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.
@@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".
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!
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.
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.
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
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
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.
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)
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!
@@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
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)
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.
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
@@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
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
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 !
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!
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
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
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
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.
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
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?
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
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 😊
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
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.
Oh my goodness precious ❤ he's so calm and he has mathematics. Dude I'm so much obsessed with his teaching style.... I hope if he will professor of my dream IIT or mit ❤
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
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.
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.
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.
Wow, very nice proof method. Loved the s Work with systems of compound inequalities. Would love to see more problem solving with compound inequalities!! Very cool.
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
@@PrimeNewtons no i ment adding the equations together Adding the x on the top and 3x on the bottom for example, and on the right and left you have +k on the top and -k on the bottom that will cancel out
Yes, but this doesn't fully solve the equation, it gives an interval that's too large. Of course it's still 100% true and valid to say x lies in that interval, it's just not the solution set.
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)
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.
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)
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
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,
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.
No you can't. For one, division by 4 doesn't tend to leave denominators of 3... This is a rather key note. Just because you can simplify something into something that *looks* like an answer doesn't mean you have the answer. If you follow all the laws of logic, you just have a true statement, in this case, that x lies between 1.5 and 2. Which is true, but we can, and must, do better to get the solution set, which spans from 5/3 to 2.
@@fahrenheit2101 Yes, I realised that you have to take into account that k is an integer, especially when solving floor equations, which further restricts the solution set. So, you have to solve for k first before using the allowed integer values of k to solve for x.
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
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
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I took 10 university math courses but never learned floor/ceiling functions. Thanks for enlightening me.
same here,
never stop learning. those who stop learning , stop living
Another name of this function is greatest integer function.
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.
@@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".
That’s the triad from R to Z : floor, cuisine and round :)
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!
The quality of his voice also helps. It's soothing but motivational!
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.
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.
He is far beyond a basic math tutor
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
He helps me not only learn of new functions, but also helps me fall asleep at night. Keep it up!
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
What an amazing teacher.
This channel has become my favorite now ❤🎉
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
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.
I like your teaching style, am slowly incorporating it into my online classes, so well explained always, NEVER STOP LEARNING....
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
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)
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!
The basics that you teach in your videos are the pillars of India's most difficult exam like JEE Advance.
Are you a IITIAN?? Or a aspirant?
the question is does it really matter? he never said this video is intended for a jee aspirant
@@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
You're my new favorite math channel. Your voice is so calming! Thank you for the video :)
Wow. six views and six likes. Everyone liked! (not surprised, love your vids)
Love your punch line “Never stop learning, those stop learning stop living”❤
Your explanation in the introduction was good enough for me to quickly solve it in the Pure Data graphing/programming environment. Great video!
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)
This was my approach as well. I named my variables n and u, but that hardly matters.
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.
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
Dang, I made some faulty assumptions when I attempted this the first time. Great video as always!
Let me guess, floor(3x) + floor(x) = floor(4x) ?
@@jumpman8282 nope, i wanted to bring the 3 to the outsideof the floor, like one would with |3x|
@@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
Yeah, I havea bad habit of not doing that. It occured to me after I should, but Ill probably do it again one day
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
Your explain is perfect!!! Congratulation!
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 !
That's new to me. Interesting to discover 🤔
In Germany we learn both ways. It is up to the teacher, which one you are supposed to use...
we also have this notation [[3; 7]] for integer interval :)
I've seen the Portuguese use outward-facing parentheses
Poles use something like >
in norway we would write [5/3,2> big > though
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!
Your style is very relaxed. I didn't think mathematics could be taught this way
THNX❤
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
You're my new favourite math RUclipsr
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
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.
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
Nice systematic approach. Not at all what I went for, but definitely one I'll bear in mind.
Thanks for considering my request and teaching another question of this type....
Let floor(x)=k, so that k
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.
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
TIL editing a hearted comment removes the heart. Whoops.
That was good reasoning.
For these equations I always write
x = n + a
where
n integer
0
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?
You are a great teacher! 👏
You’re my favorite math youtuber :).
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
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 😊
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
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.
The floor of x?!
New math 😮 how very exciting.
