Determine whether the following propositions are true or false: (a) 5 is an odd number and 3 is a negative number.
(b) 5 is an odd number or 3 is a negative number.
(c) 8 is an odd number or 4 is not an odd number.
(d) 6 is an even number and 7 is odd or negative.
(e) It is not true that either 7 is an odd number or 8 is an even number (or both).

a) false

b) true

c) true

d) true

e) false

Step-by-step explanation:

In a statement of the type:

p ∧ q

where ∧ means "and"

The statement is true only if both p and q are true

the statement is false if p, q, or both, are false.

and in the case of:

p ∨ q

where ∨ means "or"

The statement is true if at least one of p or q (or both) are true.

The statement is false if both are false.

Now that we know that, let's solve the problem:

a) "5 is an odd number and 3 is a negative number."

Here we have:

p = 5 is an odd number

We know that this is true

q = 3 is a negative number

This is false.

then the complete statement is false.

b) "5 is an odd number or 3 is a negative number."

here we have:

p = 5 is an odd number.

this is true

q = 3 is a negative number

because in this case we have an "or", with only p being true, the whole statement is true.

c) "8 is an odd number or 4 is not an odd number."

p = 8 is an odd number  (this is false)

q = 4 is not an odd number  (this is true, 4 is a even number)

Again, we have an "or", so we need only one true proposition, then the statement is true.

d) "6 is an even number and 7 is odd or negative."

p = 6 is an even number  (true)

q = 7 is odd or negative  (notice that we have an or, and 7 is odd is true, so this proposition is true)

Then both propositions are true, then the statement is true.

e) "It is not true that either 7 is an odd number or 8 is an even number (or both)."

This is most complex, this will be true if at least one of the propositions is false.

but:

7 is an odd number is true

8 is an even number is true.

Then both statements are true, which means that the statement is false.

step-by-step explanation:

let pq = qr = rs = x, and xz = y and zu = z.

if the horizontal lines are parallel, then corresponding segments are in proportion:

substitute:

therefore 3 = y = z.

xu = y + z → xu = 3 + 3 = 6