a floor equation.

We solve a nice equation involving the floor function, square roots, and cube roots.
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[@garrethutchington1663]

12:35 as a computer scientist, I approve this method of counting :D

[@MyOneFiftiethOfADollar]

Taught as Mesa State University in Grand Junction for a few years.Your name rings a bell. Did you ever do any professional development presentations for Computer Science Departments in that region?

[@rhythmmandal3377]

"case 0", you went full programmer there lol.

[@rhythmmandal3377]

@@angelmendez-rivera351 how many are likely to make that connection. Even he himself referenced "counting like a programmer" even though he is math man.

[@rhythmmandal3377]

@@mr.knight8967 too easy.

[@piper3643]

forgot the switch{}

[@rhythmmandal3377]

@@piper3643 forgot the semi-colon too, the most important thing

[@piper3643]

@@rhythmmandal3377 i’m a macos developer haha

[@JoeWolf]

"I'm counting like a computer scientist"😎

[@Kokurorokuko]

I like your way of teaching because it is so close to what I'm used to. Every good thought that I get during watching your videos is matching with yours! Thank you!

[@juanalbertovargasmesen2509]

Since n is an integer, the whole expression is integer. That means you can add 1 and turn the "less than" into a "less than or equal to" that can be easily factorized.

[@quantumgaming9180]

I don't quite understand what you mean. Can you elaborate a bit more

[@goodplacetostop2973]

15:02 ‭Daily homework... Show that whatever five integers we choose, there are two among them, whose sum or difference is divisible by 7‬

[@garvittiwari11a61]

Ok

[@pbj4184]

@@garvittiwari11a61 Haha how does it feel not being first? LOL 🤣

[@anonymoususer9837]

Homework response: Each of the five integers must be of the form 7n+k, where n and k are integers and k is on the interval [0, 6]. Specifically, if our integer is x, n=floor(x/7), and k=x mod 7. If any two of our choices have an equal k, then their difference is (7nA+k) - (7nB+k) = 7(nA-nB). nA and nB are integers, so nA-nB is also an integer, meaning the difference of our two choices is divisible by 7. Therefore, our choices must have distinct k values to not have a difference that is divisible by 7. However, if any of the sets {1, 6}, {2, 5}, or {3, 4} is a subset of the set of chosen k values.. adding our chosen integers with k values in one of those subsets gives (7nA+kA) + (7nB+kB), where kA+kB=7. That simplifies to 7(nA+nB)+7 = 7(nA+nB+1). nA+nB+1 is an integer, so that sum is divisible by 7. Now we look at the worst case scenario for distinct k values: one of our choices has k=0, and three more have values from distinct subsets. However, we chose five integers, so the fifth choice must either match a previous k value or complete a subset, thus creating a sum or difference divisible by 7.

[@Goku_is_my_idol]

Ok so the easiest method: Let the 5 integers have a,b,c,d,e mod 7 If even one of them is NOT distinct Then their difference will be divisible by 7 Eg. X=1(mod7) and Y=1(mod7) cannot be chosen bcuz X-Y=0(mod7) So next suppose we take 4 of the numbers as 0,1,2,3 mod 7 respectively Next whatever distinct remainder we choose for the last remaining number we will always have sum of two as a multiple of 7 Suppose last no. is 5(mod7) But we already have a no. which is 2(mod7) so their addition will be divisible by 7 Thus we can have at most 4 numbers where their sum/difference will not be divisible by 7

[@garvittiwari11a61]

@@pbj4184 Not here to be first, just finding good place to stop...

[@77Chester77]

Hey, very interesting equation! I didn't expect the solutions to be in such a "small" range and that the range is interrupted.

