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.

the answer   is 200

step-by-step explanation:

this is a problem about poisson probabilities.   i don't know if you've

already studied poisson distributions, or if you have a place to look

to read more about the subject.  

here's how you can see a poisson distribution coming.   when there are

a lot of events that may happen at any time (or any place), and when

you're counting how many of them happen at a particular time (or a

particular place), and each event is independent of the others, then

the result is a poisson distribution.   this is roughly true for the

chocolate chips; one chip being in a cookie has only a small effect on

the probability that another chip will be in the cookie.

once you know you have a poisson distribution, you know a lot.   there

is only one parameter to a poisson distribution, so if you know the

mean, you know everything.   the form of the distribution is

  probability of n chips = {[e^(-x)]*(x^n)} / (n! )

what is all that apply?