Question

A freelance computer consultant keeps a database of her clients, which contains the names S = {Acme, Bakers, Cores, Dual, Energy, Flavour, Global, Hilbert}. These clients owe her money:

A = {Acme, Cores, Energy, Global}.

These clients have done at least $2,000 worth of business with her:

B = {Acme, Bakers, Cores, Dual}.

These clients have employed her in the last year:

C = {Acme, Cores, Dual, Energy, Global, Hilbert}.

A subset of clients is described that the consultant could find using her database. HINT [See Example 4.]
The clients who do not owe her money or have employed her in the last year. Write the subset in terms of A, B, and C.

A. A' ∩ B
B. A' ∩ C
C. A ∪ C
D. A ∩ B
E. A' ∪ C

List the clients in that subset.

A. {Flavour, Bakers}
B. {Acme, Energy, Cores, Global, Bakers, Dual, Flavour, Hilbert}
C. {Acme, Energy, Cores, Global}
D. ∅
E. {Acme, Cores}

Answer 1

The clients who do not owe money and have not employed the consultant in the last year are found in the intersection of A' and C', which yields the subset {Flavour, Bakers}.

Step-by-step explanation:

The client subset we are looking for are those who do not owe her money (A') and have not employed her in the last year (C'). Using set notation, we are interested in finding A' ∩ C'.

First, let's find out A', which is the set of clients who do not owe money. A' = {Bakers, Dual, Flavour, Hilbert}. Then we find C', which is the set of clients who have not employed her in the last year; C' = {Bakers, Flavour}.

The clients in both A' and C' are those who meet both criteria, so we perform an intersection: A' ∩ C' = {Bakers, Flavour}

Therefore, the correct subset from the choices is:

A. A' ∩ B

And the clients in that subset are:

A. {Flavour, Bakers}

Answer 2

We want clients that do not owe her money or have employed her in the last year. In set terminology, the or operator is represented by the union of two sets.

So, we're looking for the union between the subsets of clients that don't owe money and clients that have employes in the last year.

The first subset is A', because we're looking for the negation of its condition.

The second subset, by definition, is exactly C.

So, the answer is A' ∪ C.

To list the customer, we have:

A' = {Bakers, Dual, Flavour, Hilbert}

C = {Acme, Cores, Dual, Energy, Global, Hilbert}

So, their union is composed by all elements belonging to A' or E (or both), without repetitions:

A' ∪ C = {Acme, Bakers, Cores, Dual, Energy, Flavour, Global, Hilbert} = S