In this video , I showed how solve a simple floor equation
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@JSSTyger7 ай бұрын
I took 10 university math courses but never learned floor/ceiling functions. Thanks for enlightening me.
@mohammedel-gamal34557 ай бұрын
same here, never stop learning. those who stop learning , stop living
@Mustapha.Math_at_KUSTWUDIL7 ай бұрын
Another name of this function is greatest integer function.
@thelightningwave7 ай бұрын
This my surprise you, but you were already were using it, when ever people talk about their age they always give an integer, and that integer is the floor function of their actual age.
@m.h.64707 ай бұрын
@@thelightningwave not really, it is more like the rounding function. If you are less than a month away from your birthday, you typically don't say your "old" age, you say "I'm going to be X next month".
@KannaKamui210007 ай бұрын
That’s the triad from R to Z : floor, cuisine and round :)
@rinaldo.garcia6 ай бұрын
this is so calming and soothing in some way... the clacking of the chalk on the blackboard, him speaking so quietly and calmly... I love these videos, great work!
@QFredfons6 ай бұрын
The quality of his voice also helps. It's soothing but motivational!
@boristhedestroyerofancient5087 ай бұрын
Man, you really have the most relaxing style of teaching maths ever 😅. I like the hyper-focused youtubers as well but your videos make me not want to click off with how peaceful they are.
@corneliusagu29037 ай бұрын
I have got to the peak of my studies having has a PhD. in Process and energy engineering. On my love for mathematics, which controls my daily activities at work, I often spend time looking at tutors online for basic mathematics. I have been following your videos, and at every point in time and I have developed different interests in your teaching approach beside that I learn new things like this today which I have not come across. Well done, brother, and I urge you to continue at this pace.
@valiant89877 ай бұрын
He is far beyond a basic math tutor
@RoyBoy_i3c7 ай бұрын
He helps me not only learn of new functions, but also helps me fall asleep at night. Keep it up!
@danobro7 ай бұрын
Wow. six views and six likes. Everyone liked! (not surprised, love your vids)
@harshit75106 ай бұрын
There's also a very good intuitive way of solving it By hit and trial I put x=1 3+1 = 4 6 So by analysing this 1
@seedmole6 ай бұрын
Your explanation in the introduction was good enough for me to quickly solve it in the Pure Data graphing/programming environment. Great video!
@akultechz23427 ай бұрын
Being a jee advanced aspirant this looked easy tho. Taking x = I + F 3I + [3F] + I = 6 Now 4I + [3F] = 6 Here, I must be >0 So I = 1 satisfies only. Now we have [3F] = 2 Let 3F = t Then t belongs to [2,3) So F belongs to [⅔,1) Hence x belongs to [5/3,2)
@Mothuzad7 ай бұрын
This was my approach as well. I named my variables n and u, but that hardly matters.
@melikmourali20726 ай бұрын
The thing missing in your proof is that you have to prove that x is in [1,2). How I did it is that I plugged x=1 and x=2 and used the fact that the floor value function is increasing and that is easy to prove as well, if you have to. (It was implied in your reasoning but you have to show it.) Prime Newtons' proof is cleaner though I must admit.
@magefreak93567 ай бұрын
Great video as always! It would have been cool to have seen graphs of both the floor(3x) and floor(x) and see where the sum of these adds up to 6. Never stop learning
@fahrenheit21016 ай бұрын
Think about it: floor(x) is integer valued, and steps up by 1 at each next integer, starting with floor(0) = 0 floor(3x) is similar, but steps up by 1 at each 1/3, starting again at floor(3*0) = 0 So the function on the LHS starts at f(0) = 0, but steps up by 1 at every integer, and also by 1 at every 1/3 : the combined effect is a step of 1 at ever 1/3, except every 3/3 = 1, i.e. every integer step, for which it jumps up by two. Long story short, it starts at 0, and takes jumps of size 1,1,2,1,1,2,1,1,2... etc., at every 1/3. That tells you what it looks like and if you cut off the steps after 1+1+2+1+1 = 6, then you've taken 5 steps of size 1/3 so you're at 5/3. So we need the bit of the function starting at 5/3 and going right up until 6/3 = 2, but not including that endpoint. Hopefully that made some sense.
