Q is dense in R

[@NotoriousSRG]

Mathematics, WTF: Want To Find Everyone else: What the fuck?

[@Jim-be8sj]

These density concepts are often surprising. Any real can be approximated within any given tolerance by a rational, yet the rationals are an infinitely tiny set when compared with the reals. Or another, any continuous function can be approximated in the same way by a polynomial. Sometimes I think only Cantor really understood infinite sets.

[@ayush.kumar.13907]

*headline in beginning on board:* Q is Dense *Q:* Huh

[@demidevil666]

Thank you for producing these videos Dr. Peyam. ♥️ I finished my Bachelor's in applied mathematics last year and ever since the purely mathematical courses were behind me, I missed beautiful proofs like this one here. And after I got my degree and struggled to find work I had to change careers into IT out of financial necessity. Your content brings me back to the time when I first learned analysis, linear algebra and discrete mathematics. Like a nostalgic journey back through my memory. You help many of us here in more ways than you might realize at first glance. Thank you again for doing this. :)

[@dougr.2398]

{return -1;} for many of us, IT is “it” (and so are we!)

[@mehdimemar]

Before opening up your videos, I always take a moment appreciate the thumbnails. They're deeper than simply repeating the video titles. Excellent.

[@drpeyam]

Thank you! I work really hard on them

[@JasmineAllyson]

I'm amused that you denoted "want to find" as "WTF"! Anyway, great job, Dr Peyam! You've helped me so much in mathematics.

[@drpeyam]

Thank you!!!

[@crosseyedcat1183]

Been going through these videos to learn more about analysis. I think I understand why that max has to exist. I proved a theorem about subsets of integers which stated that every subset of integers bounded above has a max and every subset of integers bounded below has a min. We can see that the set S you construct in this video is bounded above since m/n < y means m < yn. We can then use the ceiling function on y to get m < ceil(y)n. So the subset of integers is upperbounded by ceil(y)n and must have a maximum. Thank you Dr. Peyam. I really appreciate all your lectures and also taught myself linear algebra from watching them.

[@RodrigoGonzalez-zy1mf]

Thanks for these videos Dr. Peyam!

[@muskannm1342]

Your teaching style is so eloquent , loved the lectures !

[@gillesnassar8743]

I just learned that theorem but we proved it using the floor function ( also using the Archimedean property). Mainly, we used the floor function to find r (rational number) between a and b (real numbers). But that proof is interesting! keep the good work :)

[@gustavocardenas6489]

Thank you very much Dr Peyam! I finally understand the motivation/intuition behind the construction of the set S, without just constructing it and stating the well ordering principle (in the general case of positive and negative numbers), like some textbooks tend to do.

[@senhueichen3062]

I feel Q is dancing in R.

[@dougr.2398]

A waltz, cha-cha or rhumba? Ballet, perhaps?

[@gcewing]

At least the rationals are maintaining their gaps for social distancing. But those pesky irrationals are flouting all the lockdown regulations.

[@cauchy2012]

You can use this proprietary that for all a and b in IR such that b-a>1 there is some p€Z /a

[@cauchy2012]

This simple more

[@md2perpe]

Here's an even denser set with Lebesgue measure 0: Let C be the ordinary Cantor set (uncountable set with Lebesgue measure 0) and let Q be the rational numbers (countable). Then C+Q = { c+q | c in C, q in Q } = Union_{q in Q} (C+q) is a countable union of sets with Lebesgue measure 0 and so also has Lebesgue measure 0. Yet every interval of R contains an uncountable number of points from C+Q.

[@包子他爹嘻嘻哈哈]

Really interesting proof, and the ideas flows fluently.Thank u

[@barryzeeberg3672]

You could take the ( infinite , non-repeating) decimal expansion of "a" and "b". Since these are not equal to one another, at some finite number of decimal places, they will differ. Then simply truncate "b" at 2 decimal places after where "a" and "b" start to differ. Then this truncated decimal expansion is the representation of a rational number, as it is no longer non-repeating because the terminal 0 repeats forever. Because of how the truncation was performed, this rational is larger than "a" and smaller than "b".

[@TheAlx32]

Thank you. Very elegant proof

[@michaelempeigne3519]

Note that a does not have to strictly less than b in ( a / b ) to be a rational number technically

[@pandabearguy1]

R is dense in R

[@jessicapriscilacerqueiraba3493]

your explanation was soooo good

[@txikitofandango]

Totally counterintuitive, since Q is countably infinite while R is not

[@drpeyam]

True!

