Suppose R and S are reflexive relations on set A and T is a transitive relation on set A. Prove or disprove each of these statements: a)R∪T must be transitive. b)R⊕S is irreflexive. c)R◦S is reflexive

(a)R∪T is not Transitive.

(b)R⊕S is irreflexive.

(c)R◦S is Reflexive

Step-by-step explanation:

Definition

Reflexive: A relation on a set A is reflexive if (a,a)∈A for all a∈A.

Transitive: A relation on a set A is said to be transitive if (a, b) ∈A and (b, c)∈A ⇒ (a, c)∈A.

Union: For two sets A and B, A∪B means all elements are either in A or B.

Symmetric Difference A⊕B: These is the set of all elements either in A or in B but not in both.

Composite of A and B (A◦B) consists of all ordered pairs (a,c) such that (a,b)∈B and (b,c)∈A.

(a)R∪T is Transitive

Let a∈R, since R is Reflexive, (a,a)∈R.

By the definition of T, (a, b) ∈T and (b, c)∈T, (a, c)∈T.

R∪T={(a,a),(ac)}

R∪T is not transitive since there is no element b∈R∪T to satisfy the condition for transitivity.

(b)R⊕S is irreflexive

Let a∈R, since R is Reflexive, (a,a)∈R.

Let a∈S, since S is Reflexive, (a,a)∈S.

The symmetric difference contains all elements either in R or in S but not in both.

Therefore: a∉R⊕S.

Therefore R⊕S is irreflexive.

(c)R◦S is reflexive

Let Let a∈R, since R is Reflexive, (a,a)∈R.

Let a∈S, since S is Reflexive, (a,a)∈S.

By the definition of Composite, if

(a,a)∈R and (a,a)∈S, then (a,a)∈R◦S.

Therefore, R◦S is Reflexive since (a,a)∈R◦S for every a∈A.

180 square centimeters

step-by-step explanation:

connie < felicia < stew

step-by-step explanation:

stew:

we need to find the slope

m = (y2-y1)/ (x2-x1)

    = (2650-795)/(10-3)

    = 1855/7

265 words per minute

connie:

we need to find the slope

m = (y2-y1)/ (x2-x1)

we have two points (0,0) and (4,800)

m = (800-0)/(4-0)

  m = 800/4

    200 words per minute

felicia:

260 words per minute

200< 260 < 265

connie < felicia < stew