A really interesting floor problem!

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[@demenion3521]

You could also just end the first inclusion at sqrt(4n+2). This would in the end yield the inequality 0

[@Czeckie]

yep, the discussion about 4n+2, 4n+3 added nothing

[@matteoanoffo1447]

It's the second time he does this exercise in his videos, nice

[@somasahu1234]

Understood everything except the 1st equivalent statement !

[@fix5072]

It is about the rhs of the goal equation. The expression inside the sqrt must be strictly less then the next bigger perfect square (4n+4) and strictly bigger than the next lower perfect square (4n+1). Thus the lhs must also be on that intervall

[@ConManAU]

@@fix5072 Not that 4n+1 and 4n+4 are guaranteed to be perfect squares, but 4n+2 and 4n+3 absolutely cannot, so if floor(sqrt(4n+2))=k, then k^2

[@fejfo6559]

the squares mod 4 are 0 and 1, so 4n+2 isn't a perfect square. That means 4n+2 is in between 2 perfect squares a² < 4n+2 < (a+1)². If we take the square root the inequality we see floor(4n+2) = a. So the goal becomes sqrt(n)+sqrt(n+1) in [a, a+1) Because of the squares mod 4, a² < 4n+1 and 4n+4 < (a+1)². So his statement suffices.

[@fix5072]

@@fejfo6559 this is what i was trying to say, even tho the inequalites must be a^2

[@tonyhaddad1394]

Yes michael always skip many steps

[@kevinmartin7760]

I'm having a little trouble with the first step, especially the "equivalency" claim. What is not clear to me is that the goal implies the first step, which is one of the requirements for equivalency. Maybe is does (but it is not obvious to show), or maybe it doesn't, but then that doesn't matter as equivalency is not required; all that is required is that the first step imply the goal.

[@henryhowe769]

Perhaps it would help to understand this in terms of inequalities instead first a few examples of the floor function floor(3.2) = 3 because 3.2 ∈ (3,4) which is equivalent to saying that 3 < 3.2 < 4 and 3 is the lowest value in this interval floor(5.9) = 5 because 5.9 ∈ (5,6) which is equivalent to saying that 5 < 5.9 < 6 and 5 is the lowest value in this interval we are looking at the floor(√[4n +2]) so what is the integer interval that we should be working with? well for a square root to be an integer the input must be a perfect square, and the nearest values that meet this requirement relative to √[4n+2] are √[4n+1] & √[4n+4] In other words √[4n+1] < √[4n+2] < √[4n+4] our claim is that floor(√[n] + √[n+1]) = floor(√√[4n+2]) this implies that they are on the same interval therefore we can substitute and say that √[4n+1] < √[n] + √[n+1] < √[4n+4] or in other words √[n] + √[n+1] ∈ (√[4n+1],√[4n+4]) Hope that helps :)

[@E1Luch]

@@henryhowe769 but why √[4n+1] possibly being an integer would imply that it is smaller than √[n] + √[n+1] for all n? it is smaller in the end, but its not always an integer, and if understand this correctly it was only proven at the end of the video, directly, without the requirement of the original goal equality, and then the goal was implied from it, and only then we could say that these statements are equivalent because they are both true for any natural n.

[@coc235]

@@E1Luch Right. The inclusion implies the goal equation, but not vice versa

[@henryhowe769]

​@@E1Luch √[4n+1] possibly being an integer does not imply, that it is smaller than √[n] + √[n+1] if it did the rest of the proof would be unnecessary. Rather the goal equation suggests that such inequality might be true. Its more like an educated guess.

[@balthazarbeutelwolf9097]

first step is just wrong. From the original you get strict a lower bound of sqrt(4n+2)-1 and a strict upper bound of sqrt(4n+2)+1. One proves these inequations by a sequence of logical steps mostly involving squaring both sides and getting rid of common terms. Looks nothing like Michael's proof though.

[@manucitomx]

Thank you, professor!

[@Kosemagician]

2:31 if n=0 it is in fact possible for sqrt(n)+sqrt(n+1) to be a natural number, so the lower bound should be included [sqrt(4n+1), sqrt(4n+4)) All of this depends on whether you define 0 to be a natural number in the first place

[@luisaleman9512]

Michael doesn't consider 0 to be a natural number.

[@sugongshow]

Just like Michael doesn't consider 0 to be a natural number, the rest of the world should not either as natural numbers are 1, 2, 3, 4, 5, .......

[@balthazarbeutelwolf9097]

@@sugongshow In Computer Science everyone considers 0 to be a natural number. When you do number theory modulo n you simply need 0. Of course, it's just a definition, just like excluding 1 from being prime - it's just a case which "works better" in the part of maths you are interested in. But Peano arithmetic uses 0 as the base case too.