[@Tiqerboy]

I got the right answer, but with a lot less rigor than the buff math prof. Here's what I did: It's obvious that you look at breakpoints where √x is an integer and ³√x is an integer. For √x : x = 0, 1, 4, 9, 16, 25. For ³√x : x = 0, 1, 8, 27.... now evaluate the floor of √x and ³√x in the ranges of interest: [0,1], [1,4] [4,8], [8,9], [9,16], [16,25] and it's obvious you don't need to go beyond 25 since √x increases a lot faster than ³√x From this you figure out the answer the prof gave : 0

[@MrRyanroberson1]

yours could be realized as a proper proof if you just look at sqrt(x) - cuberoot(x) and ask "when is this definitely larger than 1"? but with fancier math speak; notice that the difference is increasing in size (using derivative test) past 10 since sqrt(10)-cuberoot(10) > 3.16 - 2.16 > 1 (barely) (f test), and 0.5/sqrt(10) - 10^(-2/3)/3 > 0.15 - 0.072 > 0 (f' test) are both true. therefore x can never meet or exceed 10

[@HybridTheoryXero]

I love your content. Really interesting and fun. Greetings from Argentina

[@roberttelarket4934]

I'm "floored" by your solution!

[@georgewilliams9695]

13:47 the floor of negative the square root of x is equal to the floor of the cubic root of x For the square root to exist then x must be greater or equal to zero, If it is not zero then the floor of negative square root is less than minus one, but then the floor of the cubic root would be greater or equal to zero. The only solution is zero.

[@MyOneFiftiethOfADollar]

Using the two lemmas floor(sqrt(x)) = k whenever k^2

[@quantumcity6679]

Nice problem... I will try those problems as well Thank you for explaining

[@edgardomatthies2301]

n^3 < (n+1)^2 implies n^3

[@PATRICKZWIETERING]

Nice, I had the feeling this could be done a bit more easy as well!

[@wabbasMEpern]

Dang, I've seen this guy around Lynchburg before. Great videos!

[@jarikosonen4079]

Or mix of floor and ceil, like when ceil(sqrt(x))=floor(cbrt(x)). Or 2*floor(sqrt(x))=floor(cbrt(x))... Some cases could become complicated.

[@neur303]

That's a nice one for me since you only need very basic tools, but still some thinking is involved.

[@quantabot1165]

I come to learn DE, but gets hooked on the video you post daily hahaha

[@juarezmazzucajunior9529]

Sucesso! Obrigado!

[@maruthasalamoorthiviswanat153]

Really really excellent solution

[@CaradhrasAiguo49]

13:45 for the mth versus (m+1)th root case (on the non-negative reals), the solution is, in LaTeX notation: [0, 2^m) \cup \bigcup_{k=2}^{\lfloor r_1 floor} [k^{m+1}, (k+1)^m) where \bigcup represents the "Union of all those iterated intervals", and r_1 is the largest real-valued root to the polynomial x^(m+1) - (x+1)^m. I would not know how to go about proving that this r_1 > 2 for all m >= 2, though. An example to visualise the above is if m=4, then the union of the intervals [0,16), [32,81) [243, 256) constitutes the admissible values of x.

[@mcwulf25]

Nice. I would prefer to see the max/min calculated before the guess at 3. The 3 can be seen from (1+sqrt(7))/3, which is between 2 and 3.

[@JoseFernandes-js7ep]

There are some problems here that I not even have a clue how to solve them. I'm glad I was able to solve this one mentally .

[@PlayerMathinson]

You did all of this mentally or did you have a better method? Also, what really gave you the clue as there is no way I could've thought of this solution like in the video.

[@JoseFernandes-js7ep]

@@PlayerMathinson I haven't even thought about that complicated (but sound) I method. Numbers in [0,1) have both roots in that same interval. Numbers in [1, 4) have both roots in the interval [1,2). Which n have cubic root of 2. It is 8 and its square root is 2.....so another interval begins at 8. However, the sqroot of 9 is 3, so this new interval is [8,9). The cube of 3 is 27,but the sqroot of 27 is 5..... As the cubes grow faster than the square no other intervals are possible.

[@PlayerMathinson]

@@JoseFernandes-js7ep Thank you for your insightful comment. I found a new and more elegant way to solve the problem.

[@GhostyOcean]

@@PlayerMathinson I thought about the graphs of each of the functions and imagined where they overlapped. Then I used a more handwavey explanation (but very similar) as to why there are no more intersections. The formal proof would look something like what he showed in the video

[@horseman684]

great video sir but i have a question with the first assumption, that the equation implies we are dealing with a natural number, is that necessarily true? like if we allow x

[@mrminer071166]

"Sweep away the dust after the decimal, what's left is the bare floor."

[@MichaelRothwell1]

For negative numbers, you must then subtract 1.

[@roberttelarket4934]

mark miner: very clever wording!

