Just like sets and logic, Boolean algebra is a modern concept. It is concerned with statements which are either true or false. It was George Boolen who developed this theory and its significance was realized when Claude Shanon introduced circuit algebra to deal with relay circuits in 1938. The digital computer which contains a large number of logic circuits in small space and works in switching them opened a wide field for this algebra and its recognition as significant part of modern mathematics.

Definition

A non-empty set B = (a, b, c …..) with two binary operations OR or JOIN denoted by +), AND or meet (denoted by ) and a unary operation complement (denoted by’) is a Boolean Algebra if the elements satisfy the following axioms:

B.I. commutative law,

(i)a + b = b + a

(ii) a.b = b.a

B II. Distributive law,

(i)a.(b + c) =a.b + a.c

(ii) a+(b.c) = (a + b).(a + c)

B III. Existence of identity elements,

There exist elements, 0, such that

a + 0 = a

a.1 =a

B IV. Existence of complements,

a + a’ = 1

a.a’ = 1

a.a’ = 0

we may note here that associative property has not been assumed in the above definition but will be derived from the above definition but will be derived from the above axioms. Also the distributive law (ii) is different from ordinary algebra applied to numbers and the reader is advised to familiarize himself fully with this property applied both was, the uniqueness of the identity elements and the complement shall be proved later on. There is great similarity among the set theory, logic and Boolean Algebra as shown below

	Set theory	Logic	Boolean Algebra
Union		V	+
Intersection		^
Complement	‘		‘
Universal set	S	T	1
Null set		F	0
Implication	Mapping		P’ + q
Equivalence	One-one mapping	q[/latex]	P’q’ + pq

Some properties

Now we shall prove some properties of Boolean algebra.

Theorem I : the two identity elements are unique.

Proof: suppose if possible there are two zero elements in B. THEN

And   (B II)

But (B I)

Hence

Similarly we can prove for the second identity 1.

Theorem II: a + a = a and a.a = a (indemptent laws)

Proof: a  = a + 0 = a+ (a. a’) (B IV)

= (a + a). (a +a’) (B II)

= (a + a).1

= a +a

Also a = a.1

= a.(a +a’)

= a.(a + a.a’

= a.a + 0 = a.a

Theorem III: form every element

A + 1 = 1, a.0 = 0.

Proof: 1 = a + a’

= a + (a’.1) = (a + 1).(a + a’)

= (a + 1).1 = a+ 1

Also

0 = a.a’ (B IV)

= a.(a’ + 0)  (B III)

= a.a’ + a.0   (B II)

= 0 + A.0  (B IV)

= a.0

Theorem IV: (Absorption laws)

A + (a.b) = a and a.(a + b) = a.

Proof:

a =  a.1

= a.(1 + b)

= (a.1) + (a.b) (B II)

= a + a.b

Also

a. (a + b) a.a + a.b

= a + a.b  (Th. II)

= a

And

Proof : we prove the first implication,

(a + x).(a’ + x) = a.a’ + x  (B II)

= 0 + x = x

(a + y).(a’ + y) = a.a’ + y

= 0 + y = y

(a + x).(a’ +x) = (a + y).(a’ + y)

Similarly a.x + a’.x = (a + a’).x

= 1.x = x

And a.y + a’.y = (a + a’).y

= 1.y = y

Implying x = y

Theorem VI: addition and multiplication are associative,

i.e., a + (b + c) = (a + b) + c and a.(b.c) = (a.b).c

proof: Let a + (b + c) x, (a + b) + c = y.

Then a.x = a.{a+ (b + c)}

=a’    ( Th. IV)

And

a.y = a.{(a + b) + c}

= a.(a + b) + a.c

= a + a.c = a

Also x = a’.{a + (b + c)}

= a’.a + a’.(b + c)

= 0 + a’ .(b + c)

= a’.(b + c)

And

a’.y = a’.{(a + b) + c}

=a’. (a + b) + a’. c

= {a’.a + a’.b} + a’.c

= {0 + a’.b} + a’ .c

= a’. b + a’.c = a’.(b + c)

Hence a.x = a. y and a’.x = a’.y

   (Th. V)

Also let a.(b . c) = p and (a . b) . c = q

Then  a + p = a + a(a.(b.c)}   (B II)

