In this video, I showed how to solve a ceiling over floor equation. Note that x is not an integer since floor(integer) = ceiling(integer) but x is then 3.5
I enjoyed this video very much, all of those assumptions you didn't mention in the previous video, I had to convince myself that they were true and that you were allowed to do all of those operations, good on you for making it clearer and coming back to it in this one!
I thought there was a glitch in the matrix 😅. I love these ceiling and floor function videos, because I have never learned these function in school or university. I'm learning how these types of problems are solved by watching your videos😊.
I have used ceiling() and floor() functions in programming languages to "round up or down", but I have never learned the mathematics notation for them. This video was fun for learning math syntax.
I really like how you re-did your video to explain why you did specific actions in the previous one. It shows how much you care about your viewers. I myself am studying to become a math teacher and always like it when people try to explain why they do certain actions when solving math problems. Keep up the work! ❤
Hi :) it’s my first time on your channel and I think I might be staying; the problem was fun to solve on my own, it wasn’t pointlessly trivial, and the explanation was great! One small thing I noticed: if x isn’t an integer, x is strictly not equal to floor(x), so the inequality in the video shouldn’t be k
This was the second time to watch this video. I never heard of ceiling/floor functions before, so this was an entirely new concept for me. It looked like a really good application of logic and algebra, but it sure was a long process!
7:16 Sorry for pointing out the mistake but since you already said x is not an integer, you don't need to say k ≤ x (i.e. k is less than or equal to x). k is simply less than x (k < x)! No chance it is equal to x.
5:35 I don't understand "You gonna get 2 positive numbers" but you assumed x to be negative therfore 2x must be negative as well and the quantity you labeled as "tve" could be bigger than 8. But what your result is actually correct because ceiling(x)/floor(x)
lol I just asked that question lol! I'm not sure how he justified it in the video here specifically, but I guess you can justify that the fraction is less than 8 since it's (k+1)/k = (1 + 1/k), and since -1
5:24 You floored ceil(x) / floor(x) by 0. But you also should cap it in order to have 8 - ceil(x) / floor(x) positive. Indeed, you could say that ceil(x) / floor(x) is between 0 and 2 so that 8 - ceil(x) / floor(x) > 6 so that's always positive.
My strategy was to study the range of the ceiling(x)/floor(x) ratio. For x < 0 -> c/f is between (0, 1) For x > 1 -> c/f is between (1, 2) Since x ≠ Z, x is not btw (0, 1) therefore the range of c/f is between (0, 2) except 1. Therefore he range of 2x = 8 - c/f is between (6, 8) except 7, and the range of x = 4 -0.5 c/f Is between (3, 4) except 3.5. If x is between 3 and 4, then c/f = 4/3. 2x = 8 - 4/3 = 20/3 -> x = 10/3.
I started in a similar way but also noted that x could not be greater that 4 or we’d overshoot the 8. Since only 3 cases were left ( 1..2, 2..3, and 3..4) I just tried them all to show that there was only one solution because it felt doing the quadratic equation would take more time.
That's how I did it, too. No reason to go through two quadratic formulas when there are only three integers that k can be, and once you find the right one, you can use it to solve for x.
