Answer:
a)Reflexive, not symmetric, transitive
b)Reflexive, not symmetric, transitive
c)Not reflexive, symmetric, not transitive
d)Reflexive, not symmetric, transitive
Step-by-step explanation:
a)
[tex]R=\left \{ (a,b)\epsilon \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}[/tex]
The relation R is reflexive for
[tex]a\leq a[/tex] for every real number a
it is not symmetric because 0 is less than 1, but 1 is not less than 0
it is transitive
[tex]a\leq[/tex] and [tex] b\leq c\Rightarrow a\leq c[/tex]
So if aRb and bRc, then aRc
b)
[tex]R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}[/tex]
R is reflexive because m=1.m for every integer m
R is not symmetric: 2 is a factor of 4, but 4 is not a factor of 2
R is transitive: if mRn and nRp if m=kn and n=qp, so m=(kq)p and kq is an integer , so mRp
c)
[tex]R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}[/tex]
R is obviously not reflexive because all numbers equals themselves
R is symmetric: if a different to b, then b different to a
R is not transitive: 1R2 and 2R1 (because 1 different to 2), but 1 = 1
d)
[tex]R=\left \{ A,B\mid A\subseteq B \right \}[/tex]
R is reflexive for every set A is a subset of itself
R is not symmetric {1,2} is a subset of {1,2,3} but {1,2,3} is not a subset of {1,2}
R is transitive: if A is subset of B and B is subset of C, then A is subset of C