= (a + a). (a + b.c)

= a.(a + b.c)

= a    (Th. IV)

Also a + q = a- {(a.c).c}

= (a + a.b).(a + c)   (B II)

= a.(+c)

= a . a + a.c

= x = a.c = c

a’ + p = a’ + {a, (b, c)}

= (a’ + a).(a’ + b.c)

= 1.(a’ +b .c) + a’ + b.c

= (a’ + b). (a’ + c)

= {1 .(a’ + b).(a’ + c)}

= {(a + a’). (a’ + b)}. (a’ + c)

= (a’ +a.b). (a’ +c)

= a’ + (a.b).c   (B II)

= a’ + q

Here a + p = a + q

And a’ + p = a’ + q

   (Th. V)

Note: some authors include associative law in the definition of Boolean Algebra.

Theorem VII. The complement of is uique.

Proof: let there are two complements a’ and a’’.

Then a + a’ = 1, a.a’ = 0

a + a’’ = 1, a.a’’ = 0

now a’ = 1.a’ = (a + a’’).a’

= a.a’ + a’’.a’

= 0 + a’’ . a’ = a.a’’ + a’’ .a’

=a’’.(a + a’) = a’’.1

= a’’

Here a’ is unique.

Theorem VIII (a’)’ = a

Proof : (a’)’ = a.a’ + (a’)’

= {a + (a’)’}. {a’ + (a’)’} …(B II)

= {a + (a’)’}.1

= {a + (a’)’}.{a + a’}

= a + (a’).a’ = a+ 0

=a

The theorem is direct result of B IV

Theorem IX: complements of identities

0’ = 1 and 1’ =0.

Proof: the proof follows from B III and B IV as

1 + 0 = 1 and 1.0 =0

Theorem X: de morgan’s law.

(a +b)’ = a’.b’

And (a.b)’ = a’ + b’

Proof:

(a’.b’).(a + b) = (a’.b’) . a + (a’.b’).b

= a’.(b’.a) + a’.(b’.b)  (Th. VI)

= a’.(a.b’) + a’.0

= (a’.a).b’ + 0 = 0.b’ + 0

= 0

Also (a’.b’) + (a + b) = (a’.b’ + a) + (a’ + b + b)

= (a + a’).(a + b’) + (a’ + b).(b’ + b)     (B II)

= 1.(a + b’) + (a’ + b).1

= (a + b’) + (b + a’)

= a + (b’ + b) + a’ = a  1 + a’

= a + a’ = 1

Hence a’.b’ is the complement of (a + b)

Also (a.b).(a’ + b’) = {(a.b).a’} + {(a.b).b’}

= {a’. (a.b)} + a. {(b.b’)}

= {(a’.a).b}  {a.0}

= 0.b + a.0 = 0 + 0 = 0

And  (a.b) + (a’ + b’) = (a’ + b’) + (a.b)

= {(a’ + b’) + a}.{(a’ + b’) + b}

= {a + (a’ + b’)}.{a’ + (a’ + (b’ + b)}

= {(a + a’) + b’}. {a’ + 1}

= (1 + b’). (a’ + 1) = 1.1 = 1

Hence a’ +b’ is the complement of a.b.

Duality

Any result that is obtained from the axioms of Boolean Algebra remains true if  + and are interchanged along with the interchange of 0 and 1 throughout the statement of the results.

Partial ordering

Just like inclusion of a set into another set as the case of subsets, we have the idea of partial ordering in Booleean algebra. Thus

if a.b = a,a,b

Or a + b = b

It can also be stated as

if a.b’ =0

And if

With the above notation we define as:

If ( B, + , … 0, 1) is a Boolean algebra, then is a partially ordered set with greatest element 1 and least element 0. Moreover each pair {a, b} of elemtns has a least upper bound a + b and a greatest lower bound

Switching algebra

The knowledge of Boolean algebra received recognition mainly due to its application in electrical circuits C.E. Shanon in his paper in 1938 was the first to introduce circuit algebra as method of dealing algebraically with relay circuits. The emergence of digital computers brought this system into significance as a part of pure mathematics applied to electrical engineering.

With the symbols in the beginning of the chapter, we consider the system [0,1] +, ]. The system is a Boolean with the composition tables.