Así lo resolví yo: * Como fllor(x) distinto de cero, entonces x no está en [0,1). NO EXISTEN SOLUCIONES EN EL INTERVALO [0, 1) * Si x es entero entonces ceil(x) / fllor(x) = x/x = 1 y sustituyendo en la ecuación original 1 + 2x = 8 x = 7/2 pero 7/2 entra en conflicto con la suposición de que x es entero y por lo tanto, 7/2 no es solución válida. En consecuencia. NO EXISTEN SOLUCIONES ENTERAS. * Si -1 < x < 0 entonces ceil(x) = 0 y floor(x) = -1. Por tanto ceil(x) / floor(x) + 2x = 8 0/(-1) + 2x = 8 x = 4 Pero como ya dedujimos, no existen soluciones enteras y por tanto x=4 no es una solución válida. De todo esto se deduce que NO EXISTEN SOLUCIONES EN EL INTERVALO (-1, 0) * Dado que no existen soluciones en el intervalo cerrado [-1, 1] y tampoco soluciones enteras, x debe ser un número fraccionario (no entero) mayor que 1 ó menor que -1. podemos aplicar la propiedad ceil(x) = floor(x) + 1 ceil(x) / floor(x) = ( floor(x) + 1 ) / floor(x) = 1 + 1 / floor(x) * Caso x< -1. Estudiamos 1 / floor(x) Si x tiende a menos infinito entonces floor(x) tiende a menos infinito y en consecuencia 1 / floor(x) tiende a cero con valores negativos. Si x tiende a -1 por la izquierda entonces floor(x) = -2 y por tanto 1 / floor(x) = -1/2 Con estos límites obtenemos que -1/2 < 1 / floor(x) < 0 Sumando 1 a toda la inecuación 1/2 < 1 + 1/floor(x) < 1 Pero como ceil(x) / floor(x) = 1 + 1/floor(x) pues x no es entero 1/2 < ceil(x) / floor(x) < 1 Pero despejando de la ecuación original ceil(x) / floor(x) = 8 - 2x y sustituyendo en la anterior expresión, 1/2 < 8 - 2x < 1 3.50 < x < 3.75 Pero que x deba pertenecer a ese intervalo entra en contradicción con la suposición de que x < -1. Es decir, no existen soluciones tales que x< -1. Además sabemos que tampoco existen soluciones entre -1 y 0 ni soluciones enteras. En consecuencia, NO EXISTEN SOLUCIONES NEGATIVAS. * Caso x > 1. El último caso que nos queda por comprobar. De existir soluciones están deben ser positivas y mayores que uno. Dado que x es un número fraccionario (no entero) y como ya vimos anteriormente ceil(x) / floor(x) = 1 + 1 / floor(x) Si nos enfocamos en acotar 1 / floor(x) mediante límites, tenemos; Si x tiende a 1 por la derecha entonces floor(x) = 1 y por tanto 1 / floor(x) tiende a 1 por la izquierda.. Si x tiende a +infinito entonces floor(x) tiende a +infinito y por tanto 1 / floor(x) tiende a cero con valores positivos. En definitiva 0 < 1 / floor(x) < 1 Sumando 1 a toda la desigualdad 1 < 1+1 / floor(x) < 2 1 < ceil(x) / floor(x) < 2 Pero como ceil(x) / floor(x) = 8 - 2x según la ecuación original, 1 < 8 - 2x < 2 3 < x < 3.5 que no contradice x > 1 y podemos continuar. Si x pertenece a ese rango entonces ceil(x) = 4 y floor(x) = 3 Sustituyendo en la ecuación original para estos valores de x ceil(x) / floor(x) + 2x = 8 4/ 3 + 2x = 8 multiplicando por 3 toda la ecuación 4 + 6x = 24 despejando x x = 10/3 posible solución. Test: ceil( 10/3 ) / floor( 10/3 ) + 2 • 10/3 = 8 4 /3 + 20/3 = 8 24/3 = 8 8 = 8 ok.
