Boolean Algebra

Boolean Algebra is a branch of mathematic logics  whose use of symbols and  theory, set to represent the logical operations in the form of mathematics. This is the first logic which uses algebra and different methods for combining symbols used in proofs as well as deduction.

A Boolean Algebra is defined as:

It is  a set, having two special elements i.e, 0 and 1.
Algebra having three types of operations , which are
sum of two elements ("+"),
product sum of two elements  ("*") and
complement sum of two elements (" ' " or "prime")

these operations need to satisfy the Commutative axiom, Distributive axiom , Identity axiom (not including the boundedness identities) and Complement axiom.

These above axioms are almost equal to commutative property ,distributive property, identity property and complement property. Here we call them axioms because they are assumptions.

Boolean Algebra contains:

A  set of all propositions
The special characteristic elements - True  (1) i.e T  and False (0) i.e, F.
Three operations are
AND (product),
OR (sum) and
NOT (complement).

Laws of Boolean Algebra Axioms

To do any kind of operations using real numbers, they  depends on commutative axiom, associative axiom, and distributive axiom. In algebraic form these axioms  are expressed with letters or symbols, which are used to indicate an unknown number.

Commutative axiom

The commutative axioms explains that, numbers can be used for addition or multiplication in any manner.

Commutative axiom of Addition:

a + b = b + a        ( using addition law )

Commutative axiom of Multiplication:

a(b) = b(a)           ( using multiplication law )

Associative axiom

The associative axioms explain that, numbers which are used in addition or multiplication, also it can be grouped or regrouped in anyorder.

Associative Law of Addition:

a+(b+c) = (a+b)+c      ( using addition law )

Associative Law of Multiplication:

a(bc) = (ab)c                ( using multiplication law )

Distributive axiom

The distributive axioms are used for  both addition as well as  multiplication and state the following.

Distributive axiom for addition :

a(b + c) = ab + ac        ( using addition law )

Distributive axiom for multiplication :

(a + b)c = ac + bc             ( using multiplication law )

Identity axiom

Identity axiom for multiplication :

x · x = x                 ( using multiplication law )

Identity axiom for addition :

x + x = x                ( using addition law )



Zero Property in Boolean algebra axioms

0 · x = 0                  ( using multiplication law )

0 + x = x                   ( using addition law )

One Property  in Boolean algebra axioms

1 + x = 1                  ( using addition law )

1. x = x                     ( using multiplication law )
Examples on Boolean Algebra:

1)  `5xx(8+9)` = `5xx8 + 5xx9` ( USING DISTRIBUTIVE AXIOM  FOR MULTIPLICATION )

= `40 + 45`

=` 85`

= 5 x 8+5 x 9

2)  `3+7 = 7+3` ( BY USING COMMUTATIVE AXIOM FOR ADDITION)

3) `9xx5 = 5xx9 (` BY USING THE COMMUTATIVE AXIOM FOR MULTIPLICATION)

4)` 6xx1 = 6 ` (BY USING PROPERTY FOR ONE)
Practice Problems on Boolean Algebra:

1) Prove that  C+(A×B)=(C+A)×(C+B) by using Boolean algebra axioms

2) Prove that B+(C×A)=(B+C)×(B+A) by using Boolean algebra axioms

Practical uses for Algebra

Algebra been used in all parts of our life activities either directly or indirectly.  If we want to find some interest which may recur with amount, we can apply progressions.  To see some different arrangements or combinations between available sources, we apply permutation and combination. To prove a generalized statement, we can apply mathematical Induction so on.

Let us see few problems of this kind.
Example Problem on Practical Uses for Algebra:

Ex 1: Peter buys a truck for dollar 120,000.  He pays half of the amount by cash. That is, dollar 60,000.  The remainingdollar 60,000, he pays in 12 annual installments of dollar 5000 each.  If the rate of interest is 12% and he pays with the installment the interest due on the unpaid amount, find the total cost of the truck.

Sol:  First installment = 5000 + 12% of 60,000

= 5000 + `12/100 xx` 60000

= $ 12,200.

Second Installment = 5000 + 12% of 55,000   = 5000 + `12 /100 xx` 55,000

= $ 11600.

Third Installment   = 5000 + `12/100 xx` 50,000

= $ 11,000

Since the common difference of the amount is $600,

Total cost of the shop = 60,000 + sum of the 12 installments.