Also 0’ 1 and 1’ = 0. In an electrical circuit if a switch is closed (or on) we denote it by 1 and when it is open (or off), it will be denoted by 0.

Switches in parallel are denoted by + and in series by . Thus

The above figure gives two switches in parallel and the truth table is as under:

Hence we see that the current does not flow only in the case when both the switches are open.

Switches in series are denoted by . The figure of two switches in series is

The truth table is as under:

In this case, the current flows in the case when both the switches are closed.

Some other circuits are below shown:

Draw figures to show the following equivalence of switches:

(i)x.y = y.x

(ii) x+ y = y + x

(iii) x + y .z = (x  + y). (x + z)

(iv) x.(y + z) = x.y + x.z

(v) x + x’ = 1 and x.x’ = 0

(vi) x + (y +z) = (x + y) + z

(vii) x.(y.z) = (x.y).z

In the above, we have studied cases where the switches are either open or closed independent of each other. Now we shall see same cases where x represents the closed switch and x’ the open in the same circuit.

The Two Switch Problem: we have to design a circuit such that the light of a bulb can be turned on or off by the change in state of any of the two switches in the circuit. The figure is shown below:

There are tow switches different from ordinary switches as they have two way contacts. While state closes B and opens EF, closes EF and opens AB, similar is the case with so that the current flows in the state

If the circuit is on with either a change in any switch will make the circuit off. Again any change in the state of any of the switches will make the circuit on. We can write

A three switch can be designed as

A n switch circuit can similarly be designed.

Truth table

The truth of the above Boolean relations can also be shown with the help of a truth table. If a variable x is denoted by 1, x’ is denoted by 0. For + sometimes , and similarly for are used. Thus to prove the relation a(a + b) = a, we form the table

Duality: Another definition

In every (correct) formula of Boolean algebra, we can interchange addition and multiplication. However when an equality fulfills the laws of Boolean Algebra involves the ‘special’ elements 0 and 1, then the interchange of Boolean addition and multiplication in this equality must be followed by the interchange of elements 0 and 1. For instance the validity of

(A + B) (A + 1) + (A + B) (B + 0) = A + B

Results in

(AB + A0). (AB +B1) = AB

Right hand and left hand limits

Let h be a small positive number. Left hand side limit of f(x) when , is denoted by f(a -0) and is defined as:

Right hand side limit of f(x), when , is denoted by f(a + 0) and is defined as:

exists if

Indeterminate forms

If a function f(x) takes the form , then say that f(x) is indeterminate at x=a. Other Indeterminate Forms are .

L’ hospital’s rule

If and are functions of x such that , then

The form

This form can easily be reduced either to form of .

Example:

Evaluate

Solution:

The form

This can also be reduced to the form

Example:

Evaluate

Sandwich Theorem (or Squeeze principle)

If f, g, h are functions such that for all x in the neighborhood of a and if ,

Algebra of limits

,

Evaluation of exponential limits of the form

Result:

(i) If

(ii) If

A function f(x) is said to be continuous at x=a if given , there exists such that | f(x) – f(a) | such that | x –a | < .

Alternative definition: f(x) is said to be continuous at x =a if value of f(x) = limit of f(x) at x = a

Or if lim f(x) = f(a)

Of if lim f (a+h) = lim f(a –h) = f(a) …….Equation   1

If the condition (1) is not satisfied, then f(x) is said to be discontinuous at x =a.

Note: In order to rest continuity of a function at a point, we verify the equation (1). This is the working rule of continuity.

Definition of continuity in an interval

Let ‘h’ be a small positive number. A function f(x) is said to be continuous in closed interval [a,b], i.e., if:

(i) f(x) is continuous in open interval (a ,b), i.e.,

Continuity and Differentiability

(ii) f(x) is continuous at x = 1 from right

(iii) f(x) is continuous at x=b from left

Kinds of discontinuity

(1) Removable Discontinuity: A function f(x) is said to have removable discontinuity at x=a if exists or but . In this case value of function and limit of function are not equal.

(2) Discontinuity of First Kind: If both exist and f (a -0) f (a +0), then f(x) is said to have discontinuity of first kind or ordinary discontinuity at x=a.

