Ramanujan's 723rd problem

We prove an identity from Ramanujan's 4th notebook. This identity involves the floor function as well as the square root.
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[@ilanal99]

another way: since 4k+2 is not a perfect square(even and not divisible by 4) we get floor(sqrt(4n+1))=floor(sqrt(4n+2)) and by simple algebra: sqrt(4n+1)< sqrt(n)+sqrt(n+1)

[@chenshen7959]

Your simple reasoning solved the problem beautifully.

[@williamnathanael412]

He's going to put "overkill" in the title soon.

[@MarcoMate87]

I'm not sure of this equation: floor(sqrt(4n+1))=floor(sqrt(4n+2)). It seems to be true, but at least it has to be proven more carefully. It can be easily proven that, in general, for all x > 0 then sqrt(x+1) - sqrt(x) < 1. So, the distance between the square roots of two consecutive natural numbers is less than 1, but it's not sufficient to prove that floor(sqrt(4n+1))=floor(sqrt(4n+2)). If n is a real positive number that equation is generally false: for example, sqrt (4.1) > 2 and sqrt (3.1) < 2. I'd like to see the explicit proof that floor(sqrt(4n+1))=floor(sqrt(4n+2)) for all n natural numbers.

[@MrRyanroberson1]

since 4n+1 is a natural number, we know it is of the form k^2 + m. since 4n+2 is also a natural number, we know it is NOT of the form k^2. therefore k^2

[@leif1075]

How do you know 4k plus 2 is not a perfect square...just because it's not divisible by 4 does not mean it's not..

[@arpitdwivedi9175]

Your work is so good that I almost ignored the fact that you misspelt Ramanujan's name.

[@raphaelkelly861]

I saw that too lol

[@PANKAJKUMAR-nq1qr]

Brilliant explaination on whole youtube!!! You very well demonstrated it! Especially, I am a math lover and a olympiad math lover and you brilliantly gave me a technique to prove identities involvon the floor functions! Olympiad questions involving Floor functions were a little hard to tackle to me after watching you method of proof I have cleared all my concepts about the "FLOOR FUNCTION"! Lots of thanks Professor!

[@SudarshanBaurai]

Michael Penn is a great mathematician and very humbly reproduces Ramanujan's mind. 👏👏👏

[@oliviermiakinen197]

10:20 _"1/2 < a,b,c < 1"_ It is not quite right. What was proven so far is _"1/2 < a,b < c

[@m4riel]

c is also equal to 1 whenever n is one below a perfect square, but again, the goal inequalities still hold

[@sigmainclination9483]

Nice way to target problem with clarity and simplicity ,i am from India ,keep working great job man ..

[@maypiatt3766]

Amazing problem😊

[@mathysicssaransh7939]

Love your work sir 👍👍 Great keep up the great work

[@kartiksharma7166]

Great video!! keep it up.

[@memofie]

Love your work

[@spartacus8875]

Thank you for your execercices ...

[@richardfredlund3802]

I think this is a much simpler proof: because squares are 1 or 0 mod 4, neither 4n+2 or 4n+3 are perfect squares, which guarantees that floor(sqrt(4n+1))=floor(sqrt(4n+2))=floor(sqrt(4n+3)) ----- By squaring the LHS (ignoring the floor) we get 2n+1+2*sqrt(n(n+1)) ---- and --- n < sqrt(n(n+1)) < n+1 ------ so 4n+1 < square of LHS < 4n+3. By square rooting again we see that the floors must be the same.

[@eugeniuszdymek4334]

Worth noting, that Sqrt(4 n + 1) = Sqrt(4 n + 2) =Sqrt(4 n + 3) because only 4 n and 4 n + 1 can be full squares. So more generally, if k is not divided by 4 and n = k div 4 (division with remainder), then Floor(Sqrt(k)) = Floor(Sqrt(n) + Sqrt(n+ 1))

[@richardfredlund3802]

yeah that's really the key observation: because squares are 1 or 0 mod 4, neither 4n+2 or 4n+3 are perfect squares, which guarantees that floor(sqrt(4n+1))=floor(sqrt(4n+2))=floor(sqrt(4n+3)) ----- By squaring the LHS (ignoring the floor) we get 2n+1+2*sqrt(n(n+1)) ---- and --- n < sqrt(n(n+1)) < n+1 ------ so 4n+1 < square of LHS < 4n+3. By square rooting again we see that the floors must be the same.

[@satyapalsingh4429]

Very good method of teaching by the professor .

[@hadireg]

👍👍 I use floor and ceiling functions often in programming, this is the first time I see how it's written :) Great job mate! cheers!