[@afoster1955]

In your proof that the rationals are dense in the reals. You claimed that the set S which consists of numbers of the form m/n where m/n < b is a "finite" set (which I don't agree with you here). However, I do agree that S is bounded above by b and that sup S exists in the reals. But your definition of S as all the fractions m/n that are less than b is not finite. In fact, S is infinitely countable. I think that you intended to collect in S the "number of increments" of length 1/n starting from 0 and traveling to the right along the real line until one exceeds the first point a. It is this collection of increments that is finite! But not the collection of "fractions" which precedes b.

[@camillamagi1841]

My teacher made this sound so difficult during the explanation, with this video I understood right away thanks

[@dougr.2398]

I’m wondering at the use of the word “density” here, because density is a qualitative term as employed here, but “density” (mass per unit volume, in another discipline) is a quantifiable and quantitative term. My question is then: Can the density of the rationals, (Q) be measured and quantified relative to the irrationals that are solutions of polynomials (algebraic) and the irrationals that aren’t (the transcendentals)? Are the integers then “sparse”? And if so, how sparse?

[@athelstanrex]

Life is old there.... Older than the trees.....

[@punditgi]

Nicely done, Dr Peyam. Now, please give us more videos about the point set topology of Q vs R. 😇

[@tgx3529]

I have seen this proof. If x1, then exists r from Z where x

[@drpeyam]

Well, although it might seem obvious, you still need to show the statement about nx < r < ny

[@dougr.2398]

Dr Peyam are )or “arrr!”) is greater than New Xeeland but less than New York?!?

[@prateekmourya9567]

I have really good question for you if alpha and beta are roots of equation x^2-x-1(Fibonacci equation) we define a function f(n)=alpha^n+beta^n then prove that f(n+1)=f(n)+f(n-1)

[@OtherTheDave]

Isn’t there a rule or something that says something along the lines of “there’s an infinite number of reals between adjacent rationals”? I don’t remember it exactly, but its point was to prove there are infinitely more reals than rationals. Off the top of my head, it seems like being able to find a rational between any two reals would contradict that.

[@drpeyam]

That’s much easier to prove: If a and b are rational then a + (b-a)/sqrt(2) is an irrational between a and b

[@drpeyam]

And I guess for infinite number of reals, just replace 2 above with p for any prime number p

[@OtherTheDave]

Dr Peyam But wouldn’t there be a rational between “a” and “a + (b-a)/sqrt(2)”, and another rational between that and “b”?

[@user-rf5hr6ek8x]

@@OtherTheDave It doesn't really make sense to talk about "adjacent rationals". If a and b are assumed to be so-called adjacent rationals, i.e. b is the closest rational number to a where b>a, then (a+b)/2 is between a and b and is rational, and so a and b aren't adjacent.

[@xavierplatiau4635]

Well, if you look at Q with some measure theory lenses, then I guess for exemple the probability to pick a rational number out of any real number (uniformly distributed) in [0;1] is 0. Which is interesting, rationals are both everywhere (since Q is dense in R) and nowhere in R.

[@acesarich535]

Suppose that since the limit as q->0 [ln(q)] = 0, where for all q in the Rationals, is strictly greater than 0. Then for all minima in a subset of the set {1/n: n is a natural number}, the finite argument fails to show an infinite cardinality of elements in a subset of {ln(q): q is a rational number}. Suppose we don’t restrict our solution to finite subsets of {1/n: n is a natural number}, and suppose the solution defining this Archimedean Property for Q dense in R can be solved for an infinum solution. Then the infinum of any subset of Q must be the limit argument which can be shown to be 0. Which is strictly less than the subset we want to have a least upper bound for. This contradiction without strong proof should suffice to demonstrate one of many examples by which the Rationals cannot be taken to be Dense in the Reals.

[@DancingRain]

Well done. :) One little quibble though: the audio is too quiet.

[@TheForever119]

I think for Dr Payems videos should be superb button. otherwise it is unappreciated

[@piyalikarmakar5099]

Please also make a video showing the proof of Z[√2] is dense in R

[@chillfill4866]

Just take the midpoint and use the Cauchy sequence definition of the midpoint.