[@mohamedcharioui3589]

Ok you answered my question in the video. Thank you

[@mrkorchsearch]

nice video!

[@fivestar5855]

Will there be a video with solution the last questions?

[@pratikmaity4315]

Hey Michael your videos really motivate me to pursue pure mathematics. Here is a nice Diophantine equation problem you can make a video on. Find all a,b,c belonging to natural numbers such that a^3+b^3+c^3=a^2b^2c^2. Thanks and keep going!!

[@rustemtehmezov9494]

İMO SL-2019 - N/2

[@pratikmaity4315]

@@rustemtehmezov9494 yes I got to know that this is ISL 2020 N2 after I posted the comment. But the problem is nice. I hope he will make a video on it.

[@rustemtehmezov9494]

@@pratikmaity4315 İ would like to see a İmo level nice geometry problem. But NT is nice too see here.

[@pratikmaity4315]

@@rustemtehmezov9494 Michael Penn make most videos on NT, algebra or analysis so thought of posting NT problem😄😄

[@alxjones]

First, note sqrt and cbrt are differentiable increasing functions which agree at (0,0) and (1,1), and sqrt' > cbrt' for x > 1. Further, we restrict the domain of x to [0,inf), as floor isn't defined for complex values. From this, it's clear that [0,1] satisfies the equation. By continuity and monotonicity, (1,1+e) must satisfy for some positive e. The equation reads "1 = 1" on this interval, and so the value of e will be the smallest value for which the equation reads "2 = 1". Since sqrt grows faster, we have sqrt(e) = 2 and so e = 4. Thus, [0,4) satisfies the equation while x = 4 does not. Next, we know that cbrt(8) = 2 and 2 < sqrt(4) < sqrt(8) < sqrt(9) < 3, so x = 8 satisfies the equation. Again, we have (8,8+e) also satisfies, until sqrt(e) = 3, i.e. e = 9. Thus, [8,9) satisfies the equation. Finally, we know that cbrt(27) = 3 and sqrt(27) > sqrt(16) > 4, so sqrt(27) - cbrt(27) > 1, from which we can extrapolate sqrt(x) - cbrt(x) > 1 for all x >= 27, and thus there are no solutions there. On the other hand, since 2 = 3 on that same interval, there is no solution here either. Therefore, the equation is satisfied iff x is in [0,4) U [8,9)

[@dbmalesani]

That's a nice problem and a nice solution. I have a comment though, probably nitpicking. At 2:23, when squaring and cubing the two inequalities, this operation is valid only for x ≥ 1 (otherwise x² ≤ x). @Michael Penn actually says n is positive (at 2:13). But then, the condition n³ ≤ x < (n+1)² is derived assuming x ≥ 1, and it should not be used when checking "case 0" (at 11:41). That said, obviously 0 ≤ x < 1 is a solution as [√x] = [³√x] = 0, so there is no practical change in the solution.

[@2kreskimatmy]

very cool

[@mathscornersomilpitliya4243]

What are some good books for Intermediate Number Theory , Algebra and Combinatorics ????? PLZ REPLY....

[@glenm99]

How "intermediate" are you thinking? Among my mathematically-minded friends, Hungerford's algebra books tend to be universally liked. There's an undergraduate book, usually used in third year courses, and a graduate level text, which can be understood by a bright/motivated undergrad. They have lots of examples and exposition, and that's what I like. It's not just theorem-proof, theorem-proof. You actually get an intuitive feel for the structure of things. A lot of the teaching occurs in the problems, too, which I think is good. Kenneth Bogart's introductory combinatorics book is the one I kept from my schooling. It covers a wide enough range and is slow enough to be suitable for undergrads, but it's got enough advanced topics that you could use it as reference for graduate level study. But if you want to really delve into Ramsey theory or designs or whatever, then you'd need to find something more specialized.

[@yonatanrosmarin4135]

The absolute value case seems to be similar to the original one, so its solution is the reflection of the original solution.

[@nevokrien95]

This could be solved eqsist by graphing the twp functions, even just in yoyr head. They are both monotonicpy increqsing and qre only equal to eachother at one. You quickly get that n €{0,1} now just check by hand the hqndful of cases

[@RatKillCat]

I felt a bit dumb for completely losing track halfway through the first board, but luckily my inflated sense of smarts returned soon after the clean wipe when I spotted the erroneous ≤ sign before he said anything about it.