[@hamzamsila4614]

Thank you very much prof ,great prove

[@crazyspider17]

i have a great idea for the overkill series you could show there are infinitely many primes by proving that the sum of all the reciprocals of primes, diverges.

[@MichaelPennMath]

good idea! I'll put it on the list.

[@federixiron345]

Hi Michael, congratulations from Argentina for your great work! Love your videos and hope you keep doing such effort 😁. Here are two exercises that I would like you to try: 1) Let’s call D(n) the greatest divisor of n that is ≠ n. Find all n such that n + D(n) is a power of 10 2) Find all real number non integers “x” such that: x + 2014/x = floor(x) + 2014/ floor(x) Again congratulations and good luck!

[@federixiron345]

Adam Romanov 1) 75 is the only solution 2) The only solution is - 2014/45, you are correct. Is not extreme level but is challenging as I see it (I’m only an 17 years old student interested in this topic :D)

[@federixiron345]

Pd: I know tougher problems, but I proposed only 2 examples of challenging problems that I love with really elegant solutions 😃

[@federixiron345]

Adam Romanov great job!!

[@forgetfulfunctor2986]

GREAT STUFF

[@vic88tor]

Greetings , Michael. How do you make your video pictures ??

[@darkshoxx]

Great Video. I have a follow-up question. I've plotted the functions floor(sqrt(x)+sqrt(x+1)) and floor(sqrt(4x+2)) for continuous x and it looks like the area where they differ is a collection of intervals that become smaller and smaller. I wonder if the resulting integral over the difference of those two functions converges...

[@darkshoxx]

I think it's pi^2/24-1/4

[@grzechu9751]

where i can find list of all ramanujan's problems?

[@vikaskalsariya9425]

www.pdfdrive.com/ramanujan-notebooks-d33656526.html here is pdf version of ramanujan notebooks with alot of the theorems proved(Ramanujan didn't prove alot of them in his book) by Bruce C. Berndt

[@richardfredlund3802]

@@vikaskalsariya9425 that's really awesome thanks!

[@piotrsawek3613]

In case 2 inequality between N^2+N+1/4 and n is false. If n=N^2+m and N

[@djvalentedochp]

Nice job

[@olivierbegassat851]

I'm sure this has been asked and answered before, sorry for the repeat. The colors in your videos always pop, what color chalk are you using ? Is it the famed Hagoromo chalk ? Thanks : )

[@youssefamen6872]

Pls do IMO 2006 problem 5 thx for the great work

[@StanleyDevastating]

the part at 8 mins where we have, N^2 + N + 1/4 > N^2 + m +1, with m < N, it was not clear at all to me that m was a natural number! Because we have introduced real numbers a,b and c I was thinking in the Reals where this inequality obviously does not hold.

[@arkochowdhury4325]

Great video! But the audio is a bit low. It would be better if it was a bit high.

[@vasilisr7]

Amazing Maths broo

[@bohredphysicists1850]

Such a tricky problem but solved well! :)

[@jeremy.N]

great video, have the close up camera on eye height though please. (It was kind of akward)

[@donegal79]

were you triggered? Snowflake corner, off you go.

[@tomholroyd7519]

@@donegal79 were you triggered by polite constructive criticism boo?

[@Justuy]

Sir please do some projective geometry videos

[@bijalshah9113]

*Ramanujan (for the spelling on the blackboard)

[@martinyang5596]

10:24 In case2, can c be equal to 1? (Even if c is equal to 1, the result still holds.)

[@brandonklein1]

No, m is strictly less than 2N+1, so n+1=N^2+m+1=(N+c)^2=N^2+2Nc+c^2. So m+1=2Nc+c^2=c(2N+c), the largest m can be is 2N here, so c is strictly

[@martinyang5596]

@@brandonklein1 If c = 1, then c(2N+c) is equal to 2N+1. m+1 is at most 2N+1. Why does that deduce a contradiction?

[@brandonklein1]

@@martinyang5596 Ah I didn't see that for some reason! Ended up showing what you said 😂. Yes, I do believe that's correct then:)

[@martinyang5596]

@@brandonklein1 Thanks for the reply :)

[@antoniopalacios8160]

Great. This problem appeared in The Eighth William Lowell Putnam Mathematical Competition in 1948. I'm telling you this in case you want to label it as such.

[@marisaelizabethleoncollant7509]

Is so sad im not yet ready to watch this video :(

[@binhtrancong7630]

I saw that number line at 2:40 and thought "looks like this gonna be a short video". Then after checking the full length, I was like, "oh boy..."

