Answer the following Algebraic equations

Question

Answer the following Algebraic equations

1. Prove the identity of the following Boolean equations, using basic rules of Boolean algebra. So i.e. AB +A B+AB=A+B

2. Simplify the following Boolean expression:

X Y + XY+XYZ

3. Reduce the following Boolean expression to a minimum number of literals: X Y(Z W+W) + ZW+ A9 4. And For the Boolean function E and F, as given in the following truth table:

X Y Z E F 0 0 0 1 0 0 0 1 0 1 0 1 0 1 1 0 1 1 0 0 1 0 0 1 0 1 0 1 0 1 1 1 0 1 0 1 1 1 0 1

4. Express E and F in sum-of item an algebraic form.

5. So Simplify the following Boolean expression using K-map

F(a,C) = AC + AB + ABC

6. And Simplify the following Boolean function by using K map

F(A,B,C,D) = /m(0,2,3,5,7,8,1 0,1 1,14,15)

 

 

Summary

We have given 6 different questions with algebraic equations. And we have to answer them.

Explanation

In the algebraic expression true is always represented by 1. And false is represented by 0.
So X will represent 1. And X’ will represent 0

So X and X’ are complements of each other.

1.

A B’ + A’ B’ +AB

= (A B’ + A B’) + A’ B’ + AB     ….Using A+A=A

= (A B’ + AB) + (A B’ + A’ B’)

= A(B’ + B) + (A + A’)B           …..Using X = X’=1

= A+ B’

2.

X’Y’ + X’Y + XYZ

=(X’Y’ + X’Y)+(X’Y + XYZ)

= X'(Y’Y)+ Y(X’+XZ)

= X’1 + Y (X’+X)(X’+Z)

= X’ + Y.1 (X’+Z)

= X’ + Y X’ + YZ

= X'(1+Y)+YZ

3.

X’Y(Z’W +W’) + Y(X’ ZW + X)

= X’Y (W’+Z’)(W’+W)+Y(X+X’)(X+ZW)

= X’Y(W’ +Z’).1 + Y(X+ ZW)

= Y [ (W’X’ + Z X’) + (X + ZW)]

= Y [ (W’X’ + X) + ZX’ +ZW ]

= Y [ (X + X’) (X +W’) + Z(X’ +W) ]

= Y [ X + W’ + ZX’ + ZW]

= Y [ (X + ZX’) + (W’ + ZW) ]

= Y (X+Z +W+Z)

= Y (X+W+Z)

= XY +WY +ZY

4.

E = Σ(M₀, M_2, M₄, M₆)

= ΣM(0,2,4,6)

= X’Y’Z’ + X’ Y Z’ + XYZ’

F= ΣM (1,2,5,7)

= X’Y’Z + X’YZ’ + XY’Z + XYZ

5.

F(A,B,C) = AC + AB + ABC

= AC(B+B’) + AB(C_C’) + ABC

= ABC + AB’C +ABC + ABC’ + ABC

= AB’C + ABC’ + ABC

= ΣM (5,6,7)

	B’C’	B’C	BC	BC’
A’	0	0	0	0
A	0	1	1	1

F= AC +AB

6.

F(A,B,C,D) = Σ M (0,2,3,5,7,8,10,11,14,15)

	C’D’	C’D	CD	CD’
A’B’	1	0	1	1
A’B	0	1	1	0
AB	0	0	1	1
AB’	1	0	1	1

 

F= CD+ AC + B’D’ + A’BD

 

Also read, Randomly generate 50 integers within the range of 0 – 999.