= 60,000 + `12/2` [2` xx` 12,200 + (12 – 1) (- 600)]

= 60,000 + 6 [24,400 – 6,600]

= $ 1, 66,800.
More Example Problem on Practical Uses for Algebra:

2. A father is three times a old as his son.  In 12 years time, he will be twice as old as his son.  Find the present ages of father and son.

Solution: Let x and y be the present ages of father and his son respectively.

Given:  x = 3y `implies` x – 3 y = 0 -------- (1)

Also, x + 12 = 2 (y + 12)` implies` x + 12 = 2y + 24

`implies` x – 2y = 12 -------------------------- (2)

Therefore (1) – (2) `implies` x – 3y = 0

- x + 2y = - 12

`implies` - y = - 12 `implies` y = 12.

Therefore (1)` implies` x – 3 (12) = 0

`implies` x = 36.

Therefore the age of the father is 36 and his son’s age is 12.

Steps to Solving Quadratic Equations

In mathematics, an equation with second degree and one variable is called quadratic equation. The general equation for quadratic equation is given by,

A x2 + B x + C =0

Here x represents the variable and A, B, and C are constants, with a ≠ 0. (When a = 0, the equation is named as linear equation.)

The constant A, B, and C are expressed as quadratic coefficient, linear coefficient and the constant expression or release expression.

Steps to Solving Quadratic Equations Example 1:

5x² - x – 6 = 0, solving the factor for the given quadratic equation.

Solution:

Now, we can find the factor for the given quadratic equation

5x2 + 5x - 6x - 6 = 0

Now get the value for 5x from the primary term and 6 from secondary term.

5x (x + 1) - 6(x + 1) = 0

Now we can combine the similar term (x +1)

(5x - 6) (x + 1) = 0

To get the value for x we can associate the factor to zero

x + 1 = 0   or   5x – 6 = 0

x = - 1        or   5x = 6

x = 6 / 5

Thus, the factors  x1 and x2 are -1, 6/5.
Steps to solving quadratic equations Example 2:

3x² - 6 = -3x solving the factor for the given quadratic equation and solving the sum and product of quadratic equation.

Solution:

Fetch the -3x over: 3x² + 3x - 6 = 0

Separate 3x as 6x and -3x,

3x² + 6x - 3x - 6 = 0

Take out 3x from first two terms and 3 as common from next two terms

3x(x + 2) – 3(x + 2)  = 0

Thus, the factors are: (3x - 3) (x + 2) = 0

Modulate both expressions to zero: 3x - 3 = 0 and x + 2 = 0

3x - 3 = 0            x + 2 = 0

3x = 3                   x = - 2

x =  3/3

x = 1

So, the factors for x are  1, -2

Sum of the roots:

To find sum of roots consider the factor as x1 and x2

The sum of the roots  = x1 + x2 = (1) + (-2)

Sum of the roots = -1

Product of the roots:

To find product of roots consider the factors as x1 and x2

The product of the roots is given by x1x2 = (1)(-2)

Product of roots = -2


Quadratic Equation

ax2 +bx + c = 0


Values of

‘a’, ‘b’ and ‘c’


One Root

(x1)


Other Root

(x2)


Sum of Roots

(x1 + x2)


Product of Roots

(x1x2)

3x² +3x - 6 = 0


a = 3, b = 3, c = -6


1


-2


-1


-2

Steps to Solving Quadratic Equation Example 3:

Solving the value for x, to the given quadratic equation x2 + 5x + 4 = 0.

Solution:

Steps 1: To find the factor for the given quadratic equation, find the multiplicative value for the 4 and the sum of root value for 5

x2 + 1x + 4x + 4 = 0

Steps 2: Now obtain x as similar from the primary term and 4 as similar from last two term.

x (x +1) + 4(x + 1) = 0

Steps 3: Now we can unite similar term (x +1)

(x + 1) (x + 4) = 0

Steps 4: To obtain the value of x we can associate the factor to zero

x + 1 = 0   or   x + 4 = 0

x = - 1        or   x = -4.

Steps 5: Thus, the factors are x are -1, - 4.

Solve Algebra Squared Fraction

Fractions:

In algebra, A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

In this article we are going to see how to solve algebra squared fraction and some example solved problems on algebra squared fraction.

Solve Algebra Squared Fraction :

Steps to solve algebra squared fraction:

Step 1: We know that ( a/b)^n = a^n / b^n . So we can take square for the numerator and denominator.