@ReallyAmateurPianist6 ай бұрын
Nice video! My solution for this was: Left hand side is 4 with x=1, and 8 with x=2. Therefore we know that 1
@fahrenheit21016 ай бұрын
Yep, what I did too. Another idea I had which is a little less formal is to just literally picture the function. The interesting stuff happens every 1/3, and it's just an increment by 1 each time, except at the integer values of x where you get an increment of 2 (I'm sure you can see why). Tracking that, you see that you get 6 after 5 steps, hence 5/3 up to but not including 6/3 = 2. This also has the benefit of telling you that certain values are unattainable by this function (specifically anything congruent to 3 mod 4)
@duckyoutube63187 ай бұрын
The meaning of floor is less than or equal to. Thank you Mr.Prime. Im not in college but when i get there.... ima pass calc 3 on day 1.
@MichaelAdjei-up2ce7 ай бұрын
What an amazing teacher. This channel has become my favorite now ❤🎉
@ounaogot7 ай бұрын
I like your teaching style, am slowly incorporating it into my online classes, so well explained always, NEVER STOP LEARNING....
@Niki1A_6 ай бұрын
I got to the solution much quicker looking at the thumbnail by observing that it is a (non-strictly) monotonely rising function, so you just have to find the minimum and supremum of it working (minimum and supremum because the floor function includes the lower bound, but excludes the upper bound). There can't be any negatives involved because both terms would be negative then. Also 2 is clearly at least an upper bound. Also both floor terms must be integers, due to the floor function. Therefore, you only have to consider how to split up the 6 into two non negative integers: 6+0 doesn't work because for floor(3x) = 6, you would need x >= 2, but then floor(x)>=2>0. 5+1 does work with floor(3x)=5 giving us x>=5/3 as a lower bound and floor(x)=1 giving us x
@sfenelyeah7 ай бұрын
as someone who paused and managed to solve it looking at it for less than 15 seconds, i feel very accomplished
@duckyoutube63187 ай бұрын
The floor of x?! New math 😮 how very exciting.
@KannaKamui210007 ай бұрын
12:25 in France we use different notations for that, it's great to know other notations ! We don't write [5/3, 2) in France, but [5/3; 2[ with bracket facing outwards for excluding value :) btw I dropped a subscribe here !
@PrimeNewtons7 ай бұрын
That's new to me. Interesting to discover 🤔
@m.h.64707 ай бұрын
In Germany we learn both ways. It is up to the teacher, which one you are supposed to use...
@KannaKamui210007 ай бұрын
we also have this notation [[3; 7]] for integer interval :)
@ДмитрийОсипов-м9д7 ай бұрын
I've seen the Portuguese use outward-facing parentheses Poles use something like >
@Pandorarl6 ай бұрын
in norway we would write [5/3,2> big > though
@janda12587 ай бұрын
I’m sorry my guy, I cannot fokus on the math. That chalk is way to smooth to not admire
@PrimeNewtons7 ай бұрын
Nice of you.
@JourneyThroughMath7 ай бұрын
Dang, I made some faulty assumptions when I attempted this the first time. Great video as always!
@jumpman82827 ай бұрын
Let me guess, floor(3x) + floor(x) = floor(4x) ?