I had my own way to do it, although the beginning was basically the same. checking when ceiling(x) = floor(x) when x is a non-zero integer leads to a contradiction, so our soln cant be an integer. can't divide by 0 either. ceiling(x) = floor(x) + 1 when x is a real number that is not an integer. using that in the original equation we get (floor(x) + 1)/floor(x) + 2x = 8 floor(x)/floor(x) + 1/floor(x) + 2x = 8 1 + 1/floor(x) + 2x = 8 1/floor(x) + 2x = 7 Now we can focus on this new equation: 1/floor(x) + 2x = 7 we know x = floor(x) + {x} where 0 < {x} < 1 since x cannot be an integer. Substituting this in, we get 1/floor(x) + 2(floor(x) + {x}) = 7 1/floor(x) + 2floor(x) + 2{x} = 7 since floor(x) is an integer -1
same idea but less notation problem by using "a" and "b" excuse rigor, avoided negative solutions as it is trivial it wouldn't work as 2x>ceiling(x)/floor(x) always so no need to check negative sols. Solution : write x=a+b where "a" is the integer part of "x" and "b" is the decimal part of "x" ceiling of "x"is then gonna be "a+1" and floor of "x" is always gonna be "a" 2x will become 2a+2b we now get that (a+1)/a +2a +2b=8 simplify the fraction, you get 1+ 1/a + 2a +2b=8 so 1/a + 2a +2b=7 we can clearly see that "a" cannot be bigger than 4 and therefore has to be 3 or lower ( and we know "a" is an integer) we can also clearly see that 2b is always gonna be smaller than 2, so 1/a +2a has to be bigger than 5, meaning "a" has to be 3 as if "a" was equal to 2, 1/a +2a wouldn't reach 5 therefore, "a" can only be equal to 3, put back in the equation and you get that 1/3 + 6 + 2b=7 and from that you get that "b"=1/3 and so x=3+(1/3)=10/3 (3.3333333...)
5:34 how do u know 2x is positive? I'm not sure how you justified it in the video here specifically, but I guess you can justify that the fraction is less than 8 since it's (k+1)/k = (1 + 1/k), and since -1
write x=a+b where "a" is the integer part of "x" and "b" is the decimal part of "x" ceiling of "x"is then gonna be "a+1" and floor of "x" is always gonna be "a" 2x will become 2a+2b we now get that (a+1)/a +2a +2b=8 simplify the fraction, you get 1+ 1/a + 2a +2b=8 so 1/a + 2a +2b=7 we can clearly see that "a" cannot be bigger than 4 and therefore has to be 3 or lower ( and we know "a" is an integer) we can also clearly see that 2b is always gonna be smaller than 2, so 1/a +2a has to be bigger than 5, meaning "a" has to be 3 as if "a" was equal to 2, 1/a +2a wouldn't reach 5 therefore, "a" can only be equal to 3, put back in the equation and you get that 1/3 + 6 + 2b=7 and from that you get that "b"=1/3 and so x=3+(1/3)=10/3 (3.3333333...)
Can you please make videos on Additively Indecomposable Ordinals, Multiplicatively Indecomposable Ordinals, Exponentially Indecomposable Ordinal, Tetrationally Indecomposable Ordinals, and Pentationally Indecomposable Ordinals?
That is more math than I have done. You're referring to Cantor and the gamma numbers. Haha. Not my forte. Maybe in the future after I've learned it on my own.
Hhey if x be an integer then you said that it's ceiling and floor are same but by equation K ≤ x < K + 1 if x = k as it is integer then it's ceiling should be x +1 and we know that x ≠ x +1 until x = 0 in all cases. Please explain how a ceiling of a number is equal to its floor.
You could think of the ceiling and floor functions as “rounding up to the next integer” and “rounding down to the next integer” respectively. e.g. ceil(4.5)=5, floor(4.5)=4 However when the input is already an integer, there is no point to rounding, hence both functions return the same integer, namely the input. Hope that helps! ^-^
You didn't prove that ceil(x) / floor(x) < 8 if x < 0. You just stated it. Without that you can't know for sure that x >= 0. Luckily it is true since ceil(x)
@@PrimeNewtonsI agree with @nerdatmath's comment. You never stated that it is a "proper" fraction. It might be obvious but you only insisted that it is positive which is not relevant at all for what you want to prove here.
1. Flor(x) must not be eqal 0 2. If x is integer Flor (x)= Ceiling(x) then equation has no solution so x is NOT integer. 3. As x is not integer, sa Ceiling(x)= Flor(x)+1 and we have (1+Flor(x))/Flor(x) +2x=8 1/Flor(x) +1 +2x=8 1/Flor(x) +2x=7 1/Flor(x) =7-2x 4 As x in not equal 1 1/Flor(x)x therefore 3
The explanation of why x is not negative was lacking. You really need an upper bound to the fraction to explain that. The largest this fraction can be when x is negative is 1, which then implies that x is positive, a contradiction.
I think you need to upload a new video for this,for idiots like me,need to be explained in depth for every decision. I dont get the part of K and itsi inequality.
I enjoyed this video very much, all of those assumptions you didn't mention in the previous video, I had to convince myself that they were true and that you were allowed to do all of those operations, good on you for making it clearer and coming back to it in this one!
I thought there was a glitch in the matrix 😅. I love these ceiling and floor function videos, because I have never learned these function in school or university. I'm learning how these types of problems are solved by watching your videos😊.
Me, neither. These are good practice for limit problems, which I loved.
I have used ceiling() and floor() functions in programming languages to "round up or down", but I have never learned the mathematics notation for them. This video was fun for learning math syntax.
I really like how you re-did your video to explain why you did specific actions in the previous one. It shows how much you care about your viewers.
I myself am studying to become a math teacher and always like it when people try to explain why they do certain actions when solving math problems.
Keep up the work! ❤
Hi :) it’s my first time on your channel and I think I might be staying; the problem was fun to solve on my own, it wasn’t pointlessly trivial, and the explanation was great!
One small thing I noticed: if x isn’t an integer, x is strictly not equal to floor(x), so the inequality in the video shouldn’t be k
Welcome!! Thanks for the feedback!
I almost had a deja vu moment.
😅
@@PrimeNewtons where is the older video? I saw that though
Your commitment to us is unrivaled.
Thank you
man, this one was crazy to look at in thumbnail. Olympiad-level stuff.
Wow, was this hard. I need to do an intensive review of these concepts.
To check: (4/3) + 2(10/3) = 24/3 = 8.
This was the second time to watch this video. I never heard of ceiling/floor functions before, so this was an entirely new concept for me. It looked like a really good application of logic and algebra, but it sure was a long process!
7:16 Sorry for pointing out the mistake but since you already said x is not an integer, you don't need to say k ≤ x (i.e. k is less than or equal to x). k is simply less than x (k < x)! No chance it is equal to x.
No he is right ,think again
I wish i can like this video more that two times😂, because 10/10 minus nothing. Thank so much for this.
Glad you liked it!
Excellent video, excellent logical analysis
~6:35 what about if 8 - R < 0? wouldn't that allow i didn't understand.
the largest R can be is 2 (if -2 < k < -1)
@@Parsa-je1vdthank you.
@@Parsa-je1vd I think this part didn't get explained and certainly should have.
You did so good till the last sentence: 4/3 + 2 * 4/3 IS NOT 8... because it should be 4/3 + 2 * 10/3
😅 I already signed out at that time.
5:35 I don't understand "You gonna get 2 positive numbers" but you assumed x to be negative therfore 2x must be negative as well and the quantity you labeled as "tve" could be bigger than 8. But what your result is actually correct because ceiling(x)/floor(x)
lol I just asked that question lol! I'm not sure how he justified it in the video here specifically, but I guess you can justify that the fraction is less than 8 since it's (k+1)/k = (1 + 1/k), and since -1
5:24 You floored ceil(x) / floor(x) by 0. But you also should cap it in order to have 8 - ceil(x) / floor(x) positive. Indeed, you could say that ceil(x) / floor(x) is between 0 and 2 so that 8 - ceil(x) / floor(x) > 6 so that's always positive.
If I pause at 5:49 can I say that the ceiling over the floor can never exceed 1 if x is negative? I presume that is why "positive+2x = 8" can't work.