[@neur303]

Your stare at 4:49 made me laugh 😁❤️

[@jasonzurlo1543]

If I had a nickel for every math KZbin channel with "pen" in the name, I'd have 2 nickels. Which isn't a lot, but it's weird that it happened twice.

[@lukassimanaitis2243]

First time this quick

[@anshumanagrawal346]

3:18 how'd you get n^2

[@RandomBurfness]

How can you ever get the first equation to hold for nontrivial values? If you plug in a negative number in a square root, you get an imaginary number and the floor of an imaginary number is not defined inasmuch as there is no natrual well-ordering of the complex plane.

[@mathissupereasy]

Is n^3 < (n+1)^2???

[@kristianwichmann9996]

Fun one :-)

[@jonaskoelker]

Let y = |x|. Then one of the homework problems is floor(sqrt(y)) = floor(cbrt(y)) but we just solved that: |x| in [0, 4) or in [8, 9) so x is in one of (-9, -8] or (-4, 4) or [8, 9).

[@roberttelarket4934]

Cleverly worded: the elevator takes it down.

[@xevira]

I used the fact that, after a certain point, sqrt(x) grows faster than cqrt(x). Was it rigorous? Probably not, but I got the same results.

[@holomurphy22]

Yes its rigorous. sqrt(16)=4, cqrt(16)16, so will stay >1. Thus the initial equation can't work for x>=16. Then its easy to show what works within that [0,16[ range But as you can see it is not sufficient to only use that sqrt grows faster. For example, on IR+, x grows faster than x+1/x, but you can find floor(x) = floor(x+1/x) for arbitrarily large x (for example on all integers >1).

[@followNoxville]

Why not consider the complex floor function as well floor(1.5 + 2.3i) = 1 + 2i (like how WolframAlpha defines it)?

[@MrRyanroberson1]

because the floor function depends on an objective ordering. there are many ways to define such an extension, but none of them are objective orderings.

[@IAmTheFuhrminator]

Could you show us how you might go about solving the general case of floor(x^(1/r))=floor(x^(1/s))? I have been at it for a couple hours and I can't find a good way to get around the fact that with an unknown r and s, I don't know how to solve for all n s.t. -> n^s

[@michaelcampbell6922]

Try to show that f(t)= (t)^s -(t+1)^r and f '(t)=s*t^(s-1) -r(t+1)^(r-1) are both positive for t>=3

[@GameInOne]

Can anyone try this Find all solution which satisfies the eq 1/a + 1/b + 1/c =1

[@andimandi8491]

As far i know a video whoch solves that problem already exist

[@Tiqerboy]

Are a, b, c real numbers or integers?

[@VaradMahashabde]

Do the map of (x,y,z)→(1/x,1/y,1/z) on the solution of x+y+z=1, which is simply a plane

[@dionisis1917]

It is easy . If a bigger than b bigger than c bigger than 3 the only solution is a=b=c=3 Then i will take c=2 If a bigger than b bigger than 4 only solution is a=b=4 Then i will take c=2 and d=3 then a=6

[@VaradMahashabde]

@@dionisis1917 I suppose you are talking about natural number solutions

[@JMTchongMbami]

12:10 you write the interval [0,1) that means zero included, 1 excluded. Is that the international way of writing that? I learnt to write this interval as [0,1[

[@goodplacetostop2973]

I don’t think there’s an international way for that. Some educational systems around the world use [0,1[, others use [0,1). Same thing with 0 in N, some countries (like France) include 0 in N (and use N* for N without 0), others don’t include 0 in N.

[@tompearson3604]

In the UK it's taught as [0,1). Always assumed that was "the way" if you know what I mean.

[@JoseFernandes-js7ep]

Jean-Moloud Me too. The [0,1) notation was unknown to me. But now I think it is simpler.

[@heh2393]

In India we have [] for included and () for excluded. Also here we use ]0,1[ for { -inf to 0 (incl.)} Union {1(incl.) to inf}

[@MichaelRothwell1]

[0,1[ is to be preferred over [0,1) since if you write (0,1) for ]0,1[ it could be confused with a point in R^2.