[@stewartcopeland4950]

Good Place To Stop preferred to take a nap

[@goodplacetostop2973]

I was outside for a visa renewal actually. The nap will be later 😂

[@klausg1843]

I miss an explanation of the conclusive step. So I propose the following: Since both n^0.5+(n+1)^0.5 and (4n+2)^0.5 lie in the open interval from (4n+1)^0.5 to (4n+4)^0.5 and this interval does not contain an integer, then the conclusion follows.

[@sugongshow]

I love that shirt!!

[@wesleydeng71]

First, LHS < RHS. So we only need to consider the case sqrt(n+1)=k/2 (where LHS is most close to an integer k but still less than it). In this case, 4n+2=k^2-2, which means sqrt(4n+2) is also less than k.

[@ByteOfCake]

Have you ever thought of doing videos on the derivation of the explicit formula of the prime counting function? Like finding identities for ln(zeta(s)) and then using Mellin inversion. I think it would be interesting.

[@ByteOfCake]

actually, i'll put it in your form. Didn't see it

[@minamagdy4126]

I wonder if your proof also shows that both also equals floor(sqrt(4n+3)), which makes sense given the between 2 integers thing in the top line. Also would be interesting to explore all situations for which 4n+1 or 4n+4 are not equal to the above (after the transformation. I expect low numbers like 1 to be the only exceptions.

[@jabunapg1387]

This Problem was in the first round of the german Math Olympiad a few years ago

[@dicksonchang6647]

same problem?

[@pranjalpathak214]

Wow nice job for making the complex thing more complex🤣🤣🤣

[@cyberjet5946]

Can we do it with AM >= GM property? sqrt(n) + sqrt(n+1) >= 2 * sqrt(sqrt(n).sqrt(n+1)) I have arrived till sqrt(n) + sqrt(n+1) >= sqrt(4n + 2). How to proceed next? is this possible via this method? How do I apply the floor function on this inequality logically if possible?

[@drsonaligupta75]

8:21

[@user-ur2en1zq4f]

Where is that good place to buy that good shirt?

[@Frank9412co]

In the description there is the 🔗 to the teespring site

[@user-ur2en1zq4f]

@@Frank9412co where is that good place you live?

[@ayoubabid8783]

Good

[@user-bx7rw1pt4p]

I didn't understand why the inclusion on the interval solve the problem. For example: if the sum is a little bit greater than sqrt(4n+1) then the floor of the sum is equal to the floor of sqrt(4n+1)

[@deadfish3789]

But the floor of sqrt(4n+1) is always equal to the floor of sqrt(4n+2). If not, that would be equivalent to saying there is a square number on (4n+1, 4n+2], but clearly this is impossible since they are consecutive integers and there are no squares of the form 4n+2.

[@user-bx7rw1pt4p]

@@deadfish3789 thanks

[@chessematics]

Meanwhile Ceiling: becomes jealous of his brother.

[@ZedaZ80]

You should remake all of those videos, but replace floor(x) with -ceiling(-x) XD

[@backyard282]

But why you never show any love for the ceiling function :(

[@theophonchana5025]

(Square root of n) + square root of (n+1) = Square root of (4n+2) n = undefined

[@romajimamulo]

There's something I find kinda weird about the range you picked. It's not clear to me that it will always be tighter than the range needed for the floor function

[@leif_p]

He went very quickly through that part, but his range is equivalent to asking "What are the closest numbers to 4n + 2 that _could_ be the square of an integer?" This is actually _stronger_ than needed for the floor function for large n. Plug in n=100 and you get (approximately): 20.02 < sqrt(4n+2) < 20.07, way tighter than needed.

[@romajimamulo]

@@leif_p right, I knew it was smaller, I just wasn't sure about the fact it has to be between the integers.

[@anguzman]

I saw this problem as #20 of Chapter 3 of Tom Apostol's number theory book.

[@MrDivinity22]

From the proof it looks like this equality also holds with 4n+3 on the right, doesn't it?

[@binaryagenda]

And with 4n + 1. It's kind of unsatisfying that it is not a unique expression.

[@user-bx7rw1pt4p]

Sqrt(4n+3) can t be a perfect sqr then its floor is equivalent to the floor of 4n+2

[@titan1235813]

No. Okay, yes ☺

[@Frank9412co]

8:22 is a good place to stop

[@goodplacetostop2973]

Yes, it is.

[@Gandarf_]

That problem already was on this channel more than year ago btw

[@TI5040]

Isn't it ramanujan's result?

[@idontknowwhatido3972]

Hey, he is behaving quite differently...hm maybe he is drunk?, high? withdrawal?

[@trueriver1950]

Counter example N=0 LHS = 1 RHS = 2 The failed step in the proof is where Michael assumes that consecutive integers cannot both be squares. Did I miss a comment that n has to be positive excluding zero?

[@theophonchana5025]

#undefined

[@wojteksocha2002]

Cool video bro, but I think I have to copystirke your video cause Michael Penn already made video about it.

[@trueriver1950]

I am guessing he used it with permission :) He did say at the outset that this is a problem he really likes...

[@aashutoshgupta6690]

First view