Step 2: If it is possible we can simplify it.

Let us see the example problems.
Example Problems on Solve Algebra Squared Fraction :

Problem 1:

Solve (4/5)^2

Solution:

Given , (4/5)^2

We need to squared the given fraction .

(4/5)^2

We know that ( a/b)^n  = a^n / b^n .

We can apply the above rule,

(4/5)^2 = 4^2 / 5^2

= 16 / 25

Answer: (4/5)^2 = 16 / 25

Problem 2:

Solve (3x / 4xy) ^ 2

Solution:

Given, (3x / 4xy) ^ 2

We need to find the squared the given fraction,

We know that (a/b)^n = a^n / b^n .

(3x / 4xy) ^ 2 = ((3x) ^ 2) / ( (4xy)^2)

= (9x^2) / ( 16x^2y^2)

now we can simplify it,

Divide by x^2 on both numerator and denominator,

(9x^2) / ( 16x^2y^2) = (9x^2)/x^2 / ( 16x^2y^2) / x^2

= 9 / 16y^2

Answer: (3x / 4xy) ^ 2   = 9 / 16y^2

Problem 3:

Solve (( x^2 - 4 ) / ( x + 2) )^2

Solution :

Given , (( x^2 - 4 ) / ( x + 2) )^2

We need to find the squared the given fraction,

We know that ( a/b)^n = a^n / b^n .

(( x^2 - 4 ) / ( x + 2) )^2 = ( x^2 - 4 )^ 2 / ( x + 2)^2

Before that we can simplify the given fraction,

(( x^2 - 4 ) / ( x + 2) )^2 = (((x+2)(x-2))/ ( x + 2))^ 2

= ((x - 2)^2)

= x^2 - 2x + 4

Answer: (( x^2 - 4 ) / ( x + 2) )^2 = x^2 - 2x + 4

Results and Discussion Section

Here we are going to see about the result and discussion, whether the solution to the given equation is correct or not. By simplifying the equation we find the value for the particular variable. when we substitute the value in the equation the equation balances on both sides.

For example: x – 4 = 0.

Here    x – 4 = 0

Add 4 on both sides. We get

x = 4

The solution is x = 4

According to the topic we have to check whether x = 4 is a solution to the given equation or not. By substituting the value the in equation we can see that both sides have same value.
Example Problem for Result and Discussion Section:

Given equation is discussion section  x + 7x + 5x + 3x – 5 + x = 29

Solution:

Given question is x + 7x + 5x + 3x – 5 + x = 29

Step 1:

Arrange the given question as per the x term and the numbers

x+ x + 3x + 5x + 7x – 5 = 29

Step 2:

Add the x term first

2x + 3x + 5x + 7x – 5 = 29

Step 3:

Add all the  ' x ' term in the equation

17x – 5 = 29

Step 4:

Add both sides by 5.Then we get

17x = 29 + 5

Step 5:

17 x = 34

Divide by 17 on both sides.

`(17x)/17` = `34/17`

x = 2

Discussion section about the result:

The result is 2. Here we are going to discuss about the result for given equation, whether the solution x =2 is correct or not.

Given question is x + 7x + 5x + 3x – 5 + x = 29

The result is x =2

Here we need to substitute the x value in the equation if that both side value is equal means the solution is correct. Otherwise the solution is not correct.

x + 7x + 5x + 3x – 5 + x = 29

Substitute the value x = 2 in the given equation

2 + 7( 2) + 5( 2) + 3(2) – 5 + 2 = 29

2 + 14 + 10 + 6 – 5 + 2 = 29

34 – 5 = 29

29 = 29

Both the sides are equal, so given solution is correct for the equation.
One more Example Problem for Result and Discussion Section:

Solve discussion section  16x - 12 = 20

Solution:

Add 12 on both sides. we get

16x - 12 + 12 = 20 + 12

16x  = 32

Divide by 16 on both sides.

`(16x )/ 16` = `32 / 16`

x = 2

whe we substitute x =2 in the equation 16x - 12 = 20. we get

16(2) - 12 = 20

32 - 12 = 20

20 = 20.

Thus the equation satisfies.

Raw Score Formula

Raw score is defined as the unique, unchanged transformation or calculation. When we bring together the information, in which the numbers we get hold of are known as raw score or raw data. Raw score can able to find the precise location of each data in a given distribution.

percentage of sampling error = (value of statistic-value of parameter)*100.