(3) Discontinuity of second kind: A function f (x) is said to have discontinuity of second kind at x = a if f(a-0) or f (a + 0 ) or both do not exist.

Properties of continuous function

(1) if f is continuous on a closed interval [a,b], then it is bounded in this interval.

(2) if f is continuous in [a,b] and f(a) and f(b) have opposite signs, then there is at lest one value of x=c such that f(c) = 0 and a<c<b.

(3) If a function f is continuous in closed interval [a,b], then f(x) takes at least once all values of between f(a) and f(b).

Important note: If there is no gap in the graph of the function in a certain range [a,b], then f(x) is said to be continuous in [a,b].

Some standard continuous functions

(1) Every constant function is everywhere continuous.

(2) The identity function I (x) = x is everywhere continuous.

(3) The modulus function f(x) = |x| is continuous .

(4) The exponential function f(x) = and a >0 is continuous everywhere.

(5) The logarithmic function is continuous

(6) Every polynomial function: 

Differentiability

A function f(x) is said  to be differentiable at x = a if exists and is finite. Let ‘h’ be a small positive number.

Right hand derivative of f(x) at x = a is denoted by Rf’ (a) and is defined as Rf’ (a) =

Left hand derivative of f(x) at x = a is denoted by Lf’ (a) and is defined as Lf’ (a)=

Alternate definition: A function is said to be differentiable at x = a if Rf’ (a) and Lf’ (a) both are finite and Rf’ (a) = Lf’ (a).

The common value is denoted by f’ (a).

Relation between continuity and differentiability

(1) if f(x) is differential at x = a, then it is continuous at x =a.

(2) if f(x) is continuous at x = a, then there is no guarantee that f (x) is differentiable at x=a.

(3) If f (x) is not differentiable at x =a, then it may or may not be continuous at x=a.

(4) if f(x) is not continuous at x=a, then it is not differentiable at x=a.

Some important results on differentiability

(1) Every polynomial function, constant function, exponential function is differentiable .

(2) if f(x) and g(x) are differentiable, then are differentiable functions provided in the last case g(x) .

(3) The composition of differentiable function is differentiable function.

Darboux Theorem

If f(x) is differentiable in the closed interval [a,b] anf f’ (a) and f’ (b) are of opposite signs, then there is a point , ie., a < c < b such that f’ (c) = 0.

Graphical Meaning of Differentiability

F’(a) represents slope of tangent at x=a. A function f(x) is said to be differentiable at x=a if it (tangent) has unique slope at x = a.

It is obvious from the diagram that the function y=f(x) is maximum at P and is minimum at Q.

At both these points tangent is parallel to x-axis, so that its slope is zero.

Maxima and minima

for both maximum and minimum.

Working Rule

(1) Find out from the equation f(x,y) = c and put . Solve it for x. Let, on solving x=a, x=b.

(2) If = negative for x=1, then y is maximum at x=a.

If positive for x=b, then y is minimum at x=b.

(3) If both are 0 for some value of x, we have to find . If it is 0 then f(x) is maximum if , minimum if .

However, if is zero for some x and for that x then f(x) is neither minimum nor maximum for that x. It is called the point of inflexion. At such point the curve changes from concave to convex or from convex to concave.

(4) Between two minimum there is at least one maximum. Similarly between to maxima, there is at least one minima.

Maxima and minima occur alternatively.

(5) At a maxima, changes sign from positive to negative and at a minima, changes sign from negative to positive.

Concavity and Convexity

A curve is convex and concave at some point ‘P’ to x-axis according as: is positive or negative at P.

Critical points

(1) The values of x for which f’ (x) = 0 are called stationary values or critical values of x and corresponding values of f(x) are called stationary or turning values of f(x).

(2) if f(x) is not differentiable at x=a, then also x=a is called critical points. Also f’ (x) = 0 has no solution in an interval (a,b) then (a,b) is called range of those points where critical points do not exist.

Important point:

(1) Maximum and minimum values of .

(2) Least value of x+y subject to condition xy=constant, occurs when x=y.

is defined as slope of gradient of tangent PT.

We also define

= slope of tangent,

Slope of normal =

Equation of Tangent PT is

Equation of normal PN is

PM is perpendicular from P on x-axis.

By ,

By

We define:

Sub tangent = TM =

Sub normal = MN =

Length of Tangent = PT =

Length of normal = PN =

Where is point P.

