulo]

Some UFO 🛸 appeared at 1:40 in upper right side of the board

[@deodattanewalkar7342]

Any textbooks to follow

[@davehall4075]

I have same question. I am doing your proofs course now and your videos and the recomended text are perfect combo. Any recommmended text for this course?

[@noahtaul]

Will this just be a reuploading of the course from your other channel?

[@browse101]

Probably

[@mathmajor]

We will be reuploading the course videos from the other channel, while also providing new supplementary example videos in between each lecture!

[@jacobjones8131]

@@mathmajor Is there a reason this playlist seems to have one less lecture than the playlist on the main channel?

[@tomkerruish2982]

Are you going to address nonstandard models of the naturals, even in passing?

[@schweinmachtbree1013]

this would be really cool to do, maybe as a "bonus" at the end of the series

[@Qhsjahajw]

but michael already has 2 nt playlists, why another one?

[@tomctutor]

I learnt how to add up first year at elementary school, now you have me doubting that 1+1=2, and even if it is how can I show it is, other than using my fingers?

[@2070user]

1+1=s(1)=2

[@xoppa09]

Not quite, 9 = s(s(s(...s(1)))). The peano axioms only define 1.

[@mikicar96]

The video is really quiet

[@JM-us3fr]

Not including 0 means the natural number won't be a "semi-ring," nor a monoid of addition. It also abandons a nice analogy between addition and multiplication, specifically in how we define "less than or equal" and "divides." I can't really see a good argument for starting at 1 other than: in analysis it would be nice to take 1/n over natural numbers.

[@braindead3201]

There really is no argument to start with 1 since we already have the symbol Z^+ to represent the positive integers.

[@JM-us3fr]

@@braindead3201 True

[@quanquin3822]

In my math education (15+ years ago) 0 was always considered a natural number. I didn't know other people defined N differently.

[@JM-us3fr]

@@quanquin3822 I think it is common in France? Most places (and most of my instructors) start N with 1, which I think is crazy.

[@quanquin3822]

@@JM-us3fr I went to American schools and N started with 0, always.

[@quanquin3822]

Interesting. When I took Number Theory in college we didn't go over Peano axioms. I did that in a mathematical logic course. In Number Theory we spent a lot of time on Gaussian Integers for whatever reason.

[@abrahammekonnen]

This is convenient.