3.6.5: Cardinality of a set defined by a Cartesian product. (a) What is |{0, 1}7

ID: 3143575 • Letter: 3

Question

3.6.5: Cardinality of a set defined by a Cartesian product.

(a)

What is |{0, 1}7|?

(b)

What is |{a, b, c, d}3|?

efine the sets A, B, C, D, and E as follows:

A = {x R: x < -2}

B = {x R: x > 2}

C = {x R: |x| < 2}

D = {x R: |x| 2}

E = {x R: x -2}

Use the defintions for A, B, C, D, and E to answer the questions.

(a)

Do the sets A, B, and C form a partition of R? If not, which condition of a partition is not satsified?

(b)

Do the sets A, B, and D form a partition of R? If not, which condition of a partition is not satsified?

(c)

Do the sets B, D, and E form a partition of R? If not, which condition of a partition is not satsified?

(a) Do the sets A, B, and C form a partition of R? If not, which condition of a partition is not satsified?

- No. because the x = +2 and -2 are not present in set formed by A U B U C.

A = {x R: x < -2}

B = {x R: x > 2}

C = {x R: |x| < 2}

A U B U C = R - {-2,2}

(b) Do the sets A, B, and D form a partition of R? If not, which condition of a partition is not satsified?

Yes they do, because they do have all the elements of R in their union also no 2 sets have any common element.

A U B U D = R

A B = {} i.e. null set

B D = {} i.e. null set

A D = {} i.e. null set

E U B U D = R

E D = {-2} not a null set so doesn't partition R.

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.