A Floor Value Equation (x^2-⌊x⌋-2=0)

[@aashsyed1277]

floor equations are hands down my most favorite.

[@MathElite]

I love floor equations :DD

[@leickrobinson5186]

Me too! :-D

[@MushookieMan]

Notice that x^2-⌊x⌋-2 ≥ x^2-x-2. Factoring the RHS, we can easily establish invervals for x and x^2. Then try each case. It's a bit hand-wavy, I know.

[@SyberMath]

You're right! Good thinking!

[@arshsverma]

I did it with just replacing x with n+epsilon, and you get (n+epsilon)^2-n-2=0, now you know you have 0

[@carloshuertas4734]

Another great explanation, SyberMath! I only had 1 solution to that math problem. I never knew about floor problems in math until you told me. Thanks!

[@tonyhaddad1394]

Awesome , now im familiar with floor function , they are so important and fun !!!!

[@aashsyed1277]

i love floor equations so much!!!!!!!!!!!

[@tonyhaddad1394]

4:26 in fact you can call n is positive since in this case x=sqrt(n+2) >=0 so if x >=0 then floor(x) >= 0 floor(x)= n

[@andreasavraam6898]

Actually it's not that we can say that,but we HAVE to say that,basically what he did is wrong if n isn't non-negative. For example if n=-2, then -2=n0=n+2,so it would have been better not to use n,but |n| instead since we have the inequality for both sqrt(n+2) and -sqrt(n+2), that would have lead to the same answers and actually be correct.

[@77Chester77]

A quadratic equation with three solutions, nice :-)

[@antoniopedrofalcaolopesmor6095]

For x integer, [x]=x, so the roots of x^2-x-2 are also solutions for the given floor value equation. So we already have solutions x=2 and x=-1. Note that x^2-[x]-2 >= x^2-x-2. Since for x 2 we have that x^2-x-2 > 0, so x^2-[x]-2>0 and there are no solutions less than -1 or greater than 2. We just have to still look for solutions in the intervals (-1,0) , [0,1) , [1,2) and we're done.

[@alexminsky1]

A more straightforward method is to plot x^2 and [x]+2 and see the three solutions.

[@aahaanchawla5393]

Or plot y = x² - floor(x) - 2

[@alexminsky1]

@@aahaanchawla5393 Yep, that definitely works, but I guess for most folks, it’d be much easier to plot x^2 and [x]+2 separately and looking for the intersections there.

[@aayushshende1943]

X lies in [-1 , 0] to [1,2] . I solved it as a regular quad eqn and then found it intervals by using how the floor func works.

[@davidseed2939]

here’s my way. [_x_] is an integer, so is 2 , so x^2 must be an integer, also x^2 non-negative. So, we can just go through all the values of x^2 and see whether [_x_] and x^2 are compatible. x^2 …..:……… 0 , 1 , 2 , 3 , 4 , 5 x^2-2=[_x_] -2 , -1 , 0 , 1 , 2 , 3 x"…………………* , -1 , * , 1.73, 2 , *(sqrt(5)=2.23) * indicates that there is no value x^2 such that +-sqrt(x^2) = x^2-2. since x^2 -2 rises faster than x once we have x^2-2>[_x_] then there will be no further solutions. So x={ -1, sqrt(3),2}

[@242math]

you handled this floor equation very well. great job bro

[@SyberMath]

I appreciate it, man! 🤗

[@markobavdek9450]

Floor equations are like falling down the stairs 😆

[@mrhatman675]

Hahahahahahahahaha quite literally

[@mathsandsciencechannel]

Great video sir. Love that ♥️♥️♥️♥️

[@aashsyed1277]

heelo

[@nicogehren6566]

nice question sir thanks

[@SyberMath]

You're welcome!

[@gmutubeacct]

Thanks for posting this. Why did you not work on the OR part of the 2 inequalities you defined at 3:46? It turns out that it would not give you any additional solutions. But, don't you have to work on that for thoroughness? Thanks

[@elgalvin]

Where from did you get -1

[@comingshoon2717]

hermosa solución

[@luiscrispinvargas3061]

Creo que hay un pequeño error en la resolución, las respuestas sí están bien pero la parte donde resuelves n≤√n+2, estás elevando al cuadrado suponiendo que n sea positivo, lo cuál no lo tenemos la certeza, por lo que no podrías elevar al cuadro, ahí comienza lo que estaría mal la resolución. Espero haber ayudado, saludos desde Perú.

[@3r3nite98]

We are currently in the 1st Floor rn. Also I bet this will be a good video.

[@pardeepgarg2640]

In search of Floor we lost our Ceiling 🤦🤦🤦

[@SyberMath]

😁😂

[@aashsyed1277]

lmao

[@supramitra]

Sorry,I couldn't join the premiere yesterday.But,I am watching now lol

[@aashsyed1277]

Hello

[@manojsurya1005]

Great problem, quadratic and floor are a good combo 😃

[@SyberMath]

I agree!

[@jimschneider799]

@9:05 - since: x^2 - floor(x) - 2 >= x^2 - x - 2 and x^2 - x - 2 is greater than zero for all x < -1 or x > 2, the solutions to x^2 - x - 2 = 0 bound the interval in which solutions to the original equation may be found. In fact, for any positive integer k, if x^2 - floor(x) - k = 0 has any solutions, they will be in the closed interval bounded by the roots of x^2 - x - k = 0. This does raise the question: does x^2 - floor(x) - k = 0 have any roots when k is a positive integer such that the roots of x^2 - x - k = 0 are not integers? I'll reply to this if I find proof either way.

[@jimschneider799]

I found that x^2 - floor(x) - 1 = 0 has the solution x = sqrt(2), so the answer to the question I raised is "yes". Perhaps a more interesting question: does x^2 - floor(x) - k = 0 have any integer roots that are not also roots of x^2 - x - k = 0? Due to the fact that x^2 - floor(x) - k has a jump discontinuity at every integer value of x, I suspect the answer to that question is "no".

[@jimschneider799]

A moment's reflection gave me an even easier proof; assume that x^2 - floor(x) - k = 0 has an integer root, x=n, such that n is not also a root of x^2 - x - k = 0. Since n is an integer, then n = floor(n), so that n^2 - n - k = 0. But that means that n is a root of x^2 - x - k = 0, a contradiction. Therefore, any roots of x^2 - floor(x) - k = 0 that are not also roots of x^2 - x - k = 0 cannot be integers.

[@vameza1]

Nice!!!

[@SyberMath]

Thanks!

[@mathevengers1131]

I also made a floor video today. A coincidence.

[@mathunt1130]

One of the first things you can see is that x=2 is an is a solution...

[@crykoh4082]

Nice

[@SyberMath]

Thanks

[@aishbgm5743]

once explain factorial in floor like [x/2!]

[@leecherlarry]

compi finds three real solutions heh: *Solve[x^2 - Floor[x] - 2 == 0, x, Reals]*

[@ςγτε]

what is compi?

[@leecherlarry]

@@ςγτε computer (software)

[@ςγτε]

@@leecherlarry so you are using Wolfram language (☉｡☉)!

[@leecherlarry]

@@ςγτε yeah me just beginner level. it's a beautiful language though 🙋

[@rssl5500]

Interesting ..

[@SyberMath]

Glad you think so!

[@illuminati5676]

Hey,SyberMath can we solve this in this way 31¹

[@antoine5571]

Now I know how to kill those equations thanks

[@aashsyed1277]

How can you write syber's emoji?

[@antoine5571]

@@aashsyed1277 maybe because I'm member

[@SyberMath]

Because he is a member 😉

[@aashsyed1277]

@@SyberMath can you write it? if you can then how? if you are not a member of yourself!

[@ivansakovich7653]

Почему не решили два двойных неравенства?

[@TheodoreLong1995]

Can you calculate x^i=?

[@ganda3454]

(x-2)(x+1)

[@Germankacyhay]

👍

[@threstytorres4306]

Let X is a Non-Decimal Number X^2 - X - 2 = 0 Using the Quadratic Formula We have X is equal to: 2 and -1

[@SyberMath]

What about non-integers?

[@montynorth3009]

I get the solution of x being equal to 2 or -1 which I calculated using the formula for solving quadratic equations. I don't know about the square root of 3 and also not being an integer.

[@mustafizrahman2822]

x=plus minus 2

[@hongkongsmartboy]

Divided into 2 cases: 1. x >=0， |x| = x 2. x

[@rafiihsanalfathin9479]

Floor and absolute value is not the same thing

[@aashsyed1277]

@@rafiihsanalfathin9479 he/she thought that by "floor" the absolutele value |