Math Aunty

GYHD 1

a)      x = {1} P(x) = { ∅, {1}} |P(x)| = 2

b)      x = {1 , 2} P(x) = {∅, {1}, {2}, {1,2}} |P(x)| = 4

c)       x = {1, 2, 3, 4} P(x) = {∅, {1}, {2}, {3}, {4}, {1, 2,}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}, {1, 2, 3}{1, 2, 4}, {1, 3, 4}, {2, 3, 4}, {1, 2, 3, 4}} |P(x)| = 16

 

 

GYHD 2

                Is it possible for something to be both an element and a subset of the same set?

The null set is the only thing that can be an element and a subset of the same set. The null set is a subset of ALL sets, so any set containing the null set as an element, has it as both an element and a subset.

 

GYHD 3

a)      {1, 2, 3} X {2, 4, 6}

T = {(1, 2), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 2), (3, 4), (3, 6)}

|T| = 9

 

b)      {1, 2} X {1, 2, 3} X {2, 3}

T = {(1, 1, 2), (1, 1, 3), (1, 2, 2), (1, 2, 3 ), (1, 3, 2), (1, 3, 3), (2, 1, 2), (2, 1, 3), (2, 2, 2), (2, 2, 3), (2, 3, 2), (2, 3, 3)}

|T| = 12

 

GYHD 4

a)      3x – 6y = 12

T = {(1, -9/7), (2, -6/7), (3, -3/4), (4, 0), (5, 3/7)}

 

b)      r + 2s + 3t = 17

T = {(1, 2, 4), (3, 4, 2), (5, 3, 2), (7, 4, 2), (9, 1, 2)}

 

c)       a does not equal b (Universe is {1, 2, 3, 4} X {2, 3, 4, 5})

T = {(1, 2), (2, 3) (3, 4) (4, 5) (3, 2)

 

 

 

GYHD 5

GYHD 5

If you Cartesian product any set with the null set, you get nothing! Cartesian products produce ordered pairs which pull ELEMENTS from two sets. Because the null set has no elements, you can’t produce an ordered pair with something that has elements, and something that doesn’t.

GYHD 6

A = {1, 2, 3, 5}

B = {2, 3, 6}

C = {3, 4, 5}

Universe = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

a)      S = {3}

b)      S = {1, 4, 6}

c)       S = {1, 2, 4, 5, 6}