The biggest that can be is 1.1
@@PrimeNewtonsHow can it be 1.1?
My strategy was to study the range of the ceiling(x)/floor(x) ratio.
For x < 0 -> c/f is between (0, 1)
For x > 1 -> c/f is between (1, 2)
Since x ≠ Z, x is not btw (0, 1)
therefore the range of c/f is between (0, 2) except 1.
Therefore he range of
2x = 8 - c/f is between (6, 8) except 7,
and the range of x = 4 -0.5 c/f
Is between (3, 4) except 3.5.
If x is between 3 and 4, then c/f = 4/3.
2x = 8 - 4/3 = 20/3 -> x = 10/3.
I found it easier to solve for d and go from there.
Set x = k + d, where k = floor(x) and 0
Smooth!
I did similarly, but after getting (k+1)/k + 2(k+d) =8 just expanded left 1 + 1/k + 2k + 2d = 8 or 2 k + 1/k + 2d = 7 , from there can see k
I started in a similar way but also noted that x could not be greater that 4 or we’d overshoot the 8. Since only 3 cases were left ( 1..2, 2..3, and 3..4) I just tried them all to show that there was only one solution because it felt doing the quadratic equation would take more time.
That's how I did it, too. No reason to go through two quadratic formulas when there are only three integers that k can be, and once you find the right one, you can use it to solve for x.
Así lo resolví yo:
* Como fllor(x) distinto de cero, entonces x no está en [0,1). NO EXISTEN SOLUCIONES EN EL INTERVALO [0, 1)
* Si x es entero entonces
ceil(x) / fllor(x) = x/x = 1
y sustituyendo en la ecuación original
1 + 2x = 8
x = 7/2
pero 7/2 entra en conflicto con la suposición de que x es entero y por lo tanto, 7/2 no es solución válida. En consecuencia. NO EXISTEN SOLUCIONES ENTERAS.
* Si -1 < x < 0 entonces ceil(x) = 0 y floor(x) = -1. Por tanto
ceil(x) / floor(x) + 2x = 8
0/(-1) + 2x = 8
x = 4
Pero como ya dedujimos, no existen soluciones enteras y por tanto x=4 no es una solución válida.
De todo esto se deduce que NO EXISTEN SOLUCIONES EN EL INTERVALO (-1, 0)
* Dado que no existen soluciones en el intervalo cerrado [-1, 1] y tampoco soluciones enteras, x debe ser un número fraccionario (no entero) mayor que 1 ó menor que -1. podemos aplicar la propiedad
ceil(x) = floor(x) + 1
ceil(x) / floor(x) = ( floor(x) + 1 ) / floor(x) = 1 + 1 / floor(x)
* Caso x< -1. Estudiamos 1 / floor(x)
Si x tiende a menos infinito entonces floor(x) tiende a menos infinito y en consecuencia 1 / floor(x) tiende a cero con valores negativos.
Si x tiende a -1 por la izquierda entonces floor(x) = -2 y por tanto 1 / floor(x) = -1/2
Con estos límites obtenemos que
-1/2 < 1 / floor(x) < 0
Sumando 1 a toda la inecuación
1/2 < 1 + 1/floor(x) < 1
Pero como ceil(x) / floor(x) = 1 + 1/floor(x) pues x no es entero
1/2 < ceil(x) / floor(x) < 1
Pero despejando de la ecuación original
ceil(x) / floor(x) = 8 - 2x y sustituyendo en la anterior expresión,
1/2 < 8 - 2x < 1
3.50 < x < 3.75
Pero que x deba pertenecer a ese intervalo entra en contradicción con la suposición de que x < -1. Es decir, no existen soluciones tales que x< -1. Además sabemos que tampoco existen soluciones entre -1 y 0 ni soluciones enteras. En consecuencia, NO EXISTEN SOLUCIONES NEGATIVAS.
