For each relation, indicate whether it is reflexive or anti-reflexive, symmetric or anti-symmetric,

For each relation, indicate whether it is reflexive or anti-reflexive, symmetric or anti-symmetric, transitive or not transitive. a. The domain of the relation L is the set of all real numbers. For x, y ∈ R, xLy if x < y.

b. The domain for the relation D is the set of all integers. For any two integers, x and y, xDy if x evenly divides y. An integer x evenly divides y if there is another integer n such that y = xn. (Note that the domain is the set of all integers, not just positive integers.)

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This Post Has 4 Comments

  1. (a)

    L is not reflexive, L is anti-reflexive  

    L is not symmetric.

    L is not anti-symmetric

    L is transitive.  

    (b)

    D is reflexive

    D is not symmetric.

    D is anti-symmetric

    D is transitive.

    Step-by-step explanation:

    a)

    Given that;

    domain of the given  relation L is the set of all real numbers

    For x , y ∈ R , xLy if x less than y.

    relation L, where xLy if x less than y, For x, y ∈ R

    so For every x ∈ R, it is then false that x less than x.  

    That is (x, x) does not belongs to L.

    ∴ L is not reflexive, L is anti-reflexive.

    For every x,y ∈ R, if (x,y) ∈ L (i.e. x < y), then (y, x) does not belongs to L, since it is false that y < x.

    ∴ L is not symmetric.

    For every x ∈ R, we can say  its  false that x less than x. That is (x, x) does not belongs to L.

    ∴ L is not anti-symmetric.

    For every x,y,z ∈ R, if (x, y) ∈ L(i.e. x < y) and (y, z) ∈ L(i.e. y < z), then (x, z) ∈ L, since it is true that x<z when x<y and y<z.

    ∴ L is transitive.  

    b)

    Also lets consider a relation D, where xDy if there is an integer n such that y = xn, For x, y ∈ Z.

    Now

    For every x ∈ Z, it is true that x = x × 1. That is (x, x) belongs to D.

    ∴ D is reflexive,

    For every x,y ∈ Z, if (x,y) ∈ P (i.e. y=x × n), then (y, x).

    if (x,y) ∈ D, then there exist an integer n such that y=x × n.

    Then x = y × (1/n).

    Thus, (y,x) does not belongs to D, since 1/n is not an integer, but it is a real number.

    ∴ D is not symmetric.

    For every x,y ∈ Z, if (x,y) ∈ D (i.e. y=x × n), then (y, x) also belongs to D only when x=y. where n=1.

    ∴ D is anti-symmetric.

    For every x,y,z ∈ Z, if (x,y) ∈ D (i.e. x × n1= y), and (y,z) ∈ D (i.e. y × n2 = z), then (x,z)∈ D.

    if (x,y) ∈ D, then there exist an integer n1 such that y=x × n1.

    if (y,z) ∈ D, then there exist an integer n2 such that z=y × n2.

    Thenz = (x × n1) × n2 ⇒ z=x × (n1 × n2).  

    where (n1 × n2) is an integer. Thus (x,z) ∈ Z

    ∴ D is transitive.

  2. False

    Step-by-step explanation:

    |x+y| = |x| + |y|

    To show this is false, all we have to do is find one example where it is false

    Let x = -1 and y = 4

    |-1+4| = |-1| + |4|

    |3| = |1| + |4|

    3 = 5

    This is false so we have a set of integers where the statement is not true

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