@JourneyThroughMath7 ай бұрын
@@jumpman8282 nope, i wanted to bring the 3 to the outsideof the floor, like one would with |3x|
@DatBoi_TheGudBIAS7 ай бұрын
@@JourneyThroughMathye, I had the same idea, but then I tried it with some values, and saw it wasn't true. Specifically 0.4 floor(3*0.4)=floor(1.2)=1 3floor(0.4)=0
@JourneyThroughMath7 ай бұрын
Yeah, I havea bad habit of not doing that. It occured to me after I should, but Ill probably do it again one day
@Dominus_Potatus7 ай бұрын
I love this one since it is not even taught when I was in school
@MaxPicAxe6 ай бұрын
You're my new favourite math KZbinr
@Mr.Rinfinite6 ай бұрын
An interesting thing I thought of, this means the equation: floor(3x) + floor(x) = 7 Actually has no solutions! Because if x=2, our RHS turns to 8, but anything up to 2 (1.99999 for example) would still have our RHS being 6. You can get the equation's RHS to equal 4 (x=1) and also 5 in a range of [4/3 , 5/3), as well as 6 and 8, as you showed, but never 7. Kind of interesting!
@shadrackshija28486 ай бұрын
Thank you for showing the way on how to go about
@Dalroc6 ай бұрын
3x+x>6 if x>=2 3x+x
@ReyazulislamReayal7 ай бұрын
You are a grate teacher❤❤❤
@IITianNikhil7 ай бұрын
The basics that you teach in your videos are the pillars of India's most difficult exam like JEE Advance.
@SakshamMahajanB7 ай бұрын
Are you a IITIAN?? Or a aspirant?
@ts9dream7 ай бұрын
the question is does it really matter? he never said this video is intended for a jee aspirant
@SakshamMahajanB7 ай бұрын
@@ts9dream why are you concerned for that I just want to ask him whether he is a IITIAN/nitian or aspirant if he were a IITIAN I would ask my question . Who the hell are you concerned with that
@pizza87257 ай бұрын
I calculated in my mind in like 10 seconds that x would be at least 5÷3 and smaller than 6
@DatBoi_TheGudBIAS7 ай бұрын
6÷3, but yes
@aadityavikram50307 ай бұрын
Thanks for considering my request and teaching another question of this type....
@TheRealZeaga6 ай бұрын
You're my new favorite math channel. Your voice is so calming! Thank you for the video :)
@tjipondauheua4534 ай бұрын
when it comes to the equation formation is that supposed to be 3t instead of letting it be t since tis representing 3x
@jennymarx92287 ай бұрын
Your videos are really useful for me hats off brother 😌✨ (i'm in my 12th grade gonna face my exams soon🎉)
@florianbasier6 ай бұрын
you took the long scenic road my friend. We know that no matter what x is, there is a unique (n, k, y) triplet from Nx{0,1,2}x[0,1/3[ so that x=n+k/3+y. Floor(3x)=Floor(3n+k+3y). 3n+k is a natural number and 0
@PrimeNewtons6 ай бұрын
I've seen comments saying something similar. I have to digest it and see how I can tell viewers to consistently look for such pattern in all similar problems. Thanks 😊
@estebansalgado47087 ай бұрын
I took a slightly different approach. I wrote x as the sum of its integer part k and its fractional part f. Then the equation becomes 4k + floor(3f) = 6 Given that 0
@fahrenheit21016 ай бұрын
Nice systematic approach. Not at all what I went for, but definitely one I'll bear in mind.
@slavinojunepri7648Ай бұрын
Well done and explained.
@edvardm43486 ай бұрын
Subscribed. Who could not appreciate the style of teaching here? I'm huge fan of more "traditional" math videos like Mathologer and 3b1b, but following this guy is like being taken to an adventure, or detective story. Just love this
@lesheq856 ай бұрын
i love how happy you are explaining this. even tho i was familiar with floor and ceiling from programming i have never solved equality containing them. thanks for that!
@vitotozzi19727 ай бұрын
Your explain is perfect!!! Congratulation!
