- #1

cronuscronus

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The question is:

Determine whether the relation R on the set of all real numbers is reflexive, symmetric, anti-symmetric, and/or transitive, where (x,y) ∈ R if and only if

(a) x+y=0

(b) x = ± y

(c) x – y is a rational number (d) x=2y

(e) xy≥0

Here are my answers, and my reasons:

a)

reflexive: true (if x = 0, then x+x = 0)

symmetric: true (if xRy = 0 then yRx = 0)

antisymmetric: false (it is symmetric, so it cannot be antisymmetric)

transitive: true (if xRy = 0 and yRz = 0 then z = x)

b) x = +- y

reflexive: true (xRx can equal positive or negative x)

symmetric: true (if x is positive or negative y, then y can be positive or negative x)

antisymmetric: false (even though xRy and yRx can be true, it doesn't mean x=y)

transitive: true (if x is positive or negative y, and y is positive or negative z, then x is positive or negative z)

c) x-y is rational

reflexive: true (x-x=0, and 0 is rational)

symmetric: false ( just because x-y is rational doesn't mean that y-x will be )

antisymmetric: false ( if x-y is rational, and y-x is rational, y != x in all cases )

transitive: true (if x-y is rational, and y-z is rational, then x-z will be rational )

d) x = 2y

reflexive: false ( x does not equal 2x )

symmetric: false ( if x = 2y, y != 2x )

antisymmetric: true ( if x = 2y and y = 2x then y = x )

transitive: false ( if x = 2y and y = 2z then x != 2z )

e) xy >= 0

reflexive: true ( some number times itself will be >= 0 )

symmetric: true ( if x * y >= 0 then y * x >= 0 )

antisymmetric: false ( it is symmetric )

transitive: true ( if x * y >= 0 and y * z >= 0 then x*z >= 0 )

Thanks for any guidance!