* Caso x > 1. El último caso que nos queda por comprobar. De existir soluciones están deben ser positivas y mayores que uno. Dado que x es un número fraccionario (no entero) y como ya vimos anteriormente
ceil(x) / floor(x) = 1 + 1 / floor(x)
Si nos enfocamos en acotar 1 / floor(x) mediante límites, tenemos;
Si x tiende a 1 por la derecha entonces floor(x) = 1 y por tanto 1 / floor(x) tiende a 1 por la izquierda..
Si x tiende a +infinito entonces floor(x) tiende a +infinito y por tanto 1 / floor(x) tiende a cero con valores positivos.
En definitiva
0 < 1 / floor(x) < 1
Sumando 1 a toda la desigualdad
1 < 1+1 / floor(x) < 2
1 < ceil(x) / floor(x) < 2
Pero como ceil(x) / floor(x) = 8 - 2x según la ecuación original,
1 < 8 - 2x < 2
3 < x < 3.5
que no contradice x > 1 y podemos continuar.
Si x pertenece a ese rango entonces ceil(x) = 4 y floor(x) = 3
Sustituyendo en la ecuación original para estos valores de x
ceil(x) / floor(x) + 2x = 8
4/ 3 + 2x = 8
multiplicando por 3 toda la ecuación
4 + 6x = 24
despejando x
x = 10/3
posible solución.
Test:
ceil( 10/3 ) / floor( 10/3 ) + 2 • 10/3 = 8
4 /3 + 20/3 = 8
24/3 = 8
8 = 8
ok.
I had my own way to do it, although the beginning was basically the same.
checking when ceiling(x) = floor(x) when x is a non-zero integer leads to a contradiction, so our soln cant be an integer.
can't divide by 0 either.
ceiling(x) = floor(x) + 1 when x is a real number that is not an integer.
using that in the original equation we get
(floor(x) + 1)/floor(x) + 2x = 8
floor(x)/floor(x) + 1/floor(x) + 2x = 8
1 + 1/floor(x) + 2x = 8
1/floor(x) + 2x = 7
Now we can focus on this new equation:
1/floor(x) + 2x = 7
we know x = floor(x) + {x} where 0 < {x} < 1 since x cannot be an integer.
Substituting this in, we get
1/floor(x) + 2(floor(x) + {x}) = 7
1/floor(x) + 2floor(x) + 2{x} = 7
since floor(x) is an integer
-1
same idea but less notation problem by using "a" and "b"
excuse rigor, avoided negative solutions as it is trivial it wouldn't work as 2x>ceiling(x)/floor(x) always so no need to check negative sols.
Solution :
write x=a+b where "a" is the integer part of "x" and "b" is the decimal part of "x"
ceiling of "x"is then gonna be "a+1" and floor of "x" is always gonna be "a"
2x will become 2a+2b
we now get that (a+1)/a +2a +2b=8
simplify the fraction, you get 1+ 1/a + 2a +2b=8
so 1/a + 2a +2b=7
we can clearly see that "a" cannot be bigger than 4 and therefore has to be 3 or lower ( and we know "a" is an integer)
we can also clearly see that 2b is always gonna be smaller than 2, so 1/a +2a has to be bigger than 5, meaning "a" has to be 3 as if "a" was equal to 2, 1/a +2a wouldn't reach 5
therefore, "a" can only be equal to 3, put back in the equation and you get that 1/3 + 6 + 2b=7
and from that you get that "b"=1/3
and so x=3+(1/3)=10/3 (3.3333333...)
