Let C = D = {3, 2, 1, 1, 2, 3} and define a relation S from C to D as follows: For all

Chapter 1, Problem 2

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QUESTION:

Let C = D = {−3,−2,−1, 1, 2, 3} and define a relation S from C to D as follows: For all \((x, y) \in C \times D\), \((x, y) \in S\) means that \(\frac{1}{x}-\frac{1}{y}\) is an integer.

a. Is 2 S 2? Is −1S − 1? Is \((3,3) \in S\)? Is \((3,-3) \in S\)?

b. Write S as a set of ordered pairs.

c. Write the domain and co-domain of S.

d. Draw an arrow diagram for S.

Text Transcription:

(x, y) in C times D

(x, y) in S

1/x -1/y

(3,3) in S

(3,-3) in S

Questions & Answers

QUESTION:

Let C = D = {−3,−2,−1, 1, 2, 3} and define a relation S from C to D as follows: For all \((x, y) \in C \times D\), \((x, y) \in S\) means that \(\frac{1}{x}-\frac{1}{y}\) is an integer.

a. Is 2 S 2? Is −1S − 1? Is \((3,3) \in S\)? Is \((3,-3) \in S\)?

b. Write S as a set of ordered pairs.

c. Write the domain and co-domain of S.

d. Draw an arrow diagram for S.

Text Transcription:

(x, y) in C times D

(x, y) in S

1/x -1/y

(3,3) in S

(3,-3) in S

ANSWER:

Step 1 of 4

Let and a relation is defined from to as follows:

is an integer.

For ,

As 0 is an integer, therefore, .

Hence, is true.

For ,

As 0 is an integer, therefore, .

Hence, is true.

For ,

As 0 is an integer, therefore, .

For ,

As  is not an integer, therefore, .

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