[@vinayaklahoti]

At 2:04, √(n+1/2) need not necessarily be less than √N+1 supposedly n = N^2+2N+3/4 Then √(n+1/2) would be greater than N+1 Same goes for √n+1. So m has to be less than 2N..

[@bencheesecake]

Love the video, as usual. However, I found the new camera angles and the panning to be very distracting to the content. One of the things that I love about your videos normally is that I can pause and review relevant information that you have saved on one side or the top of the board, and it was very difficult to do with the new angles.

[@raphaelkelly861]

Quite cinematic, but maybe not as practical...

[@ChefSalad]

I think it would have been more enlightening to write the initial problem as Floor(sqrt(n+0)+sqrt(n+1))=Floor(sqrt(n+1/2)+sqrt(n+1/2)). That way makes it so much more clear why it's interesting.

[@modern_genghis_khan0393]

Where did you take these Ramanujan's questions?

[@jitendrachanna5838]

Unable to understand at 9:05 how a,b,c are less than 0.5

[@ThAlEdison]

I don't see how when 0

[@BlazingshadeLetsPlay]

M must be a natural number, because n is an element of the natural numbers. So no N-1/4 allowed. That tripped me up for a little too.

[@ThAlEdison]

@@BlazingshadeLetsPlay OK, it follows but I definitely missed it in the argument.

[@KostasOreopoulos]

@@ThAlEdison m m+1 n+1 = (by definition) N^2 + m + 1

[@bttfish]

the solution requires a lot of attentions on details

[@vigvigorish8838]

is sqrt(n+1/2) < N+1 ? Because if we get m = 2N+1 then sqrt(N^2+2N+1 + 1/2) is larger than N+1

[@Aeropig_]

We can't get m = 2N +1 because m is strictly less than 2N + 1

[@hybmnzz2658]

m is an integer

[@angeloluisrocattojunior3425]

723rd??

[@prunodagen]

Too complicated ... For n = 0, it's obviously true. We suppose n > 0. Easy to show that Vn + V(n+1) < V(4n + 2) (compare their square) The diophantine equation k² = 4n + 2 has no solution because k² is always = 0 or 1 mod 4. Then, to show the identity, it suffices to show that there is no intenger k such that Vn + V(n+1) < k < V(4n + 2) 2n + 1 + 2V(n² + n) < k² < 4n + 2 But, for every natural n > 0, 4n + 1 < 2n + 1 + 2V(n² + n) Then, 4n + 1 < k² < 4n + 2 There is no intenger between two consecutive ones, then there is no such k. QED.

[@Pacuvio25]

Why is it enough to show that there is no such k?

[@BlazingshadeLetsPlay]

Pacuvio25 he’s saying imagine there’s some number that is bigger then the left side but smaller than the right side. Then he contradicted that such number exists. Therefore there’s no number between them and they are equal. My only question tho is would u have to flip the equality and prove it saying that the right hand side is less than k less than the left hand side, because if there wasn’t equality then either side could be larger than the other, but I’m not sure.

[@igml1145]

@@BlazingshadeLetsPlay he already established that RHS>LHS (without the floors)

[@prunodagen]

@@Pacuvio25 Let x < y two reals and N = floor(y). Then N

[@Bob-hc8iz]

Your proof is identical to the one that I proposed! Nice job!

[@eshnar]

at 14:20 shouldn't it be x less or equal to z?

[@sharpnova2]

no. floor of 0 is -1

[@eshnar]

@@sharpnova2 what? That sounds new to me. floor(0) returns 0 in every tool I have...

[@Joseph2302]

Without loss of generality, you may assume that pi equals 6 for this proof

[@jiaming5269]

Nicee

[@azhakabad4229]

Hello Sir, Hope you are fine! Sir, I want to ask please start lectures series on discrete mathematics?

[@jebbush3130]

why do indians call everybody sir?

[@darkseid856]

How do mathematicians think like this ?

[@lucaswilkins9217]

The u in Ramanujan has gone to the same place as your right parentheses ;)

[@SanketGarg]

I am lost. How can a, c be between 0 and 1 when 'n' is a natural number?

[@geethaudupa8930]

well let's say n=3 then N=1 and a=0.732....

[@SanketGarg]

@@geethaudupa8930 yeah but n = (N+a)^2 = (1+0.732)^2 != a natural number

[@IoT_]

You're right - N is not always a natural number.

[@ThAlEdison]

@@SanketGarg a=sqrt(3)-1, N=1, (N+a)^2=(1+sqrt(3)-1)^2=sqrt(3)^2=3, a natural number.

[@SanketGarg]

@@ThAlEdison that clears it up! Thanks!!

