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)
![R=\left \{ (a,b)\epsilon \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}](https://tex.z-dn.net/?f=R%3D%5Cleft%20%5C%7B%20%28a%2Cb%29%5Cepsilon%20%20%5Cmathbb%7BR%7D%20%5Ctimes%20%5Cmathbb%7BR%7D%20%5Cmid%20a%20%5Cleq%20b%5Cright%20%5C%7D)
The relation R is reflexive for
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
and ![b\leq c\Rightarrow a\leq c](https://tex.z-dn.net/?f=%20b%5Cleq%20c%5CRightarrow%20a%5Cleq%20c)
So if aRb and bRc, then aRc
b)
![R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}](https://tex.z-dn.net/?f=R%3D%5Cleft%20%5C%7B%20%28m%2Cn%29%5Cepsilon%20%20%5Cmathbb%7BZ%7D%20%5Ctimes%20%5Cmathbb%7BZ%7D%20%5Cmid%20%5Cexists%20k%5Cin%20%5Cmathbb%7BZ%7D%20%5Cni%20m%3Dkn%20%5Cright%20%5C%7D)
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)
![R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}](https://tex.z-dn.net/?f=R%3D%5Cleft%20%5C%7B%20%28m%2Cn%29%5Cepsilon%20%20%5Cmathbb%7BZ%7D%20%5Ctimes%20%5Cmathbb%7BZ%7D%20%5Cmid%20m%5Cneq%20n%5Cright%20%5C%7D)
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)
![R=\left \{ A,B\mid A\subseteq B \right \}](https://tex.z-dn.net/?f=R%3D%5Cleft%20%5C%7B%20A%2CB%5Cmid%20A%5Csubseteq%20B%20%5Cright%20%5C%7D)
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