For each value of x, f(x) assumes the value of the greatest integer, less than or equal to x.

It is also called the Floor function and step function

Symbol of Greatest Integer Function is []

[2.56] =2

[.5]=0

[-6.45]=-7

[-.65]=-1

This can be generalized as

$[x] = –1 , –1 \leq x < 0$

$[x] = 0 , 0 \leq x < 1$

$[x] = 1 , 1 \leq x < 2$

$[x] = 2 , 2 \leq x < 3 $

Solution

For $f : R \rightarrow R , y = f(x) = [x]$ for each $x \in R$

Domain = R

Range = Z (integer set) as it only attains integer values

The graph of the Greatest Integer Function is shown below. It coincide with the graph of identity function y=x for $x \geq 0$ and for x< 0, it is a graph of linear function y=-x

[1.56] =1

[.25]=0

[-5.45]=-6

[-.15]=-1

[-5]=-5

[1]=1

This can be written as

$ [x] \leq x$

Fractional part of x is defined as the difference between [x] and x. It is denoted as {x}

So,

{x}=x -[x]

Symbol of Fractional part of x is {} Example

x=4.65 ,then [x]=4 ,{x}=.65

x= -.65, then [x]=-1 ,{x}= .35 ( This is very important thing to remember for negative number)

x =-1.45 ,then [x]=-2 ,{x}= .55 ( This is very important thing to remember for negative number)

x=3 ,then [x]=3 ,{x}=0

We know that

$ x-1 < [x] \leq x$

$ -x \leq -[x] < 1-x$

Adding x

$ 0 \leq x -[x] < 1$

$ 0 \leq \left \{ x \right \} < 1 $

So Fractional part {x} is always non-negative and lies between (0,1).

So we can define Fractional part function as

y=f(x) = {x} =x -[x]

This is good for $x \in R$

Range of the fractional part function is [0,1)

Graph of the fractional part function is given below

Solution

1. If x is an integer

[x]=x

{x} =0

2. For $x \in R$ [[x]]= [x]

[{x}]=0

{[x]}=0

3. for $k \in Z$

a. $[x] \geq k \Rightarrow x \geq k , x \in [k,\infty)$

b. $[x] > k \Rightarrow x \geq k +1 , x \in [k+1,\infty)$

c. $[x] \leq k \Rightarrow x \leq k+1 , x \in (-\infty,k+1)$

d. $[x] < k \Rightarrow x < k , x \in (-\infty,k)$

e. $[x + k]=[x] +k$

4.

a. $y=[x]+ [-x]=\begin{cases} 0 & \text{ if } x \in Z \\ -1 & \text{ if } x \notin Z \end{cases}$

b. $y=[x]- [-x]=\begin{cases} 2[x]+1 & \text{ if } x \notin Z \\ 2[x]] & \text{ if } x \in Z \end{cases}$

c. $y=\left\{ x \right\}- \left\{-x\right\}=\begin{cases} 0 & \text{ if } x \in Z \\ 1 & \text{ if } x \notin Z \end{cases}$

3. For real number x and y

a. [x +y] = [x] + [y+x - [x] ]

b. $[x+y]=\begin{cases} [x] + [y] & \text{ if } \left\{x \right\} + \left\{y \right\} < 1 \\ [x]+[y]+1 & \text{ if } \left\{x \right\} + \left\{y \right\} \geq 1 \end{cases}$

a. $y =\sqrt {[x] -1|}$

b. $y = \frac {1}{\sqrt { 2- [x]}}$

a. $y =\sqrt {[x] -1|}$

Clearly This is defined for $[x] -1 \geq 0$

or

$[x] \geq 1$

$ x \in [1,\infty)$

So Domain is $[1,\infty)$

b. $y = \frac {1}{\sqrt { 2- [x]}}$

Clearly This is defined for $2- [x] > 0$

or

2 > [x]

or [x] < 2

$ x \in (-\infty,2)$

So Domain is $(-\infty,2)$

2. Find the domain and range of function

$y= \frac {1}{\sqrt {x- [x]}}$

We have ,

$y= \frac {1}{\sqrt {x- [x]}}$

We know that

$0 \leq x -[x] < 1 , x \in R$

for $x \in Z ,x -[x]=0$

So Domain is R - Z

$0 < x- [x] < 1 , x \in R -Z$

$ 0 < \sqrt {x -[x]} < 1$

$ 1 < \frac {1}{\sqrt {x- [x]}} < \infty$

So Range is = $(1,\infty)$

3. Let $f : R \rightarrow R$ , $f(x) = x^2 + 2[x] -1$ for each $x \in R$

Find the values of f(x) at x= 1.2 ,-.5 ,-2.1

$f(x) = x^2 + 2[x] -1$

$f(1.2) = (1.2)^2 + 2[1.2] -1 =1.44 +2-1=2.44$

$f(-.5) = (-.5)^2 + 2[-.5] -1 =.25 -2-1=-2.75$

$f(-2.1) = (-2.1)^2 + 2[-2.1] -1 =4.41 -6-1=-2.59$

**Notes****Assignments**