[@vecter]

Is there a way to prove a < r without relying on proof by contradiction?

[@drpeyam]

I don’t think so

[@justwest]

R is per definition the completion of Q, and therefore Q is dense in R ^^

[@drpeyam]

Depends on your definition of R ;)

[@GlorifiedTruth]

Can someone help me? At 11:13, we know a >= r (original assumption), and r = m/n, so how do we justify a >= m/n + 1/n? How do we justify the addition of 1/n? Thanks.

[@alexhells2367]

Nice explanation

[@txikitofandango]

Is there a way of proving this fact with decimal strings? Like, you have two real numbers with infinite decimal expansions. It should be pretty easy to prove that if you chop off the bigger real number (B) somewhere, the resulting rational number R will still be bigger than A.

[@drpeyam]

Well, you’d have to define then real numbers via decimal expansions, but then it’s harder to show that the least upper bound principle is true

[@sebastianmata3425]

I had this same thought and what occurred to me was this problem. Say a=1.437298... and b=1.4373000.... then clearly you can see that if you “chop off” every digit after the 4th decimal digit on b, you still get b and so you don’t just easily get a rational between a and b. Idk though, another person said they solved it with the floor function so maybe it’s possible 🤔

[@barryzeeberg3672]

@@sebastianmata3425 In this example, "b" is already rational since it ends in repeating 0's

[@rajivdixitbhaiifollower5055]

Excellent sir❤

[@tgx3529]

I think, It's only the result "two spoons"

[@dougr.2398]

?? Reported for being cryptic!! (Kidding!)

[@dgrandlapinblanc]

Subtle. I notice that his results born of an assumption, an assumption not evident to figure out. In a certain sense it's like if the searcher in mathematics was knowing the truth before to make the demonstration which verifies the intuition with the grammar of mathematics and his language the real fact. Thank you very much.

[@ahlamouldkhesal5562]

Can you please proof that D is dense in R where D={d=a/(10^n))

[@bourgrghia1360]

i have a conjecture the density of prime number in N it is the density of real number in R

[@vinlebo88]

The density of real numbers in R is 1. Every number in R is a real number.

[@bourgrghia1360]

@@vinlebo88 i mean irrational number

[@torment808]

hey doctor peyam cn you do a video on i to the i th power i times

[@drpeyam]

Bprp already did one

[@hyperboloidofonesheet1036]

So between any two non-identical real numbers there is a rational number, and between any two non-identical rational numbers there's a real number. And there are many more reals than rationals. So where are they hiding?

[@atlaslife3800]

Whenever one deals with infinite sets, one has to let go of the idea that things will make sense. For example, it can be proven that I can theoretically fill an infinite amount of hotels, each with an infinite number of guests, with the contents of a single hotel with an infinite number of guests.

[@takudzwamukura7172]

Where can I find your notes?

[@drpeyam]

sites.uci.edu/ptabrizi/math140asp20/

[@takudzwamukura7172]

@@drpeyam Thank you

[@Kdd160]

The thumbnail was awesome. Wasn't it?? -BlackPenRedPen :)))) Edit: the video has 0 dislikes. Keep it up dr.Peyam or dr.πm!!!!!

[@ChienChiWang]

謝謝！

[@drpeyam]

Thank you so much, I really appreciate it!!

[@VainCape]

Maan you look like Kyle from Nelk boys

[@jeremy.N]

blackpenredpen

[@Feds_the_Freds]

I find this so unintuitive :) I guess, in my intuition, there are infinitesimals Element of R that are smaller than any Number 1/n with n Element in N.

[@mikhailmikhailov8781]

Why don't you just use the definition of the reals via Cauchy Sequences of rationals to do this in one line?

[@drpeyam]

But then prove the least upper bound property :)

[@mikhailmikhailov8781]

@@drpeyam Ye, the proof for that via Cauchy is quite clever.

[@KatieRoseine]

So many black pens :) You should call yourself blackpenblackpenblackpen...

[@michellauzon4640]

13 minutes clip for that! Just consider the subsets { k/2**n , k in Z } , n fixed.

[@kelvinella]

What does WTF stand for? lol

[@drpeyam]

Want to find

[@drpeyam]

No, want to find, not what the function 😆

[@brylemorga3262]

Where's The Fish

[@rogerkearns8094]

When he wrote WTF I thought WTF?!