@freya287337 ай бұрын
it's a very rigorous proof but imo way too overkill, is this done on purpose to teach how to work with more complicated floor functions or is this how you would solve it if you just wanted an awnser? no hate btw just curious
@east98766 ай бұрын
As a programmer I've grown quite familiar with using floor and ceiling functions. But despite years of math and computer science classes I've never come across such a simple yet in depth explanation of the mathametical properties behind them. Excellent video! Looking forward to checking out what else your channel has to offer
@gdmathguy7 ай бұрын
There's a way easier way to solve: Try 1 3+1=4 x>1 Try 2 6+2=8 x
@jasonryan25457 ай бұрын
Welcome to another video ... Lets find x!
@hydropage28556 ай бұрын
Bob ross of math
@a.lollipop6 ай бұрын
Never dealt with an equation including a floor yet, so I tried solving myself before watching. I did the following: first i noticed there will definetely be a range of values, so I should try to find a maximum and a minimum value that satisfies the equation so i know the range. i started by assuming x is an integer (i tried writing that the fancy way but there is no symbol for the set of integer numbers or for 'exists' on mobile :c), and realized 1 is too small and 2 is too big, therefore 1
@a.lollipop6 ай бұрын
TIL editing a hearted comment removes the heart. Whoops.
@PrimeNewtons6 ай бұрын
That was good reasoning.
@migssdz72876 ай бұрын
This question becomes realy easy if you realise that floor(x) can only be 1. If it's 2 or more, floor(3x) would be at least 6 and the sum would be at least 8. If it's 0 or less, floor(3x) would be at most 3 and so would the entire sum. After realising that, the question is just floor(3x) = 5 -> 5 ≤ 3x < 6
@m-yday6 ай бұрын
I really like how you teach!!
@bogusawsroda37477 ай бұрын
I like your personality
@omendrakumar34166 ай бұрын
It is a part of Greatest integer function
@nimaalz45137 ай бұрын
How to solve floor(2x)+floor(3x)+floor(5x)=14 I know the general way for all kind of these questions Dm me for explanations
@Misha-g3b5 ай бұрын
The answer is [1.5, 1.6).
@nimaalz45135 ай бұрын
@@Misha-g3b how
@jacobgoldman57807 ай бұрын
Before watching the video, x clearly cannot be an integer as that would mean 4x=6 or x=1.5 which is not an integer therefore a contradiction. Since the coefficient of one of the terms is 3 we should set x=n+(r/3) with 0
@LimeLogan26 күн бұрын
I did this a different way by proving x is not an integer, rewriting x as an integer plus a decimal, finding 3 cases for k in which only one is an actual integer and finding the lowest value for the decimal Here's the full thing: floor(3x)+floor(x)=6. Assume x is an integer: 3x+x=6 4x=6 x=6/4=3/2=1/5 Therefore x is not an integer. Let x=k+d where k=an integer, d=the decimal value floor(3(k+d))+floor(k+d)=6 floor(3k+3d)=k=6 3d can, at most, be 3 (because of the case d=0.99999...) 3k+1=6 OR 3k+2=6 OR 3k+3=6 3k=6-1 OR 3k=6-2 OR 3k=6-3 3k=5 OR 3k=4 OR 3k=3 k=5/3 OR k=4/3 OR k=3/3 k=5/3 OR k=4/3 OR k=1 Since k is an integer, k=1. Since d≠0.99... since if d=0.99..., k=2. Let's find the lowest value for d: floor(3k+3d)+k=6 floor(3+3d)+1=6 3+floor(3d)=6-1 floor(3d)=5-3 floor(3d)=2 3d≥2 d≥2/3 1+2/3≤x
@TheFinagle6 ай бұрын
And this video reminded me why I didn't like grade school math, even though I do actually love math. I figured this one out in my head in a few seconds (kinda using a limit like approach) BUT in grade school teachers expect to show ALL this work to get full marks even if you didn't need it.
@andreaparma72013 ай бұрын
Let floor(x)=k, so that k
@AmeNoTenshi3 ай бұрын
that smile always gets me xDD thank you so much for the simplification great work
@PrimeNewtons3 ай бұрын
Glad you like it!
