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This function is known as step function as values of function steps by "1" as we switch values of “x” from one interval to another. We see that there is no restriction on values of "x" and as such its domain has the interval equal to that of real numbers. On the other hand, the step function or greatest integer function evaluates only to integer values. It means that the range of the function is set of integers, denoted by "Z". Hence,

D o m a i n = R

R a n g e = Z

GIF is not a periodic function. Though function is defined for all real x, but graph is not continous. It breaks at integral values of x.

Problem : Find domain of function given by :

f x = 1 π [ x ]

Solution : The denominator of function is positive. This means :

π [ x ] > 0 [ x ] < π

The value of π is 3.14. Here, [x] returns integral value. Clearly, it can assume a maximum value of 3. But, GIF returns integer value “n” for x<n+1. The inequality, therefore, has solution given by :

x < 4

Domain = - , 4

Important properties

Certain properties of greatest integer function are presented here :

1: If and only if “x”is an integer, then :

[ x ] = x

2: If and only if at least either “x” or “y” is an integer, then :

[ x + y ] = [ x ] + [ y ]

For example, let x = -2.27 and y = 0.63. Then,

[ x + y ] = [ -2.27 + 0.63 ] = [ -1.64 ] = -2

[ x ] + [ y ] = [ -2.27 ] + [ 0.63 ] = -3 + 0 = -3

However, if one of two numbers is integer like x = -2 and y = 0.63, then the proposed identity as above is true.

4: If “x” belongs to integer set, then :

[ x ] + [ - x ] = 0 ; x Z

For example, let x = 2.Then

[ 2 ] + [ - 2 ] = 2 2 = 0

We can use this identity to test whether “x” is an integer or not?

3: If “x” does not belong to integer set, then :

[ x ] + [ - x ] = - 1 ; x Z

For example, let x = 2.7.Then

[ 2.7 ] + [ - 2.7 ] = 2 3 = - 1

Problem : Find domain of the function :

f(x) = 1 [ x - 2 ]

Solution : Given function is in rational form having GIF as its denominator. The denominator should not evaluate to zero for real values of x. The domain of GIF is real number set R. But, we know that GIF evaluates to zero in an interval which is spread over unit value. In order to know this interval, we determine interval of x for which [x-2] is zero.

[ x - 2 ] = 0

We can write this function as :

[ x + ( - 2 ) ] = 0

Using property [x+y] = [x]+ [y], provided one of x and y is an integer. This is the case here,

[ x + ( - 2 ) ] = [ x ] + [ - 2 ] = [ x ] - 2 = 0

[ x ] = 2 2 x < 3 x [ 2 , 3 )

Hence, domain of given function is :

Domain = R - [ 2 , 3 )

Fraction part function

We define a fraction part function (FPF) denoted by “{x}” as :

{ x } = x [ x ]

This function returns fraction part of the number, when “x” is not an integer. This exception of non-integral “x” is important. Zero is not a fraction. For integer "x", the function evaluates to zero :

{ 5 } = 5 [ 5 ] = 5 5 = 0

{ 5 } = 5 [ 5 ] = 5 + 5 = 0

Though zero is not a fraction, but FPF evaluates to zero for integral values. We should keep this exception in mind, while working with FPF. Let us, now, work out with numbers that we earlier used for evaluating greatest integer function :

{ 0.23 } = 0.23 [ 0.23 ] = 0.23 0 = 0.23

{ 1.7 } = 1.7 [ 1.7 ] = 1.7 1 = 0.7

{ - 0.54 } = - 0.54 [ - 0.54 ] = - .54 - 1 = - 0.54 + 1.0 = 0.36

{ - 2.34 } = - 2.34 [ - 2.34 ] = - 2.34 - 3 = - 2.34 + 3.0 = 0.66

We can see that interpretation of fraction for the negative number is consistent with what has been explained earlier.

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Source:  OpenStax, Functions. OpenStax CNX. Sep 23, 2008 Download for free at http://cnx.org/content/col10464/1.64
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