Dedekind Cuts - Constructing the Real Numbers (Step 5 Part 2) #4.3.1.4f

Constructing the real numbers with Dedekind Cuts. Here we continue step five, beginning the proof of the multiplicative inverse.

At last we are going to prove that real numbers exist! Using Dedekind Cuts. In this video we begin the proof that every cut has a multiplicative inverse. We define the set we want to examine, and in this video verify that it is in fact a cut.

After constructing the real number field, we will move on to the climax of this mini-series, proving the uniqueness of the real number field. We will then lay the foundation of multivariable calculus before moving on to complex numbers.

It turns out Elliptic Curves over Complex Numbers are central to Wiles's proof of Fermat's Last Theorem. With the first videos on this channel, we managed to keep the math very simple, almost exclusively nothing more advanced than algebra. Now, to complete our brief tour of elliptic curves, we have to take an ENORMOUS leap forward in the level of the mathematics involved.

I want to make sure we are all on the same page before we begin learning the more advanced stuff in earnest. I am going to assume we are all familiar with high school math and college math through first or second year calculus. I'm going to introduce complex numbers and complex integration in some detail with the next several videos. We need to examine multivariable calculus as well.

This series will culminate with the Weierstrass Equation, which is the thing that connects elliptic curves to complex numbers, and thus allows us to connect them to modular forms, which is what Wiles's proof is all about.

Here is the outline of this series on complex numbers:
0. Introduction
1. The Real Number System (We are here)
2. Complex Numbers
3. Complex Functions
4. Exponential and Trigonometric Functions
5. Complex Integration
6. Cauchy's Integral Theorem
7. Cauchy's Integral Formula
8. Laurent Series
9. Complex Residues
10. Lattices and Doubly Periodic Functions
11. Lattices and Tori and Groups
12. The Weierstrass p-Function
13. The Weierstrass Equation: Complex Functions and Elliptic Curves

Table of Contents This Video:
0:00 - Introduction
0:13 - (M5) Multiplicative Identity
1:04 - Discussion
4:29 - Verify β is a cut
4:37 - Property (I)
8:25 - Property (II)
9:59 - Property (III)
18:00 - Outro

Please leave any questions, comments, or suggestions in the comments below!

Credits:
Music: "Oberkorn" and "St. Jarna" by Depeche Mode.
Bibliography: Principles of Mathematical Analysis, by Walter Rudin

[sm-qhzp]

16:59 Is it sufficient to show that beta belongs to R? What I think is that we also need to show that 0*<beta so that our beta belongs to R+? If I m write then can you please provide the hint for what steps to follow next once after consider any element of 0* ??

[shar]

Hello, I've been looking at your multiplicative inverse proof, and I reckon there might be a little problem with your definiton of beta. You defined it as the set of all p for which exists r > 1: (1/p)(1/r) not in alpha. The problem is that the set beta cannot be a cut: it doesn't contain zero (you divided one by zero) and it doesn't contain negatives. In your proof of beta being a cut, in step (II), you also stated that if q < p then (1/q) > (1/p), which only holds for positive rationals. For example, it fails here: -3 < 2, but -1/3 < 1/2.

However, it probably can be solved by manually adding zeroes and negatives to beta, but I think there might be an easier way to it. I will keep watching.

[JeffyBurglarcat]

At 1.53 did you mean to say 'Some rational less than 1/p' rather than 'Some rational less than p'?

[zcahhf]

In which part have you shown 𝛼𝛽 ⊆ 1* ? Thank you!

[sabouedcleek]

16:15 Why does x need to be substituted here? I think it is possible to use the identities of ordered fields to obtain the final result of 1/t*1/x=1/p*1/r from t = p*r/x, without the need to know the value of x.

[billhalprin]

In your diagram at 3:37, the rationals descend from left to right. If the order were reversed, it would conform to the normal number line, but the inverse, beta, would be greater than the original integer, alpha. How is this to be understood, please.

[amartyarai]

but i might wanna point out some inconsistency here in regards to defining beta set here, if we regard beta to be a negative rational say ( -m) then number (1/-m)(1/r) dne( does not belong to) alpha cut, which is a contradiction given 0* < alpha cut and (1/-m) (1/r) is a negative number

i suggest modifying this definition a bit " beta be the set of all m with property m</= p where 0<p and there exist r> 1 such that (1/p)(1/r) de ( does belong to ) alpha cut " with addition of m variable in definition takes care of negative rational that were left out in previous definition

also with this definition of mine we might need to modify our proof that beta is indeed a cut, for condition(III) its fine but for (I) and (III) there is a slight of bit of work involved

if s de alpha cut and p=(1/s)(1/2) then (1/p)(1/2) dne alpha cut hence p de beta . so beta is not empty ( all the same, no problem till here ) IF q de alpha cut, and q>0 ( we need to mention this fact as q plays the role p plays in definition of beta and p>0 in our definition this is a slight bit of change ), then 1/q dne beta, so beta is not equal to Q. with this our condition (I) is satisfied

pick p de beta, p>0 and r>1, so that (1/p)(1/2) dne alpha cut, If 0 <q<p, then (1/q)(1/r)>(1/p)(1/r)hence (1/q)(1/r) dne alpha cut thus for all m</= q, m de beta ( this expression by default also includes q) and (II) holds

mechanics of proof that beta is a cut is all the same just slight bit of adjustment with our new definition of beta .