@AmeNoTenshi3 ай бұрын
@@PrimeNewtons I always appreciate the good work
@沈博智-x5y6 ай бұрын
The way I did it. If x was an integer 3x + x = 6 => 4x = 6 => x = 6/4 => x = 3/2 but 3/2 is not an integer so x can't be an integer. 3/2 = 1.5 So as a guess, check x \in [1,2) since we are getting x = 1.5 from 'integer contradiction' Let x = 1 + epsilon where epsilon \in (0,1) This means 3x = 3 + 3epsilon floor(3x) = floor(3 + 3epsilon) Notice 3epsilon >= 2 iff epsilon >= 2/3 Also, epsilon has been defined to be strictly less than 1. so 2/3 = 5 Also epsilon < 1 iff 3epsilon < 3 iff 3 + 3epsilon < 6 so 5 6 in this case so x \in [5/3, 2) is indeed the only interval of x values that works.
@cparks10000007 ай бұрын
I'd just break the problem into cases. Every real number x can be written in the form x=k + r where r is between 0 and 1 (possibly equal to zero). We then break into 3 cases: r= 2/3, or neither. Case 1 is immediately ruled out because this implies that 4k=6. Similarly, case 3 implies 4k=5. This leaves case 2 as being the only possible. In case 2, we have k=1, so x is in [5/3, 2). Alternatively, you can use the fact that floor(x+k) = floor(x) + k for any integer. Thus floor(3x) + floor(x) = floor(3k + 3r) + k = 3k + floor(3r) + k = 4k + floor(3r). Using the fact that r is in [0,1), we know that floor(3r) can only be 0, 1, or 2. One then concludes that floor(3r) must be 2 and proceeded to calculate x.
@fahrenheit21016 ай бұрын
Haven't watched it, but here's my quick answer: We can exclude solutions greater or equal to 2 for obvious reasons: floor is an increasing function, and floor(3*2) + floor(2) = 8 so for x = 2 and beyond, the LHS is at least 8 Similar reasoning excludes solutions less than or equal to x = 1. This guarantees that floor(x) = 1, and we only need floor(3x) = 5, which gives the inequality 5
@Pramit11567 ай бұрын
This one was pretty easy
@crazypantaloons6 ай бұрын
This problem has a trivial solution. I think the video solution glosses over the best property of floor math: the floor of an integer is itself. Find the integers, and remove them. Step 1: on quick inspection, x must be between 1 and 2, so let x = 1 + y (where 0 < y < 1), then floor (3x) + floor (x) = 6 reduces to 3 + floor (3y) + 1 + floor(y) = 6, or floor (3y) + floor (y) = 2. Step 2: we purposely picked y such that floor (y) is 0, so floor (3y) = 2, and 2/3
@pedroguilherme8686 ай бұрын
What i found before watching: The values need to be between one and two X=1 results in 3+1=4 X=2 results in 6+2=8 The floor of anything bwtween 1 and 2 is 1 So the floor of x is always 1 (Floor of 3x) +1=6 The floor of 3x needs to be 5 The lowest value is when 3x=5, or x=5/3, and the upper limit is when x approaches 2, since 3x would then approach but not equal 6, and thus the floor would still be 5 Thus, x is bigger than or equal to 5/3 and smaller than 2
@kragiharp7 ай бұрын
Thanks a lot for this video and all the others! If you go on with them then soon we will all be math professors. 😊 ❤️🙏
@sida89727 ай бұрын
how is it possible that both floor(3x)+floor(x)=6 and 5/3
@BorisNVM6 ай бұрын
For these equations I always write x = n + a where n integer 0
@GayAnnabeth6 ай бұрын
before I actually watch this, I'm gonna drop my idea for a solution (might not be the only solution, idk) for the thumbnail: 5/3
@أبوشاهين-ت6ك6 ай бұрын
Your style is very relaxed. I didn't think mathematics could be taught this way THNX❤
@m.h.64707 ай бұрын
Solution: x has to be 1 < x < 2, otherwise the left side is too small or too big. therefore floor(x) = 1. this means, that floor(3x) = 5. if x < 5/3, floor(3x) < 5. therefore 5/3 ≤ x < 2.
