Let S = {O, 1}, and define operations + and · on S by the following tables:
a. Show that the elements of S satisfy the following properties:
(i) the commutative law for +
(ii) the commutative law for ·
(iii) the associative law for +
(iv) the associative law for ·
(v) the distributive law for + over ·
(vi) the distributive law for · over +
b. Show that 0 is an identity element for + and that 1 is an identity element for ·.
c. Define
(a)
To prove:
Show that the elements of S satisfies:
Given information:
Let
Proof:
(i) To proof:
PROOF:
First case: x = 0 and y = 0
Second case: x = 0 and y = 1
Third case: x = 1 and y = 0
Fourth case: x = 1 and y = 1
Thus, we note that the commutative law
(ii) To proof:
PROOF:
First case: x = 0 and y =0
Second case: x = 0 and y = 1
Third case: x = 1 and y = 0
Fourth case: x = 1 and y = 1
Thus we note that the commutative law
(iii) To proof:
PROOF:
First case: x = 0 and y = 0 and z = 0
Second case: x = 0 and y = 0 and z = 1
Third case: x = 0 and y = 1 and z = 0
Fourth case: x = 0 and y = 1 and z = 1
Fifth case: x = 1 and y = 0 and z = 0
Sixth case: x = 1 and y = 0 and z = 1
Seventh case: x = 1 and y = 1 and z = 0
Eight case: x = 1 and y = 1 and z = 1
Thus we note that the associative law
(iv) To proof:
PROOF:
First case: x = 0 and y = 0 and z = 0
Second case: x = 0 and y = 1 and z = 0
Third case: x = 0 and y = 1 and z = 0
Fourth case: x = 0 and y = 1 and z = 1
Fifth case: x = 1 and y = 0 and z = 0
Sixth case: x = 1 and y = 0 and z = 1
Seventh case: x = 1 and y = 1 and z = 0
Eight case: x = 1 and y = 1 and z = 1
Thus we note that the associative law
(v) To proof:
PROOF:
(b)
To prove:
0 is an identity element for + and that 1 is an identity element for
(c)
To prove:
For every a in S,