Note: this is a multi-part . once an answer is submitted, you will be unable to return to this part determine whether the relation on the set of all people is reflexive, symmetric, antisymmetric, and/or transitive, where (a. b) e rif and only if ind b have a common grandparent. (check all that apply) reflexive symmetric transitive antisymmetric

A. This relation is Reflexive and Symmetric

Step-by-step explanation:

A. Before starting to test this relation, let´s remind the definition of these properties. Let R be a relation defined as R⊆A×A, where A is an arbitrary set:

R is Reflexive ↔ (∀a ∈ A)((a,a) ∈ R)

R is Symmetric ↔ (∀a,b ∈ A)("(a,b) ∈ R"⇒"(b,a) ∈ R")

R is Anti-symmetric ↔ (∀a,b ∈ A)("(a,b) ∈ R"∧"(b,a) ∈ R" ⇒ a=b)

R is Transitive ↔  (∀a,b,c ∈ A)("(a,b) ∈ R"∧"(b,c) ∈ R" ⇒ "(a,c) ∈ R")

Where (a,b)∈R ↔ a is Related to b

The question defines R as follows: "a is Related to b" ↔ a and b have a common grandparent. We use the definition and some logic to prove whether R fulfill the definitions above or not:

R is Reflexive: (we try to show that  (∀a ∈ A)("(a,a) ∈ R") where A is the set of people we work on)

the elements are people, so we suppose that a has a grandparent. Because a has the same grandparent as a (they are the same person, so they have the same family) and it happens for all people you consider. We conclude that a is Related to a ((a,a) ∈ R), and because a was arbitrary we also shown that this happens to everyone (∀a ∈ A)("(a,a) ∈ R"). This proves that R is Reflexive.

R is Symmetric: (Remember the definition above)

Suppose (a,b) ∈ R, then, "a has a common grandparent with b" and because the grandparent that they share is fixed, it´s true that "b has a common grandparent with a" so we have (b,a) ∈ R. Because we are able to do this whenever we have (a,b) ∈ R we conclude  (∀(a,b) ∈ A)("(a,b) ∈ R"⇒"(b,a) ∈ R")  and that R is Symmetric.

R is Anti-symmetric: (R does not satisfy this definition)

It´s valid to give an example that does not fulfill the definition to prove definitely that R is not Anti-symmetric. Suppose a and b are brothers, because of this (a,b)∈R, and (b,a)∈R (because we proved that R is symmetric) but a≠b. Now we now that ("(a,b) ∈ R"∧"(b,a) ∈ R" ⇒ a=b) does not happen to every pair in R, so we conclude that R is not Anti-symmetric.

R is transitive: (R does not satisfy this definition either)

Suppose (a,b) ∈ R and (b,c) ∈ R, a and b have a grandparent in common, and the same happens with b and c. It´s not hard to think that, for example, a has the same grandfather as b but b has the same grandmother with c. We show that a could not necessarily be related to c. so (∀a,b,c ∈ A)("(a,b) ∈ R"∧"(b,c) ∈ R" ⇒ "(a,c) ∈ R") it´s not true, and R is not Transitive.

tell you what. i'll find the length of the medians and you can fill in the blanks where they belong.

br

b = (2a,0)

r = (a/2,b/2) remember these things are medians. they go to the 1/2 way point of the line opposite the vertex they face.

br^2 =   (2a - a/2)^2 + (0 - b/2)^2

br^2 = (3/2 a) ^2 + b^2 / 4

br^2 = 9/4 a^2 + b^2 / 4   we need to find some relationship between a and b.

let's try ab = bc  

ab = 2a

bc = (2a - a)^2 + (b - 0)^2

bc = sqrt(a^2 + b^2)

ab = bc

2a = sqrt(a^2 + b^2) square both sides.

4a^2 = a^2 + b^2 subtract a^2 from both sides.

sqrt(3a^2) = sqrt(b^2)

sqrt(3)a = b

let's put b into br^2

br^2 = 9/4 a^2 + 3b^2 / 4

br^2 = 12 a^2 / 4

br^2 = 3a^2 

br = sqrt(3) * a

cp

c = (a,b) ; p = (a,0) this is another application of the distance formula, and it is a good one.

cp^2 = (a - a)^2 + (b - 0)^2

cp^2 = 0 + b^2

cp^2 = b^2 which you can see from the diagram. 

cp = sqrt(b^2)

cp = b but b = sqrt(3) * a 

cp = sqrt(3)*a

aq

a = (0,0)

q = (3/2) a, b/2

aq^2 = (3a/2 -0)^2 + (b/2 - 0)^2

aq^2 = 9a^2/4 + b^2/4

but b = sqrt(3) * a 

aq^2 = 9a^2 / 4 + (sqrt(3)*a)^2 /4

aq^2 = 9 a^2/ 4 + 3 a^2 / 4

aq^2 = 12 a^2/4

aq^2 = 3 a^2

aq = (sqrt  3) * a which agrees with the other two.

the answer is -3/4. i just took the test!

step-by-step explanation: