MHB Solving the Floor Function Equation: $[x]=2x+1$

[solakis1]

Solve the following equation:

$[x]=2x+1$,where [x] is the floor function

 

[HOI]

We can write x as [x}+ r where r is the "fraction part". The equation is [x]= 2[x]+ 2r+ 1 so r2+ 1= -[x]. 2r= -[x]- 1. Since -[x]-1 is an integer, r must be 1/2 or -1/2.

 

[solakis1]

So what are the values for x??

 

[Opalg]

2r= -[x]- 1. Since -[x]-1 is an integer, r must be 1/2 or -1/2.

... or 0.

 

[solakis1]

So what are the values of x

 

[DaalChawal]

Is the answer x belongs to [-1 , 0 ) ?

 

[Opalg]

Solve the following equation:

$[x]=2x+1$,where [x] is the floor function

Let $x = n+r$, where $n = \lfloor x\rfloor$ is the integer part of $x$, and $r$ is the fractional part with $0\leqslant r<1$. Then the equation $\lfloor x \rfloor = 2x+1$ becomes $n = 2(n+r)+1$, so that $n = -2r-1$. The right side of that equation is negative, therefore $n$ must be negative. But $n$ is an integer, which means that $r$ must be $0$ or $\frac12$. If $r=0$ then $n = -1$, and if $r = \frac12$ then $n=-2$. The two solutions for $x$ are therefore $x=-1$ and $x = -1.5$. Those are the only solutions.

 

[DaalChawal]

You're right thanks I made a silly mistake.This is my approach.
We know that {x} + [x] = x
Now $x-{x} = 2x + 1 $
so $-{x}=x + 1$
Now just draw graph :)

 

[jonah1]

Beer inspired graph follows.

You're right thanks I made a silly mistake.This is my approach.
We know that {x} + [x] = x
Now $x-{x} = 2x + 1 $
so $-{x}=x + 1$
Now just draw graph :)

https://www.desmos.com/calculator/0cpj2izgfi

 

[DaalChawal]

Hey sorry I meant x - {x} = 2x + 1
-{x} = x + 1
It was a typo.

 

[solakis1]

[sp]Definition of floor value of real no x denoted by [x]

$\forall A\forall B( [A]=B\Leftrightarrow B\leq [A]<B+1 \wedge( B\in Z))$

In the above definition put A=x and B =3x+2 and we have:

[x]=3x+2 $\Leftrightarrow 3x+2\leq x< 3x+3\wedge 3x+2\in Z$ , where Z is the set of integers

Which is equivalent to :

$\frac{-3}{2}<x\leq -1\wedge 3x+2\in Z$

Which is equivalent to :

$\frac{-5}{2}<3x+2\leq -1\wedge 3x+2\in Z$

Which is equivalent to :

3x+2=-1 or 3x+2=-2

Which is equivalent to :

x=-1 or x= -(4/3)

Note since all the steps in the proof are equivalent we can say these are the only solutions to the above equation[/sp]

 

[jonah1]

Beer induced observation follows.

... Which is equivalent to :

x=-1 or x= -(4/3) ...

Typo perhaps?

 

[solakis1]

The whole problem is a typo I solved the equation [x]= 3x+2 instead the one which was in OP ,[x]=2x+1

 

[Opalg]

The whole problem is a typo I solved the equation [x]= 3x+2 instead the one which was in OP ,[x]=2x+1

Other than that, your solution looks fine. :D

 

[solakis1]

Since i solved the wrong problem let me solve the right one using the n substitution.
S0 let $[x]=n=2x+1 $ this implies that $ n\leq x <n+1$ and $x=\dfrac{n-1}{2}$ and combining the two we have"

$n\leq\dfrac{n-1}{2}<n+1$ which implies $3<n\leq -1$ which implies $n=-1$ or $n=-2$ since n is integer

Hence the original equation gives :$ -1=2x+1$ or $-2=2x+1$ which implies $x=-1$ or $x=-(3/2)$