Let [a] = greatest integer less than or equal to a (so far example [3.14] = 3, [

ID: 3110612 • Letter: L

Question

Let [a] = greatest integer less than or equal to a (so far example [3.14] = 3, [3], [- 3.14] = - 4). This is also called the floor of a. likewise [a] is the ceiling of a, which is defined as the smallest integer greater than or equal to a (so far example [3.14] = 3, [3], [- 3.14] = - 4) a) Consider the function f (x, y) = [x + y] Find the invers image f^-1 (M) of the set M = {0, 1} in each of the following cases. Support your answer with brief reasoning. (a) Domain N times N, co - domain: Z (b) Domain Z times Z, co - domain: Z (c) Domain R times R, co - domain: Z Let g: R rightarrow Z times Z be defined by g (x) = ([2x], [x + 1] i) Find image g (s), where S = {x element R | 0

Explanation / Answer

b. Given g:×, g(x) = ([2x], [x+1]). [.] represents the Greatest Integer Function (GIF).

i) s = {x , 1 < x < 10}. In order to find the image g(s), when GIF is involved, we divide the given interval

1 < x < 10 into a number of integer intervals as follows and calculate the subsequent values as tabulated below:

Range of x

Value taken for calculation

(2x, x+1)

([2x], [x+1])

1 < x 1.5

1.4

(2.8, 2.4)

(2, 2)

1.5 < x 2

1.7

(3.4, 2.7)

(3,2)

2 < x 2.5

2.4

(4.8, 3.4)

(4,3)

2.5 < x 3

2.7

(5.4, 3.7)

(5,3)

3 < x 3.5

3.2

(6.4, 4.2)

(6,4)

9 < x 9.5

9.1

(18.2, 10.1 )

(18, 10)

9.5 < x < 10

9.6

(19.2, 10.6)

(19, 10)

Intermediate values are omitted for the sake of convenience, however, after only a few calculations the trend becomes clear, the first value keeps on increasing in steps of 1 from 2 to 19, while the second value increases after every two intervals like 2,2 and then 3,3 and then so on. Hence, the image g(s) is

g(s) = {(2,2), (3,2), (4,3), ..., (19,10)}

ii. It can be clearly seen that, only (2,2) g(s), hence for n 3, (n, n) Rg.

Hope that helped.