K

Properties and Representation of Relation S

Relations and Functions: Defining Relation S

These notes cover the definition and properties of a relation S from set C to set D, as provided in the transcript, focusing on identifying ordered pairs and representing the relation.

Defining the Sets and Relation

  • Sets:
    • C = D = {-3, -2, -1, 1, 2, 3}
  • Relation S:
    • For all (x, y) \in C \times D, (x, y) \in S means that \frac{1}{x} - \frac{1}{y} is an integer.
    • This condition can also be expressed as \frac{y - x}{xy} must be an integer.

Part a: Checking Specific Ordered Pairs

To determine if a specific pair is in S, we check if the expression \frac{1}{x} - \frac{1}{y} evaluates to an integer for that pair.

  • Is 2S2? (i.e., is (2, 2) \in S?)

    • \frac{1}{2} - \frac{1}{2} = 0
    • Since 0 is an integer, YES, 2S2 ((2, 2) \in S).

  • Is -1S-1? (i.e., is (-1, -1) \in S?)

    • \frac{1}{-1} - \frac{1}{-1} = -1 - (-1) = -1 + 1 = 0
    • Since 0 is an integer, YES, -1S-1 ((-1, -1) \in S).

  • Is (2, 2) \in S?

    • This is the same question as "Is 2S2". As calculated above, YES, (2, 2) \in S because \frac{1}{2} - \frac{1}{2} = 0, which is an integer.

  • Is (2, -2) \in S?

    • \frac{1}{2} - \frac{1}{-2} = \frac{1}{2} - (-\frac{1}{2}) = \frac{1}{2} + \frac{1}{2} = 1
    • Since 1 is an integer, YES, (2, -2) \in S.

Part b: Writing S as a Set of Ordered Pairs

We systematically check all possible pairs (x, y) from C \times D to see which ones satisfy the condition \frac{1}{x} - \frac{1}{y} \in \mathbb{Z} (is an integer).

  • Pairs where x=y:

    • For any x \in C, \frac{1}{x} - \frac{1}{x} = 0, which is an integer. Thus, all pairs (x, x) are in S.
    • (-3, -3), (-2, -2), (-1, -1), (1, 1), (2, 2), (3, 3) \in S

  • Pairs where y=-x:

    • This implies \frac{1}{x} - \frac{1}{-x} = \frac{1}{x} + \frac{1}{x} = \frac{2}{x}. We need \frac{2}{x} to be an integer.
    • For x=1: \frac{2}{1} = 2 (integer). So, (1, -1) \in S.
    • For x=-1: \frac{2}{-1} = -2 (integer). So, (-1, 1) \in S.
    • For x=2: \frac{2}{2} = 1 (integer). So, (2, -2) \in S.
    • For x=-2: \frac{2}{-2} = -1 (integer). So, (-2, 2) \in S.
    • For x=3 or x=-3: \frac{2}{3} and \frac{2}{-3} are not integers. So, (3, -3) and (-3, 3) are not in S.

  • Other pairs: We need \frac{y-x}{xy} to be an integer. Let's list \frac{1}{x} values: \left{ -\frac{1}{3}, -\frac{1}{2}, -1, 1, \frac{1}{2}, \frac{1}{3} \right}

    • Any combination \frac{1}{x} - \frac{1}{y} outside the above cases (where x=y or x=-y) will result in a non-integer value given the specified domain (e.g., \frac{1}{1} - \frac{1}{2} = \frac{1}{2}; \frac{1}{1} - \frac{1}{3} = \frac{2}{3}; \frac{1}{2} - \frac{1}{1} = -\frac{1}{2}; \frac{1}{-1} - \frac{1}{2} = -\frac{3}{2}).

  • The complete set S as ordered pairs is:
    S = { (-3, -3), (-2, -2), (-2, 2), (-1, -1), (-1, 1), (1, -1), (1, 1), (2, -2), (2, 2), (3, 3) }

Part c: Domain and Co-domain of S

  • Domain of S (Dom(S)): The set of all first elements of the ordered pairs in S.

    • From the set S identified in Part b, the first elements are {-3, -2, -1, 1, 2, 3}.
    • So, Dom(S) = {-3, -2, -1, 1, 2, 3} (which is equal to C).

  • Co-domain of S: By definition, the co-domain of a relation from C to D is the set D.

    • Given that D = {-3, -2, -1, 1, 2, 3}.
    • So, CoDom(S) = {-3, -2, -1, 1, 2, 3}.

Part d: Arrow Diagram for S

An arrow diagram visually represents the relation by drawing arrows from elements in the domain to elements in the co-domain for each ordered pair in S. Since C=D, we list the elements once and draw arrows connecting them.

  • Elements: -3, -2, -1, 1, 2, 3
  • Arrows based on S:
    • -3 \to -3
    • -2 \to -2
    • -2 \to 2
    • -1 \to -1
    • -1 \to 1
    • 1 \to -1
    • 1 \to 1
    • 2 \to -2
    • 2 \to 2
    • 3 \to 3

(Note: A visual diagram would typically consist of two columns of labeled points for C and D, with arrows drawn between them as listed above. In a text-based format, this bulleted list serves as the description for an arrow diagram.)