A Floor and Ceiling Equation, Floor(x-Ceil(x/2))=3

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FLOOR, CEILING and FRACTIONAL PART EQUATIONS:
#ChallengingMathProblems #FloorAndCeilingEquations

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[marcozarantonello]

Another possible value of n is 3, which gives x=6 as a solution

[leickrobinson]

Here’s a different method that I often use to solve these floor/ceiling problems that you might want to consider exploring in one of your future videos:
First, to make things easier, I substituted x=2y, to get:
floor(2y - ceil(y)) = 3
Now, y can be written as n - r, where n is an integer, and 0 <= r < 1.
Then, ceil(y) = n, and our equation becomes:
floor(n - 2r) = 3.
This immediately decomposes into 3 solutions:
If r=0, n=3 (and y=3, x=6).
If 0 < r <= 0.5, floor(n - 2r) = n-1 = 3, so n=4 (and since y = n-r, {3.5 <= y < 4}, {7 <= x < 8}).
And if 0.5 < r < 1, floor(n - 2r) = n-2 = 3, so n=5 (and {4 <= y < 4.5}, {8 < x < 9}).

S.S.: {x | x=6, 7 <= x < 8, or 8 < x < 9}

[hirokitokuyama]

There is mistake, which made you missed solution "x=6"
ceil(x)=[x]+1 only when {x}≠0, but when {x}=0 ceil(x)=[x]

[MathElite]

Floor function is super interesting and it isn't taught to students that much
Btw @SyberMath, you missed 6 as a solution
*I see your mistake ceil(x/2) = n implies that 2n-2 less than x less than or equal to 2n*

[hsjkdsgd]

So nice to see the active participation of audience in pointing a mistake when they find one. Great moment for the sybermath and of course cheers for the guys who didn't let the X=6 to pass by so easily

[qhrynxx]

n can be 3 and that gives x equal to 6, the problem is in 2n-2<x<2n the upper bound should be less than or equal to.

[stpat]

Error at 2:35. It should be 2n - 2 < x <= 2n.

[tarunmnair]

Nice video...
@2:38 I think it is 2n-2 < x <= 2n.
So taking the first lower bound and second upper bound, n+3 <= 2n which gives 3 <= n.
Thus the values of n are n={3, 4, 5}.
You have done for the other 2 solutions, but When n=3, 6 <= x < 7 & 4 < x <= 6 which when combined gives only 1 solution x = 6.
Thus, solution set of x = { x : (x = 6) or (7 <= x < 9 and x =/= 8) ; x € R }

[snejpu]

I always skipped your excercises with floor and ceiling functions, because I've never solved problems like this. After some previous videos from you, I decide to try something new, so I give it a try. I got a solution: x from 6 to 8 (including 6, but not 8).
Edit: I must've made a mistake somewhere. Maybe better luck next time. : )

[tamarpeer]

If x is an integer:
x-ceil(x/2)=3
1)x-x/2=3, x=6
2)x-(x+1)/2=3, x=7
Otherwise:

ceil(x/2)=floor(x/2)+1
floor(x/2)=t, frac(x/2)=y
x=2t+2y
floor(2t+2y-t)=3
floor(t+2y)=3
1)y<1/2, t=3, 6<=x<7
2)1/2<y<1, t=2, 7<=x<8

[aashsyed]

Nice video 😊😊😊😊 thanks for your channel. Your channel is growing fast!!!!

[mcwulf]

Clumsy error with the inequalities misses n=3 and x=6.

[mriel]

2:35
should be:

2n - 2 < x <= 2n

thus giving the possibility of n=3 => x=6

[זאבגלברד]

Your technic is very nice. If one of us wants to explain it to a kid that started to learn functions, I suggest to use desmos and graph minus ceiling of x/2 , graph 3-x and 4-x and see it very nice.

[mjones]

I reduce my ceiling functions to floor functions because they seem easier to solve for me:
⌈x⌉ = x if x ∈ ℤ,
⌈x⌉ = ⌊x⌋ + 1 if x ∉ ℤ.
By rewriting ⌈x⌉ piecewise like this, you don't miss the solution of x = 6 for this equation.

[manojsurya]

I wrote this problem along with u and found that u made a mistake, instead of <=, u had put only <, so it results in missing of 1 solution (6), but anyways math elite had already pointed it out, and good video 👍

[luiscrispinvargas]

Buenas, al parecer has cometido un error con la definición de máximo entero (piso), si estamos poniendo que el máximo entero de x/2 es n entonces n≤ x/2 < n+1
--> 2n ≤ x < 2n + 2.
Ya desde ahí el ejercicio estará mal resuelto, espero haber ayudado, saludos desde Perú.

[MINEXKILLER]

these inequality systems always get me do you an explanation/video about them? that'll be great

[sayanjitb]

At timestamp 2:55, it can be seen from the two simultaneous inequality that here n can not be a negative integer.
Please comment if this is not the case.

[ramaprasadghosh]

putting x = n + f
Here n is an integer
f is a proper fraction
i e 0<=f < 1
one reaches the following situations
case I a ( f is nonzero)
n =2m
ceiling (x/2)
= ceiling ( (2m+f)/2)
= m +1 ( for nonzero f)
Putting this value of ceiling (x/2) in
floor ( x - ceiling (x/2)) = 3, one gets
m -1 = 3
or floor (x) = 2m = 8 (for nonzero f)
case Ib ( f is zero )
n =2m
ceiling (x/2)
= ceiling ( (2m)/2)
= m
Putting this value of ceiling (x/2) in
floor ( x - ceiling (x/2)) = 3, one gets
m = 3
or floor (x) = 2m = 6
case II
n =2m+1,
Whatever be the value of f,
ceiling ( x/2) = m+1
Putting this value of ceiling (x/2) in
floor(x - ceiling (x/2)) = 3, one gets
m = 3
or floor (x) = 2m +1= 7
Hereby either x= 6
or 7 <= x < 8, 8 < x < 9