Great video! I had another method where I said floor(x) =m therefore floor(x^2) = m^2 +n for non negative integers m,n. It follows n =0 or 1 and we can solve for x in each case. It was great to see another way of thinking from you! :)
@@angelmendez-rivera351 if you plug it back into the original equation you get m^2 + n - 4m +3 = 0. This gives (m - 2)^2 + n-1=0. Or (m-2)^2=1-n. Because LHS is a perfect square, RHS>=0 giving n=0 or 1 as only integer solutions.
@@renatsamur9099 Yes, but that follows from floor(x) = m AND the given equation, not floor(x) = m alone. This needs to be clarified, because otherwise, your comment gives the impression that n = 0, 1 is a property of floor functions, especially to those who are not the most familiar with the subject.
@@angelmendez-rivera351 I was giving a brief explanation of how I solved the question. If anyone needed clarification they could have asked. I’ll be sure to mind that next time though. :)
@@SyberMath Wait, if the floor function always gives an integer right..the nearest integer less than the x value..Why can't you plug in 3 and 1 since those are answers are integers..then x can be 3. some decimal or 1. some decimal then the floor of x squared could still be 3 squared or 1 squared as long as the decimal part of x isnt so high that squared gives you the next integer up like 10 or 2..
@@leif1075 x=1 and x=3 definitely work but for greater numbers there is an upper bound. For example if x is greater than sqrt(2), then it will not work!
Thank you for showing that. If I want to explain it to a 9th grade good student, I will draw a nice y=x^2+3 graph, and y=4x graph. Show that they intersect at 1 and 3. Make lines parellal to x axis , length 1, from (0,0) , (1,4) , (2,8) , (3,12) , (4,16) , (5,20). Lets start with the line from (1,4). The parabulla changes to a line from x=1 to x=2^0.5 . So these lines are "together". Next two lines that are "together" are from x=5^0.5 to 6^0.5 ... you got the idea. At x=15^0.5 and on, they do not overlap any more.
i wish the question was (floorx)^2-4floorx+3 so we would factorize it like (floorx-3)(floorx-1)=0 that would have made the answer x (0,1),(3,4) but this intersting aswell and harder also
When you have found that x is between n and n + 1 and x^2 is between n^2 and (n+1)^2, you just have to use substitutation : floor value of x is u so the floor value of x^2 is u^2. Now, you just have to use discriminante : delta = - 4^2 - 4*3 = 4 so u1 = 1 and u2 = 3. Of course, x needs to be (or not to be 😂) between 1 and 2 or between 3 and 4. No?
@@tonyhaddad1394 Well, actually, there is a better, simpler explanation as to why it's always positive: n^2 - 2·n + 4 = n^2 - 2·n + 1 + 3 = (n - 1)^2 + 3 > 0, and since (n - 1)^2 >= 0 is trivially true, (n - 1)^2 + 3 >= 3 is always true, for any n. But 3 > 0 is trivial, so there is no information to be gained from this.
i am upvoting every time i post a computer solution, like now: *𝚁𝚎𝚍𝚞𝚌𝚎[𝙵𝚕𝚘𝚘𝚛[𝚡^𝟸] - 𝟺 𝙵𝚕𝚘𝚘𝚛[𝚡] + 𝟹 == 0, 𝚡, 𝚁𝚎𝚊𝚕𝚜]* this way i know when my post was deleted when i revisit the video thanks! :P btw please doht feel annoyed when i post computer solution. i actually post the working input only. it is interesting to see how we can formulate the problem in code. i doht post the output. hope you doht mind!
Ashish, I announced in yesterday's video that I will be doing fewer geometry puzzles and more algebra, number theory, and trigonometry problems. I will focus more on the quality of the puzzles!
@@SyberMath ohh I see. I was in a bit hurry yesterday, so I have left the vedio just after the solution so I didn't notice it. Fine! 😁😁😁 Thankyou for your response. 😄😄😄
Conjecture:
floor(x**a) = floor(x)**a
for values of x,a where
0
Great video! I had another method where I said floor(x) =m therefore floor(x^2) = m^2 +n for non negative integers m,n. It follows n =0 or 1 and we can solve for x in each case. It was great to see another way of thinking from you! :)
I don't think it follows that n = 0, 1. What does follow is that 0 =< n =< 2·m
@@angelmendez-rivera351 if you plug it back into the original equation you get m^2 + n - 4m +3 = 0.
This gives (m - 2)^2 + n-1=0.
Or (m-2)^2=1-n. Because LHS is a perfect square, RHS>=0 giving n=0 or 1 as only integer solutions.
@@renatsamur9099 Yes, but that follows from floor(x) = m AND the given equation, not floor(x) = m alone. This needs to be clarified, because otherwise, your comment gives the impression that n = 0, 1 is a property of floor functions, especially to those who are not the most familiar with the subject.
@@angelmendez-rivera351 I was giving a brief explanation of how I solved the question. If anyone needed clarification they could have asked. I’ll be sure to mind that next time though. :)
Thanks for sharing!
Very good and satisfying solution as well as the problem 😁
Thank you!
@@SyberMath 3 IS ALSO A SOLUTION!!!!
Interesting ceiling/floor problems. Btw sqrt(2) at the end can be avoided, we have (n-2)^2
Nice!
