Question

Express the following relations in the set builder notation. Then, determine whether it is reflexive, symmetric, transitive. Please show work.
a.) One number is less than or equal to another.

b.) One integer is a factor of another.

c.) Two integers are unequal.

d.) One set is a subset of another.

Answer 1

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)

$R=\left \{ (a,b)\epsilon \mathbb{R} \times \mathbb{R} \mid a \leq b\right \}$

The relation R is reflexive for

$a\leq a$ 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

$a\leq$ and $b\leq c\Rightarrow a\leq c$

So if aRb and bRc, then aRc

b)  

$R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid \exists k\in \mathbb{Z} \ni m=kn \right \}$

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)

$R=\left \{ (m,n)\epsilon \mathbb{Z} \times \mathbb{Z} \mid m\neq n\right \}$

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)

$R=\left \{ A,B\mid A\subseteq B \right \}$

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