[@mroracle464]

12:41 lol

[@juarezmazzucajunior9529]

Maravilha! Mas retifico: até quase 3 e não 4.

[@sugongshow]

For floor(nth root of x) = floor((n+1)st root of x), my conjecture is [0, 2^n) U [2^(n+1), 3^n). Check it out! :)

[@AlbertoSaracco]

In your first generalized question, only 0 is a solution, since -sqrt(x) is negative (or 0) and cubicroot of x is positive (or zero)

[@ianmathwiz7]

The cube root of a negative number is negative.

[@AlbertoSaracco]

@@ianmathwiz7 yeah, but if x is negative, sqrt(x) is not defined/not a real number...

[@oida10000]

Hey for a change up swap some floors with ceils.

[@HagenvonEitzen]

Why so complicated? If n => 3 then (n+1)² = n²+2n+1 < n² + 3n + 9 = n²+n²+n² = 3n²

[@supergamer1972]

You say that n is greater than 3 and then say that n² + 3n + 9 = n²+n²+n² = 3n²

[@RockBrentwood]

You even don't all of that or much of anything at all; even your approach is too complicated. The function f(x) = √x - ³√x is increasing, for all x > 1 since f'(x) = (½√x - ⅓³√x)/x > 0 (because √x > ³√x, when x > 1). So, once you get past f(x) = 1, it's over ... and that's just before 10, since f(10) = 1.00784.... So, the only floors that are relevant are: [0,1): ([√x],[³√x]) = (0,0), [1,4): ([√x],[³√x]) = (1,1), [4,8): ([√x],[³√x]) = (2,1), [8,9): ([√x],[³√x]) = (2,2), [9,10): ([√x],[³√x]) = (3,2), which means the range is [0,4) ∪ [8,9) - the numbers from 0 to 4 and 8 to 9, excluding 4 and 9.

[@muckchorris9745]

While hearing a numerical "ODE" lecture i was scared by the thumbnail and can't sleep for 3 days now. WTF.

[@nullplan01]

Erm... dumb question, but at the "n³

[@aweebthatlovesmath4220]

Math is about proof there's nothing obvious and dumb question btw are math major you doesn't look like because proof is not important for you. Take a look at this "i know there are infinitely many prime number i mean obviously what do you mean prove it?"

[@sugongshow]

I did floor(sqrt(x)) = floor(4th root of x), and the answer is [0, 4).

[@CM63_France]

Hi, For fun: 2 "so on and so forth", including 1 "and so on and so forth", 3 "great", including 2 "ok, great", 1 "now, next what I want to do", 1 "let's go ahead and", 2 "so let's may be go ahead and", including 1 "so let's may be go ahead and do that", 1 "I'll go ahead and".

[@NgocThinhPham-yv1pj]

Can you help me to prove that [x]+[x+1/2]=[2x] ? This is one question of my test ( I am a student from VIETNAM and study in mathematics class.Thanks)

[@MrRyanroberson1]

i would suggest you try looking at x as n + r where n is integer and r is real in [0,1). then look at that equation and apply the rules of floor/round/ceiling {the rule is f(n+r) = n + f(r) for all of those}

[@MegaJohnhammond]

Andrew Wiles would be so proud

[@guidosalescalvano9862]

What is the floor of 1.99999999 (repeating 9 infinitely)

[@polychromaa]

It’s 2 1.9999999….. is equivalent to two

[@guidosalescalvano9862]

@@polychromaa I think I remember a proof of that yeah... but thinking about that... it is very interesting that his implies that you can map any number with a finite number of digits to a number with an infinite number of digits...

[@colemanliam1]

this is very easily solved by graphing but this solution is interesting

[@nombreusering7979]

Am I on time to be here before A Good Place To Stop?

[@neur303]

apparently not 😋

[@wise_math]

Whats "good place to stop"?

[@goodplacetostop2973]

@@wise_math its me :)

[@alimoulaei9962]

n is an integer

[@InDstructR]

Why not just take the simple route? Clearly x≥0 or √x is imaginary. Both sides will round to 0 if 0≤x

[@paulmyrin5028]

It's to rigorously show us that's true; is it true it will always fail for numbers bigger than 3? This is one way to prove that .

[@InDstructR]

@@mr.knight8967 i swear these links are all bugged to track who clicks on them...

