Quadratic ceiling floor inequality

In this video, I soleved an inequality of the floor of a quadratic ceiling function. Yeah, it sounds confusing.

[@happyhippo4664]

Chem Eng here. Took dozens of math classes in high school and college. Old man now. Never heard of floor and ceiling functions. Never stop learning, indeed.

[@user-gf4eg4rm5r]

This explanation is easy to understand❗️👏👏

[@andreasjoseph144]

Thank you sir,atleast now i understand better then before,🎉🎉 Be blessed sir

[@WhiteGandalfs]

Interesting to see how the curve plot translates into pure formula handling. While the curve plot is very simple and easy (and secure) to understand, the translation into formulas provides for automation with programming.

[@JourneyThroughMath]

When I solved, i dropped the floor first and then added 1. Is that okay, or will that not be true in all cases? It feels like it would be true.

[@glgou4647]

clean new pfp!

[@surendrakverma555]

Thanks Sir 🙏

[@JourneyThroughMath]

I worked the solution to get the ceiling of x is between -1 and 2. But honestly Im not sure if more is necessary

[@JourneyThroughMath]

I see what I did wrong

[@tura361]

The easiest question solved by a 15 years old child in Turkey. Geography is destiny.

[@suryamgangwal8315]

Wont k be equal to -2,0,2 ?becasue we taking the ceiling of 1.732

[@m.h.6470]

No, we are talking about the ceiling of x, not the ceiling of √3.

[@sadeqirfan5582]

Please take out that dramatic part after you say “let’s get into the video” that comes with a music. It is excruciating.

[@m.h.6470]

Solution: ⌈x⌉² - 1 is already an integer, therefore the floor function can be ignored: -1 ≤ ⌈x⌉² - 1 ≤ 2 |+1 0 ≤ ⌈x⌉² ≤ 3 x is therefore a real number. The lower bound becomes redundant: ⌈x⌉² ≤ 3 Replacing ⌈x⌉ with (k + 1), with k being integer (k + 1)² ≤ 3 |√ |k + 1| ≤ √3 Substitute k + 1 with t case t ≥ 0: 0 ≤ t ≤ √3 only t = 0 and t = 1 satisfy this inequality, as t is an integer case t < 0: 0 < -t ≤ √3 |*-1 0 > t ≥ -√3 only t = -1 satisfy this inequality, as t is an integer t ∈ {-1, 0, 1} as t = k + 1, we can say that k ∈ {-2, -1, 0} Since x = k + d, with 0 < d ≤ 1, we can say that -2 < x ≤ 1 Side note: It is 0 < d ≤ 1 and not 0 ≤ d < 1, because we used ⌈x⌉ = k + 1.

[@comdo777]

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