Thank you,sir
5:34 how do u know 2x is positive? I'm not sure how you justified it in the video here specifically, but I guess you can justify that the fraction is less than 8 since it's (k+1)/k = (1 + 1/k), and since -1
Since you're assuming x isn't an integer, you should have k < x
What is that symbol
How it works
write x=a+b where "a" is the integer part of "x" and "b" is the decimal part of "x"
ceiling of "x"is then gonna be "a+1" and floor of "x" is always gonna be "a"
2x will become 2a+2b
we now get that (a+1)/a +2a +2b=8
simplify the fraction, you get 1+ 1/a + 2a +2b=8
so 1/a + 2a +2b=7
we can clearly see that "a" cannot be bigger than 4 and therefore has to be 3 or lower ( and we know "a" is an integer)
we can also clearly see that 2b is always gonna be smaller than 2, so 1/a +2a has to be bigger than 5, meaning "a" has to be 3 as if "a" was equal to 2, 1/a +2a wouldn't reach 5
therefore, "a" can only be equal to 3, put back in the equation and you get that 1/3 + 6 + 2b=7
and from that you get that "b"=1/3
and so x=3+(1/3)=10/3 (3.3333333...)
Can you please make videos on Additively Indecomposable Ordinals, Multiplicatively Indecomposable Ordinals, Exponentially Indecomposable Ordinal, Tetrationally Indecomposable Ordinals, and Pentationally Indecomposable Ordinals?
That is more math than I have done. You're referring to Cantor and the gamma numbers. Haha. Not my forte. Maybe in the future after I've learned it on my own.
@@PrimeNewtons Thank you. I would be glad if you need me to help you on those.
Hhey if x be an integer then you said that it's ceiling and floor are same but by equation
K ≤ x < K + 1 if x = k as it is integer then it's ceiling should be x +1 and we know that
x ≠ x +1 until x = 0 in all cases. Please explain how a ceiling of a number is equal to its floor.
You could think of the ceiling and floor functions as “rounding up to the next integer” and “rounding down to the next integer” respectively.
e.g. ceil(4.5)=5, floor(4.5)=4
However when the input is already an integer, there is no point to rounding, hence both functions return the same integer, namely the input.
Hope that helps! ^-^
By brute force, we know that 3 < x < 4
∴ x = 3.b = 3 + b/10
the equation becomes:
4/3 + 6 + b/5 = 8
b = 3.3333..
∴ x = 3.33333…
You didn't prove that ceil(x) / floor(x) < 8 if x < 0. You just stated it. Without that you can't know for sure that x >= 0. Luckily it is true since ceil(x)
If x
@@PrimeNewtonsI agree with @nerdatmath's comment. You never stated that it is a "proper" fraction. It might be obvious but you only insisted that it is positive which is not relevant at all for what you want to prove here.
❤❤❤❤
nice
Now i feel guilty you had to do this video. I was one of the guy asking why you excluded the integers numbers
Oh no. I needed to redo it. Thanks for the feedback! It was necessary.
Woah i just solved it by trial and error method... i am proud of myself btw
1. Flor(x) must not be eqal 0
2. If x is integer Flor (x)= Ceiling(x) then equation has no solution
so x is NOT integer.
3. As x is not integer, sa Ceiling(x)= Flor(x)+1 and we have
(1+Flor(x))/Flor(x) +2x=8
1/Flor(x) +1 +2x=8
1/Flor(x) +2x=7
1/Flor(x) =7-2x
4 As x in not equal 1
1/Flor(x)x
therefore 3
oof mind kinda blown 🤯
The explanation of why x is not negative was lacking. You really need an upper bound to the fraction to explain that. The largest this fraction can be when x is negative is 1, which then implies that x is positive, a contradiction.
I think you need to upload a new video for this,for idiots like me,need to be explained in depth for every decision. I dont get the part of K and itsi inequality.
There are two formulas. If x in an integer. There is no inequality. Floor(x) = x = ceilin(x). If not sure, k
It is not obvious that R is less than 8, that requires a bit of explanation.
2.5
x=10/3
x=3½-1/2[x]=3(½-⅙)=3⅓
matematiklar asosan jiddiy bo'ladi.
siz ham ozroq jiddiy bo'ling.
That, I can not be.
3.5
But k is x itself . in K+1 / k