[@12wholepizzas13]

Wait why can't we use -1

[@Kokurorokuko]

Small mistake: you wrote and said n is from natural numbers set but 0 is not in it yet it important for the case

[@geonalugala]

Find the floor integers that match this function: √x = ³√x Let √x = n; and ³√x = n. Let n be an integer. (n)² = (n)³ ≥ 0, (since n² is always positive.) This integer is strictly greater than (n-1) and less than (n+1): So, n² < (n+1)² (n-1)² < n² = n³ < (n+1)² ... [ as n² < (n+1)² ] n³ - (n+1)² < 0 n³ - (n² + 2n + 1) < 0 n³ - n² - 2n - 1 (- 1) < (-1) to fully factorize. n²(n - 1) - 2(n - 1) < -1 (n² - 2)(n - 1) < -1. Integer roots: (n - 1) < -1. n ≈ 0 (n² - 2) < -1. n² < 1 n ≈ ±1 Since n = √x = ³√x, The pairs, of (√x, ³√x): (-1, -1), (0, 0) and (1, 1).

[@axelperezmachado3500]

floor gang!

[@mattmatt9324]

1

[@MichaelRothwell1]

I like that 0 is a now natural number, but am surprised at the change of status.

[@MichaelPennMath]

That is mostly me trolling everyone a bit!

[@MichaelRothwell1]

@@MichaelPennMath It seems to be a cultural thing. In England, where I grew up, 0 is considered to be a natural number; in Portugal, where I live, 0 is *not* considered to be a natural number.

[@KiLLJoYYouTube]

Today I learned that negative numbers get bigger with a floor function

[@larswilms8275]

further away from zero is a better term to use here, more negative is also good. Getting bigger depends on your definition. getting bigger on a number line is moving to the right (towards infinity). On a numberline flooring a number moves it to the closets whole number on the left away from positive infinity. if you define "getting bigger" as moving away from zero, without regards to the direction, then yes, flooring a negative number will make it bigger. If you define "getting bigger" as moving towards positive infinity than no, flooring a negative number does not get bigger.

[@matthewsheeran]

First answer is 1. ;-) maybe 0 too

[@iridium8562]

Anyone figured out the generalisations he gave as homework?

[@sugongshow]

Yes, I think so. I posted a comment about one of them.

[@mithutamang3888]

AND THAT IS BAD PLACE TO STOP!!! 😡😡😡😡😡😡😡

[@Gamiboi612]

Cool explanation! Surprised that I actually understood. xd

[@eliasmai6170]

Equations involving floor functions are not trivial exercises

[@johnny_eth]

Just eyeballed the thumbnail and though x in [0,4). Because sqrt and cbrt of 4 are 2 and 1.something. Now watching to see if I'm correct. Edit: eh, missed [8,9)

[@frentz7]

zero is a natural number now?

[@aweebthatlovesmath4220]

Depends on definition you are using I don't consider it and i think he doesn't consider it either

[@frentz7]

@@aweebthatlovesmath4220 see 10:53

[@mohamedcharioui3589]

Why are you taking n to be only positive?

[@benjamingutman9496]

the first assignment makes no sense. from LHS it can be inferred that x>=0

[@f1f1s]

This solution is unnecessarily over-complicated. I wish you continued with the growth rates of functions instead of going through those clunky and lengthy intervals and polynomial equations. It is much more useful, in my opinion, to find the moment when the gap between a slower-growing and a faster growing function becomes 1, and handle the integer values of x separately.

[@arekkrolak6320]

why real numbers? go full pro and solve it for complex :)

[@aweebthatlovesmath4220]

There's no ordering in complex numbers :(

[@mondainicorradi]

i don't know why the youtube send me here, but i don't care.. nice video!

[@naveengantyala940]

First view right...sir

[@onderozenc4470]

Don't fake out terminology. The "solution 😄is obviously is 1^3.

[@aweebthatlovesmath4220]

PrO0f Is ImPORtaNt

[@rateeshk8175]

x= 0,1,2,3 are the obvious solutions A noob math fan will go from basics Like the comment if you agree

[@MrRyanroberson1]

x = 8 you never saw

[@charlesbromberick4247]

i don´t like this kind of math

[@roberttelarket4934]

I'm "floored" by your solution!