Oh my goodness precious ❤ he's so calm and he has mathematics. Dude I'm so much obsessed with his teaching style.... I hope if he will professor of my dream IIT or mit ❤
Your videos are really useful for me hats off brother 😌✨ (i'm in my 12th grade gonna face my exams soon🎉)
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
3x+x>6 if x>=2
3x+x
You are a grate teacher❤❤❤
Muchas gracias por el vídeo, acabo de aprender una nueva forma de ver el ejercicio, saludos desde Perú.
This is pretty cool, never seen a floor equation before
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.
that smile always gets me xDD
thank you so much for the simplification
great work
Glad you like it!
@@PrimeNewtons I always appreciate the good work
Thanks a lot for this video and all the others!
If you go on with them then soon we will all be math professors. 😊
❤️🙏
Beautiful handwriting!
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.
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.
@@PrimeNewtons Thank you for your reply, I sent you an Email under the caption Math Problem.
Please check again. I didn't get your email.
@@PrimeNewtons I sent it again
I love this one since it is not even taught when I was in school
Wow, very nice proof method. Loved the s
Work with systems of compound inequalities. Would love to see more problem solving with compound inequalities!! Very cool.
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
when it comes to the equation formation is that supposed to be 3t instead of letting it be t since tis representing 3x
Well done and explained.
I ain't never heard of these, you really make some interesting videos.
nice, the problem solved beautifully and as well as u
I really like how you teach!!
Great teaching!
but i noticed that in 4:58 you could add the sections in both equations to get rid of k instently, and you are only left with x
You'd have to add k to 3x too
@@PrimeNewtons no i ment adding the equations together
Adding the x on the top and 3x on the bottom for example, and on the right and left you have +k on the top and -k on the bottom that will cancel out
Yes, but this doesn't fully solve the equation, it gives an interval that's too large.
Of course it's still 100% true and valid to say x lies in that interval, it's just not the solution set.
Thank you for showing the way on how to go about
Отличный пример!
i love you man, thank you so much for what you are doing you really helping.
Very interesting problem. I have never had to work with math on rounding functions like floor or ceiling, but this is a good intro.
I like your personality
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)
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.
as someone who paused and managed to solve it looking at it for less than 15 seconds, i feel very accomplished
This is so cool
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)
Yup same thinking
I love your channel, makes me bored in my normal maths lessons though haha
The Bob Ross of math right here.
It is a part of Greatest integer function
Keep going
Great video! Very well explained
Excelente. Excelent. 🇧🇷🇧🇷🇧🇷🇧🇷
i love this guy
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
There's a way easier way to solve:
Try 1
3+1=4
x>1
Try 2
6+2=8
x
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,
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.
Can you solve the problem celling (x) over floor (x) + 2x =8? It is possible? Thank you Prof.
Please write the problem on paper and send me an email of the picture so I can better interpret what you wrote.
Yes, thank you. I will do it in the next few days. Sorry for my English.
Instead of solving for the k, you can just eliminate the k by adding the 2 inequalities together and then dividing throughout by 4 to solve for x.
No you can't. For one, division by 4 doesn't tend to leave denominators of 3...
This is a rather key note. Just because you can simplify something into something that *looks* like an answer doesn't mean you have the answer. If you follow all the laws of logic, you just have a true statement, in this case, that x lies between 1.5 and 2. Which is true, but we can, and must, do better to get the solution set, which spans from 5/3 to 2.
@@fahrenheit2101 Yes, I realised that you have to take into account that k is an integer, especially when solving floor equations, which further restricts the solution set. So, you have to solve for k first before using the allowed integer values of k to solve for x.
fantastic video
Came for the floor equation, stayed for the asmr chalk on board sounds
Bob ross of math
I would never think that you can solve these kinds of problems
You're great!
Thanks so much . ❤❤
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
Hush! A lil bit of maths still makes me happy😄
at 4:30 why can you not simply add the inequalities to get 6 ≤ 4x < 8 => 3/2 ≤ x < 2 ?
I noticed while editing.
I’m sorry my guy, I cannot fokus on the math. That chalk is way to smooth to not admire
Nice of you.
how is it possible that both floor(3x)+floor(x)=6 and 5/3
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?
Add K to both sides.
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
Good job
at 4:59 just add the inequalities doesn't work
6
Or 1.5, you could've just tested 1.5.