@luiscrispinvargas30613 ай бұрын
Muchas gracias por el vídeo, acabo de aprender una nueva forma de ver el ejercicio, saludos desde Perú.
@aquss337 ай бұрын
I ain't never heard of these, you really make some interesting videos.
@Icephoenix03445 ай бұрын
Hi, I saw a nice equation from the national math olympics from germany: What x satisfys floor(x/2) * floor(x/3) * floor(x/4) = x^2? I didnt ever see a floor equation where you have a division in the floor operater, so I would really enjoy a video about it.
@PrimeNewtons5 ай бұрын
Interesting! I already have 3 solutions, but I need to see the original problem to know what type and how many solutions I need. Please email to me. Primenewtons@gmail.com. Thank you.
@Icephoenix03445 ай бұрын
@@PrimeNewtons Thank you for your reply, I sent you an Email under the caption Math Problem.
@PrimeNewtons5 ай бұрын
Please check again. I didn't get your email.
@Icephoenix03445 ай бұрын
@@PrimeNewtons I sent it again
@VinodKumar-zn4fd27 күн бұрын
Love your punch line “Never stop learning, those stop learning stop living”❤
@張茗茗-y9i7 ай бұрын
5/3
@PixlatedGalaxy7 ай бұрын
Yeah its correct
@paulfoss53857 ай бұрын
x∈[5/3,2) if you want to be silly.
@oclati83137 ай бұрын
Or we can take each value of both and look when the sum equals 6 ( sry for probable english mistake french here) 👍 but ur technique seems better anyways
@OLEGEK237 ай бұрын
Отличный пример!
@janmesh23326 ай бұрын
This is so cool
@spikedskull1376 ай бұрын
I would never think that you can solve these kinds of problems
@KannaKamui210007 ай бұрын
7:00 why k < 1.75 and k positive can't mean k=0 ?
@complainer4066 ай бұрын
My guess from the thumbnail: 5/3
@marcusorban24396 ай бұрын
I solved it in a more intuitive way: 1. Set x=1: the result is too small, set x=2 and the result is too big. That means the solution is something along the line of 1,... 2. That means that floor(x) will always be 1 and floor(3x) must equal 5 Therefore, our lower bound for the solution is 5/3, the upper bound is 6/3=2 => x=[5/3, 2)
@harshit75106 ай бұрын
Yup same thinking
@apteropith6 ай бұрын
seeing this thumbnail before bed and giving it a shot before watching through ... so, when x = 1 we get 4, but when x = 2 we get 7, and can't get lower than 7 from here on so, we know floor(x) = 1, therefore the value will have to be any value between 1 and 2 where floor(3x) = 5 the lower limit of this range should be when x = 5/3, such that 3x = 5 5/3 is less than 2, so we have no contradiction, and floor(3x) doesn't increase until x = 6/3 = 2 so this equation is true for any x in the range [5/3, 2), from 5/3 (inclusive) to 2 (exclusive) and now potentially being soothed asleep listening, it turns out
@PrimeNewtons6 ай бұрын
Sleep well!