In this article we are going to see about raw score formula.
Raw Score Formula:

The raw score formula for variance can be given as follows,

Raw score formula

Where `sigma` and X indicates mean.
Raw Score Formula Related to Z Score Formula:

Z-score formula can be given as follows

Z = (x) - (X) / (S.D)

where x indicates a data value

X indicates mean (average of the given values)
S.D indicates standard deviation.

Using the formula of Z score we can find the raw score formula

But to find the raw score formula we have to solve this formula for x

hence the raw score formula for x can be given as follows


Z* (S.D) + X= x

Where x indicates a data value or raw score

X indicates mean (average of the given values)
S.D indicates standard deviation.

Example problem for raw score formula:

Let us take an example where the mean value X has to be 280 and the standard deviation is let to be 10.The data values are given as follows x1 = 300 and x2 = 285.Find the raw score value using raw score formula.

Solution:

Z = [(300) - (280)] / (10)


Z1= 2

Z= [(285) - (280)]/ (10)



Z2 = .5
Finding Raw Score using raw score formula.

Z = 2.2 hence


2.2 = (x-280)/10

(2.2) * 10 + 280 = x

x = 302

Z = 1.1

1.1 = (x- 280) /10.

(1.1) * 10 + 280 = x

x = 291

Solve a Parabola with Fraction

Parabola is a half circle and a curve formed by an intersection of right circular cone. A set of all points are same distance from a fixed line is called as directrix and also fixed point called as focus not in the directrix. The midpoint between the directrix and focus of the parabola is called as vertex and the line passing through the vertex and focus is called as axis. Let us see about how to get the value of vertex, latus rectum, focus and axis of symmetry in parabola equation. Let us see about solve a parabola with fraction in this article

The general form the parabolic curve is y = ax^(2) + bx + c or  y^(2) = 4ax . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.

Example 1 for Solve a Parabola with Fraction – Vertex:

Find the vertex of a parabola where y = x^ (2) + 7/2 x + 3

Solution:

Given parabola equation is y = x^ (2) + 7/2 x + 3

To find the vertices of a given parabola, we have to plug y = 0 in the above equation, we get,

0 = x^ (2) + 7/2 x + 3

Now we have to factor the above equation, we get,

So x^ (2) + 3/2x + 2x + 3 = 0

x(x + 3/2) + 2 (x + 3/2) = 0

(x + 3/2) (x + 2) = 0

From this x + 3/2 = 0 and x + 2 = 0

Then x = - 3/2 and x = - 2

So, the vertices of given parabola equation is (-3/2, 0) and (-2, 0) .

Example 2 for Solve a Parabola with Fraction – Focus:

What is the focus of the following parabola equation where y^ (2) = 10x

Solution:

Given parabola equation is y^ (2) = 10x is of the form y^ (2) = 4ax

To find the focus of a parabola use the formula to find the value of focus value.

We know that the formula for focus, p = 1 / (4a)

Now compare the given equation y^(2) = 10x with the general equation y^ (2) = 4ax . So, that 4a = 10

From this p = 1 / (4a) = 1 / 10

So, the focus of a parabola equation is (0, 1/10) .

Other Example Problems to Solve a Parabola Fraction

Example 3 for Solve a Parabola with Fraction – Axis of Symmetry:

What is the axis of symmetry of the parabola where y = 4x^ (2) + 8x + 14 ?
Solution:

Given parabolic curve equation is y = 4x^ (2) + 8x + 14

From the above equation, a = 4 and b = 8

So the axis of the symmetry of the given parabola is -b/ (2a) = -8/ (2 xx 4) = -8/8 = - 1

Therefore, the axis of symmetry for a given parabolic curve equation is -1 .

Example 4 for Solve a Parabola with Fraction – Latus Rectum:

Find the latus rectum of the given parabola equation y^ (2) = 6x

Solution:

The given parabola equation is y^ (2) = 6x

To find the latus rectum, we have to find the value of p.

The parabola equation is of the form y^ (2) = 4a

Here 4a = 6

So, p = 1/ (4a) = 1/6

The formula for latus rectum is 4p .

From this, the latus rectum of the parabola is = 4p = 4 (1/6) = 4/6 = 2/3

Therefore, the latus rectum for the parabola equation is 2/3 .