@@SyberMath also I wanted to tell you that 2-sqrt(2)~0.5858 not 0.6
We can also use floor (x^2)=x^2-{x^2}; floor(x)=x-{x}. We have then (x^2-4x+3)+(4{x}-{x^2})=0. But (4{x}-{x^2} )is defined on interval
Amazing, just loved it, ❤️☺️☺️☺️☺️,thank you very much
My pleasure 😊
@@SyberMath Wait, if the floor function always gives an integer right..the nearest integer less than the x value..Why can't you plug in 3 and 1 since those are answers are integers..then x can be 3. some decimal or 1. some decimal then the floor of x squared could still be 3 squared or 1 squared as long as the decimal part of x isnt so high that squared gives you the next integer up like 10 or 2..
@@leif1075 x=1 and x=3 definitely work but for greater numbers there is an upper bound. For example if x is greater than sqrt(2), then it will not work!
Marvelous solution, thank you very much sir.
You are most welcome
Excelente video, me salió a la primera 🥹, un saludo desde Perú.
Thank you for your beautiful Floor Eqn
Np. Thank you for watching
Thank u nice problem !!
You're welcome!
Thank you! Please shoot a video about gcd properties
I will take a look at it! Sounds like a good idea
@@SyberMath Thank you!
@@SyberMath Yes, please..
Thank you for showing that. If I want to explain it to a 9th grade good student, I will draw a nice y=x^2+3 graph, and y=4x graph. Show that they intersect at 1 and 3. Make lines parellal to x axis , length 1, from (0,0) , (1,4) , (2,8) , (3,12) , (4,16) , (5,20). Lets start with the line from (1,4). The parabulla changes to a line from x=1 to x=2^0.5 . So these lines are "together". Next two lines that are "together" are from x=5^0.5 to 6^0.5 ... you got the idea. At x=15^0.5 and on, they do not overlap any more.
Nice!
The best way to solve this is on the floor not on the roof!!
😁
i wish the question was (floorx)^2-4floorx+3 so we would factorize it like (floorx-3)(floorx-1)=0 that would have made the answer x (0,1),(3,4) but this intersting aswell and harder also
When you have found that x is between n and n + 1 and x^2 is between n^2 and (n+1)^2, you just have to use substitutation : floor value of x is u so the floor value of x^2 is u^2. Now, you just have to use discriminante : delta = - 4^2 - 4*3 = 4 so u1 = 1 and u2 = 3. Of course, x needs to be (or not to be 😂) between 1 and 2 or between 3 and 4. No?
Nice video!!!
Thanks for the visit!
we know frac(x) can be up to 1, so x^2 can be up to (round(x)+1)^2, and at least round(x)^2
round(x)^2-4round(x)+3
8:49 beacaus this a continous function If its not then we have to proof why its always positive
If im wrong pleaz tel me beacaus im not sure !!!!!!!!
It's monotonically non-decreasing, and it's only discontinuous at countably many points.
@@angelmendez-rivera351 thank u!!!
@@tonyhaddad1394 Well, actually, there is a better, simpler explanation as to why it's always positive: n^2 - 2·n + 4 = n^2 - 2·n + 1 + 3 = (n - 1)^2 + 3 > 0, and since (n - 1)^2 >= 0 is trivially true, (n - 1)^2 + 3 >= 3 is always true, for any n. But 3 > 0 is trivial, so there is no information to be gained from this.
@@angelmendez-rivera351 yes bro thank u that a better proof 👍👍👍
Great discussion! Thanks to you both!
i am upvoting every time i post a computer solution, like now:
*𝚁𝚎𝚍𝚞𝚌𝚎[𝙵𝚕𝚘𝚘𝚛[𝚡^𝟸] - 𝟺 𝙵𝚕𝚘𝚘𝚛[𝚡] + 𝟹 == 0, 𝚡, 𝚁𝚎𝚊𝚕𝚜]*
this way i know when my post was deleted when i revisit the video thanks! :P
btw please doht feel annoyed when i post computer solution. i actually post the working input only. it is interesting to see how we can formulate the problem in code. i doht post the output.
hope you doht mind!
x^2-4x+3 = {x^2}-4{x} < 1 so
(x-2)^2 < 2 and after a little algebra
0
waiting.and eager!!!
Liked it?
Good video
Thanks
Yeap! I solved the problem 💖☺💖
Thanks a Zettabyte, Boss!! 🌿🌿🌿
Np. Nice work!
Thank you. Very good video.
Please can you look at this one
|X²|- |X|²=100,
No solutions because |X²| = |X|²
Square is always positive so square can also be taken outside mod just solve the quadratic u will get 4 solutions ±3,±1
good!
Thanks!
Best chnl,,,
Thanks, Mahesh!
Wow this one was tougher than expected actually:
\x^2/-4\x/+3=0
A: x=n+f with n integer and f in R, 0>=f
(x-3)(x-1)
👍
Today was the turn of geometry. Isn't it? 😟😟😟 Well. Today's equation was also interesting. Thankyou BTW.🙃🙃🙃
Ashish, I announced in yesterday's video that I will be doing fewer geometry puzzles and more algebra, number theory, and trigonometry problems. I will focus more on the quality of the puzzles!
@@SyberMath ohh I see. I was in a bit hurry yesterday, so I have left the vedio just after the solution so I didn't notice it. Fine! 😁😁😁
Thankyou for your response. 😄😄😄
x=3
Fuvest 2024
👍
(x-3)(x-1)