@mayankvardani98493 ай бұрын
You should have added the initial two inequalities to make this solution shorter
@shreyesveer80737 ай бұрын
My dumbass did the calculation like this |_x_| = k; |_3x_| = 3k 3k + k = 6 4k = 6 k = 1.5 |_1.5_| = 1 therefore, x = 1
@PrimeNewtons7 ай бұрын
No longer a dumbass
@aryahan49767 ай бұрын
nice, the problem solved beautifully and as well as u
@varno6 ай бұрын
There is a simpler way to see the first part here. The function is monotonic increasing. Further, the function is based on the floor value, so it only changes when either 3x is an integer or x is an integer, which coincide on the fractions k/3, being the same on half-open intervals. Further, on these critical values this function is no greater than the equality of the linearised version of the function I.e. 4x=6, which can be solved as x=1.5/2 thus we only have to test the critical points and above 1.5, these are, 5/3 and 6/3, as the functions are equal on the integers. As the 5/3 value equals 6 and the 6/3 value is greater then this function must be equal to 6 over the half open interval [5/3,6/3=2)
@fahrenheit21016 ай бұрын
You lost me when you said "linear function 4x = 6" which isn't true. That's an equation, only the LHS is a function. You might say "you know what I mean" but... I actually don't. You're probably right in what you're saying, I'd just appreciate some precision. The function in the question is no greater than the function 4x. It doesn't make sense to be no greater than 4x = 6... But you can conclude that since the function is at most 4x, and 4x = 6 at x = 1.5, we can rule out x < 1.5 on those grounds. Which is probably what you meant, but not even close to what you said.
@sheikhAbdelrahman6 ай бұрын
The last domain had an error x € [ 5/3, 2 [ .. not [ 5/3, 2 )
@davidseed29396 ай бұрын
at 4:59 just add the inequalities doesn't work 6
@fahrenheit21016 ай бұрын
Or 1.5, you could've just tested 1.5.
@KipIngram4 ай бұрын
By inspection, this will be 5/3
@kacperkrul6 ай бұрын
This is pretty cool, never seen a floor equation before
@editvega8036 ай бұрын
You are a great teacher! 👏
@redmask60857 ай бұрын
So before watching the video I tried solving it myself and got the same solution, but I used a diffrent method (also I'm going to use E as 'is contained in'): first I declered x=k+d, k E Z, d E [0,1) => floor(x)=k and after substitiusion I got: floor(3(k+d))+k=6 floor(3k+3d)+k=6 now k E Z => 3k E Z => floor(3k+a)=3k+floor(a) as the only thing that the floor changes is the >=0,
@rssl55007 ай бұрын
I’m in 11th grade and we have these problems very frequently especially in competition math problems but still we solve this differently Notice : [kx]=[x]+[x+1/k]+…+[x+k-1/k] where k is an integer Hence : [x]+[x]+[x+1/3]+[x+2/3]=6 notice if x = [x]+ p then we can partition this into the following cases : 1: 0
@remsalt38827 ай бұрын
never got this in school, how did 3k turn into 4k? where did the +1 come from if it wasn't in the original equation?
@BlueSapphyre7 ай бұрын
Add K to both sides.
@rutgerdeboom74244 ай бұрын
You’re my favorite math youtuber :).
@danielc.martin7 ай бұрын
Ez but cool!
@1ne2woPandemonium7 ай бұрын
Hush! A lil bit of maths still makes me happy😄
@duckyoutube63187 ай бұрын
I see i missed the last 4 videos. Wth youtube?!
@ronaldjensen29487 ай бұрын
at 4:30 why can you not simply add the inequalities to get 6 ≤ 4x < 8 => 3/2 ≤ x < 2 ?
@PrimeNewtons7 ай бұрын
I noticed while editing.
@antonionavarro10007 ай бұрын
Magnífico tutorial. Vídeos muy didácticos, claros, sencillos y que van al núcleo del problema. Buen trabajo con el uso de la función 'floor'. La función 'ceiling' debe funcionar de manera similar. ¿Algún vídeo previsto sobre ella?
@Refleksifdusunur6 ай бұрын
isn’t 3/2 and -3/2 the solution?
@omendrakumar34166 ай бұрын
Come back to India and prepare for iit Jee because this is like to cut a butter cake for jee advance aspirants.
@PrimeNewtons6 ай бұрын
I am not from India
@HenrikMyrhaug6 ай бұрын
Very interesting problem. I have never had to work with math on rounding functions like floor or ceiling, but this is a good intro.