Q is dense in R

Rational numbers are dense in the real numbers

In this video, I present a classic proof that the rational numbers are dense in the real numbers. Enjoy!

[Jim-besj]

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.

[NotoriousSRG]

Mathematics, WTF: Want To Find

Everyone else: What the fuck?

[ayush.kumar.]

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

[crosseyedcat]

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.

[demidevil]

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. :)

[mehdimemar]

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

[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.

[muskannm]

Your teaching style is so eloquent, loved the lectures !

[cauchy]

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

[gillesnassar]

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 :)

[gustavocardenas]

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.

[barryzeeberg]

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".

[包子他爹嘻嘻哈哈]

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

[mdperpe]

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.

[michaelempeigne]

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

[prateekmourya]

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)

[dougr.]

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?

[punditgi]

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

[afoster]

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.

[camillamagi]

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