[@Bob-hc8iz]

I think the following is a much easier proof. Let A = n^.5 + (n+1)^.5 and B = (4n+2)^.5. Then one can easily show that: 4n+1 < A^2 < 4n+2 = B^2 Let M = Floor(B). If Floor(A) not= M, then A < M < B. Note M can’t equal B since M is B^2 is congruent to 2 mod 4 and hence is not a perfect square. But this would imply that M^2 is and integer lying strictly between 4n+1 and 4n+2 which is clearly a contradiction. For completeness I will establish the row of inequalities as follows: A^2 = n + n+ 1 + 2 root(n^2+n). But n + .5 > root(n^2+n) > n so 4n+1 < A^2 < 4n+2 as claimed.

[@Bob-hc8iz]

@@craig4320 Yes, you are correct. That was a typo - I meant to say M can't equal B since B^2 is congruent to 2 mod 4 and therefore not a perfect square!. Thanks for the correction. I will edit the text. Other than that did you follow the logic?

[@sohelzibara8166]

hi bob, my proof is something like yours, though with a different approach. i am trying just now to KZbin publish it. when done please tell me what you think.

[@richardfredlund3802]

because squares are 1 or 0 mod 4, neither 4n+2 or 4n+3 are perfect squares, which guarantees that floor(sqrt(4n+1))=floor(sqrt(4n+2))=floor(sqrt(4n+3)) ----- By squaring the LHS (ignoring the floor) we get 2n+1+2*sqrt(n(n+1)) ---- and --- n < sqrt(n(n+1)) < n+1 ------ so 4n+1 < square of LHS < 4n+3. By square rooting again we see that the floors must be the same.

[@craig4320]

@@Bob-hc8iz sorry, I deleted the post you replied to.

[@richardfredlund3802]

I think my proof is much simpler.

[@backyard282]

Great video, however I dislike constant camera angle changes, it's a little distracting.

[@craftexx15]

3rd

[@edskev7696]

Does your proof show the statement for all n>=0, not just integers?

[@Czeckie]

it does not, he uses the fact that m is an integer when he's proving the cases 1,2,3. The proofs are correct and work even for non-integer m, but the argument is not valid when m is close to N - I think the m in interval (N-3/4, N+1/4) needs special care. The case 3 (m=N) falls in this range, I'll try to check if Michael's method extends to the region, when I will have more time. I have a hunch that there will be quite a few cases to consider

[@subpopulations]

Its where 4n+1< sqr(N)< 4n+2

[@darrenhundt]

please no to the new camera angles

[@armacham]

Prove by contradiction. If it's NOT true there would exist some n for which sqrt(n+1) + sqrt(n) + 1

[@magicodabola10]

you gotta stop pointing things at the board...

[@tonyennis1787]

Camera changes are distracting and unhelpful

[@eeee69]

i got 723 problems

[@sohelzibara8166]

hello. very interesting way of doing it although you ve complicated it a bit. there is a much easier solution to this problem. good work

[@sohelzibara8166]

@Philosopher's Dream i do have two different solutions that are as I said before much more easier. i ll get back to you

[@Bob-hc8iz]

@@sohelzibara8166 Could you review the proof that I proposed? I think this must be one of those that you know. I would like to see a different one.

[@sohelzibara8166]

@@Bob-hc8iz where is the proof that you proposed. just found it

[@sohelzibara8166]

i just put an another solution on my channel on KZbin. for those who are interested please check it out

[@peterdriscoll4070]

Um. Really confusing. Too many steps not explained. No clear direction, or meta explanation of where this is going. No way this can be followed without a lot of work.

[@berndmayer3984]

I don't like a orientated line like the real number line with 2 orientations. All americans do that. And "one" is the same as "under the condition of... " like M={x | 2x-5 > 0}

[@enricolucarelli816]

I don’t understand. You say this is an identity? It just isn’t true. Just grab a calculator and check it out. For n=1 you get 2,41421=2,44949 (6 digit precision). Now, for n getting bigger and bigger the difference between the two results gets smaller and smaller, but it definitely isn’t an identity. It is an approximation.

[@enricolucarelli816]

Actually, if you square both sides of the identity, move the remaining square root to one side, and square again, you get a contradiction, namely n^2 + n = n^2 + n + 1/4

[@jebbush3130]

that's why the floor function is there

[@craig4320]

You were going too fast for me, and making too many mistakes. If it were written in a textbook, it would have been less frustrating to follow. (Of course you corrected your mistakes, but that does not solve the problem that the mistakes made it hard to follow you.) (Too fast in the logical deductions, not in the algebraic multiplications.)