The tangent is parallel to x-axis if:

The tangent is parallel to y –axis if:

Important Note:

Tangent at the origin is obtained by equating to zero the lowest degree terms, provided the curve passes through origin.

Definition of Angle of Intersection

Suppose two curves cut at P. Let be gradient of the two tangents to the two curves at the point of inserction. Angle between the two curves at P is defined as angle between the two tangents at P.

The two curves cut orthogonally if

If f is a function of several variables , then the derivative of f w.r.t. keeping other variables constant is called partial derivative of f w.r.t. and is denoted by or by and is defined as:

provided the limit exists.

3. if f=f(x,y) and partial derivates are continuous then

4. if f=f(x,y), then . if g=g (x,y,z), then .

if f=f(x,y) and , then:

And so

Homogenous Function

If f is a homogenous function of x and y of degree n, then it may be put in the form .

Example:

Euler’s theorem

If f is a homogenous function of x and y of degree n , then

Deduction: From this, we get some important results as follows:

Note: The above results are very important for doing problems.

Note:

(1) if , then degree of f is and f is homogenous.

(2) If

Then is homogenous degree 3 -4 = -1.

Formulas for Area of a Triangle:

The area of a triangle is denoted by the symbol delta ( )

We shall appeal to the formula:

And the half angle formula:

etc.

Where “s” is the semi circumference of the triangle or,

We shall now derive different formula for the area of triangle:

First formula for the area of a triangle:

Second formula for the area of a triangle:

Now , as 2s = (a+b+c)

Third formula for the area of a triangle:

Now using sine law:

The Half angle formulas are stated below:

If ABC is a triangle , A , B and C are the three angles of the triangle and a , b , c are the sides opposite to the corresponding angles and

“s” is the semi perimeter or ,   , Then:

Proof of Half angle formula:

First of all let’s prove the half angle formula for

Using the cosine law:

Now using the trigonometric sub-multiple angle formula:

Now , let us prove the half angle formula for

Using the cosine law:

Lastly , Dividing by we get :

Similarly we can also prove the half angle formula of angle B and C.

Projection law states that in any triangle:

Where , A , B , C are the three angled of the triangle and a , b , c are the corresponding opposite side of the angles.

Projection law or the formula of projection law express the algebraic sum of the projection of any two side in term of the third side.

Proof of Projection law:

To prove the projection law we shall take the help of sine law which states that:

Thus ,  using the above formula for sine law we can easily deduce the formula for projection law as:

Now using the trigonometric addition formulas we can replace by

And again in a triangle A+B+C = 180, so:

Now we can re write the original formula as:

And we can also deduce the second and third formula for projection law similarly.

Sine law states that in any triangle ABC:

And also with some mathematics we can also prove the following :

Which is also closely related to the sine law.

Where , a , b and c are sides of a triangle , A , B and C are angles opposite to sides a , b, and c correspondingly and  R is the circum-radius of the triangle as shown in following figure:

sine law

Proof of Sine Law:

Let us consider a Triangle ABC  , placed in the standard position with the vertex A at the origin and side AB along the positive x-axis as in the figure below:

Now,

* The co-ordinates of A are ( 0 , 0 )

* The co-ordinates of B are ( c , 0 )

* The co-ordinates of C are ( b cosA , b sinA )

Now ,

Area of the triangle ABC is:

Similarly if we place the other angles B and C in the standard position or origin the we also get the following by similar method:

Area of triangle =

By combining these three formulas we get:

Hence:

Now , we shall prove the sine law with another approach to find the sine relation or connection between the radius of circum-circle with the sine of angles and their opposite sides.

Or now we shall prove:

To prove this we denote the circum-centre of the triangle ABC by “O”.

There might be three cases in which angle “A” is either acute (fig. a) , obtuse (fig. b) or right angled (fig. c).

In any case let us join the circum-centre of the triangle “O” to B and produce it in another side up to “D” as shown in figures blow:

sine law

In first two cases above:

angle BDC = 90 degrees ( Because angle made on circumference from diameter is always 90 degrees)

Thus:

Or ,

And in the third case it is obvious that : because in this case angle A is 90 degrees and sine 90 = 1.

Thus we can now conclude: