Prime Factorization of the Denominator

Introduction 

In number theory, the prime factors of a number are the set of prime numbers which can divide the number without a remainder. Finding the prime number factors of a number is called as prime factorization.

For example, the prime factorization of 15 is 3, 5. Here, 3 and 5 are prime numbers and the product of 2 and 5 gives 10.

Prime factorization of the denominator is mainly used for reducing fractions. If the numerator and denominator of a fraction has common terms, then to reduce the fraction prime factorization of numerator and denominator are used.

Prime Factorization of the Denominator

Here we will see an example for the prime factorization of the denominator.

To reduce the fraction, `16/26`first we have to find the prime factorization of the numerator and the denominator.

Prime factorization of the numerator (16) = 2 * 2 * 2 * 2

Prime factorization of the denominator (26) = 2 * 13

Now the fraction can be written as,

`16/26` = `(2 xx 2 xx 2 xx 2)/(2 xx 13)`

= `(2 xx 2 xx 2)/13`

= `8/13`

So, `8/13` is the reduced form of `16/26`

Example Problems on Prime Factorization of the Denominator

Here we will see some example problems on prime factorization of the denominator.

Example 1

Reduce the fraction, `152/456`

Solution

To reduce the given fraction, first we have to find the prime factorization of the numerator and the denominator.

Prime factorization of the numerator (152) = 2 * 2 * 2 * 19

Prime factorization of the denominator (456) = 2 * 2 * 2 * 3 * 19

So, the given fraction can be written as,

`152/456` = `(2 xx 2 xx 2 xx 19)/(2 xx 2 xx 2 xx 3 xx 19)`

= `1/3`

So, `1/3` is the reduced form of `152/456`

Example 2

Reduce the fraction, `12/46`

Solution

To reduce the given fraction, first we have to find the prime factorization of the numerator and the denominator.

Prime factorization of the numerator (12) = 2 * 2 * 3

Prime factorization of the denominator (46) = 2 * 23

So, the given fraction can be written as,

`12/46` = `"(2 xx 2 xx 3)/(2 xx 23)`

= `(2 xx 3)/23`

= `6/23`

So, `6/23` is the reduced form of `12/46`

Odd Natural Numbers

Introduction of odd natural numbers:

In mathematics, there are two conventions for the set of natural numbers: it is either the set of positive integers {1, 2, 3,...} according to the traditional definition; or the set of non-negative integers{0, 1, 2, ...} according to a definition first appearing in the nineteenth century.


Odd Natural Numbers:

An odd number is an one kind of integer value that is not consistently divisible by 2. A proper description of an odd number is that it is an integer of the appearance n = 2k + 1, where k is an integer. An even number has the appearance n = 2k where k is an integer.

Properties of odd Natural numbers:

1. Associative law for odd natural numbers:

(x + y) + z = x + (y + z). – Associative law of addition.

(x * y) * z = x * (y * z). - Associative law of multiplication.

2. Commutative law for odd natural numbers:

x + y = y + x. – Commutative law of addition.

x * y = y * x. - Commutative law of multiplication.

3. Cancellation law for odd natural numbers:

If x + y = y + a, then x = y. - cancellation law of addition.

If x * z = y * z, then x = y. - cancellation law of multiplication.

4. Distributive law with multiplication on the left odd natural numbers:

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

5. Distributive law with multiplication on the right odd natural numbers:

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

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Examples for Odd Natural Numbers:

Example 1: closure property in addition:

If 1, 3 are natural numbers, and then 1+ 3=4 is also a natural number

And then,

If 1, 3 are natural numbers, and then 1 * 3 = 3 is also a natural number.

Example 2: commutative property in addition:

If 1, 5 are natural numbers, and then 1 + 5 = 5 +1=6 is also a natural number.

And then,

If 1, 5 are natural numbers, and then 1 * 5 = 5 *1= 5 is also a natural number.

Example 3: associative property in addition

If 1, 3, 5 are natural numbers, then 1 + (3 + 5) = (1+ 3) + 5=9 is also a natural number.

And then,

If 1, 3, 5 are natural numbers, then 1 * (3 * 5) = (1* 3) * 5=15 is also a natural number

Linear Algebra Friedberg Solutions

Introduction :

Fried berg linear algebra is mainly used to find the solution of linear equations in some unknowns. Fried berg Linear algebra has a demonstration in analytic geometry and is generalized in operator theory. Fried berg linear algebra associates with the families of vectors called vector spaces, and with functions contain one input vector and output another vector, according to rules and regulations. Fried berg Linear algebra deals with the theory of systems of linear equations, vector spaces and matrices.

Fried berg Linear Algebra Example Problems with Solutions:-

Ex 1:

Solve the linear equation -4(-x - 4) = 5x - 35

Sol:

Given -4(-x - 4) = 5x - 35

Multiply factors in left term

4x + 16 = 5x - 35

Subtract 16 to both sides

4x + 16 - 16 = 5x - 35 - 16

Group like terms

4x = 5x - 51

Subtract 5x to both sides 4x - 5x = 5x - 51 -5x

Group like terms

-x = - 51

Now divide it by - on both sides

We get

x = 51 is the solution for the given question.

Ex 2: 

Solve the linear equation -2(2x - 0) - 4x  +8 = (x - 4) - 20x

Sol

Given

-2(2x - 0) - 4x + 8 = (x - 5) - 20x

Multiply factors

-4x + 0 - 4x + 8 = x - 5 - 20x

Group like terms

-8x +8 = -21x - 5

Subtract 8 to both sides

-8x + 8 - 8 = -21x -5 -8

Group like terms

-8x = -21x -13

Add 21x to both sides

13x  = -13


13x = -13

Now divide it by 13 on both sides

13x/13 = -13/13

x = -1

Fried berg Linear Algebra Practice Problems:

1) Solve the linear equation -5(-x ) = 5x – 4

Answer:- no solution

2) Solve the linear equation -(x - 4) - 5x - 6 = (6x + 6) - 6x

Answer:- x = -4/3

3) Solve the linear equation -2(2x - 5)  = (8x + 6) - 6x - 6

Answer: - x = 5/3

Two Step Variable Equations

Introduction to two step variable equations:

Two step variable equation is an equation which should be solve in two steps. In this, the equation will have one variable and the steps to solve in two steps leads to find the value of  that variable. In this addition, subtraction, multiplication and division involve to solve the equation for a variable.

Steps to Solve the Equations in Two Step Variable Equations:
Step 1: Write the equations.

Step 2: Adding or subtracting is the first step in two step equations.

Step 3: Dividing or Multiplication is the second step in two step equations.

Step 4: Simplify it.

First write the equation which is given .Then add or subtract the given two step of equations .Then divide and multiply is the second step of the equations and the final step is to simplify the given equation

Examples on Two Step Variable Equations:

1.Solve the equation 5x - 2 =8 in two steps.

Solution:

Given, 5x - 2 =8

Add 2 on both sides,

5x – 2 = 8

2    2

----------------

5x       =10

After adding 2 on both sides we have 5x=10

Now divide by 5 on both sides,

5x  =10

---    ----

5        5

X=2

The value of x is 2

2.Solve the equation 5x + 2 =12 in two steps.

Solution:

Given, 5x + 2 =12

Subtract 2 on both sides,

5x + 2 =12

-2    -2

----------------

5x       =10

After subtracting 2 on both sides we have 5x =10

Now divide by 5 on both sides,

5x  =10

---    ----

5        5

X=2

The value of x is 2

3.Solve the equation 2/3x - 2 =8 in two steps.

Solution:

Given, 2/3x - 2 =8

Add 2 on both sides,

2/3x - 2 =8

2   2

-----------------

2/3x      =10

After adding 2 on both sides we have 2/3x=10

Now multiply by 3/2 on both sides

3/2 × 2/3 x = 10 × 3/2

X=15

The value of x is 15

4.Solve the equation 2/3x + 2 =8 in two steps.

Solution:

Given, 2/3x + 2 =8

Subtract 2 on both sides,

2/3x + 2 =8

-2   -2

------------------

2/3x        =6

After subtracting 2 on both sides we have 2/3x = 6

Now multiply by 3/2 on both sides

3/2 × 2/3 x = 6 × 3/2

X=9

The value of x is 9.

Foundations of Set Theory

Introduction to Foundations of Set Theory

Foundations of set theory are the branch of mathematics that learned about the sets, which are the collections of objects. Even though any type of objects can be collected into a set, set theory is applied most often to objects that are related to mathematics.Now we will see the examples of set theory.

Examples for Foundations of Union and Intersections
1)Proof the A∪(B∩C)=(A∪B) ∩(A∪C) for the following sets.
A={4,6,7,9,10} B={3,4,5,6,7} C={1,4,5,6,8}

Solution

The given sets are A={4,6,7,9,10} B={3,4,5,6,7} C={1,4,5,6,8}
A∪(B∩C)=(A∪B) ∩(A∪C)
Take the left side condition.
A∪(B∩C)
B∩C

Take the common values of the sets B and C.

B∩C={4,5,6}
A∪(B∩C)
Now joining the A set values.
A∪(B∩C)={ 4,5,6,7,9,10}

Take right side condition.

(A∪B) ∩(A∪C)

A∪B={3,4,5,6,7,9,10}

A∪C={1,4,5,6,7,8,9,10}

(A∪B) ∩(A∪C)={4,5,6,7,9,10}

So A∪(B∩C)=(A∪B) ∩(A∪C).

2) Proof the A∪(B∩C)=(A∪B) ∩(A∪C) condition for the sets A={1,2,3,4,5} B={2,4,7,9,11} C={1,5,6,8,9}

Solution

The given sets are A={1,2,3,4,5} B={2,4,7,9,11} C={1,5,6,8,9}

Take left hand side condition.

A∩(B∪C)

B∪C={1,2,4,5,6,7,8,9,11}

A∩(B∪C)={1,2,4,5}

Now take the right hand side condition.

(A∩B) ∪ (A∩C)

A∩B={2,4}

A∩C={1,5}

(A∩B) ∪ (A∩C)={1,2,4,5}

So A∩(B∪C)=(A∩B) ∪ (A∩C).

3) What is the A∪(B∪C)=(A∪B) ∪C for the following sets? A={13,17,18,20,22} B={12,13,14,15,16} C={11,12,13,15,17}

Solution

Given sets are A={13,17,18,20,22} B={12,13,14,15,16} C={11,12,13,15,17}

A∪(B∪C)=(A∪B) ∪C

Take left hand side condition

A∪(B∪C)

B∪C={11,12,13,14,15,16,17}

Grouping the set of A values with this B∪C

A∪(B∪C)={11,12,13,14,15,16,17,18,20,22}

Right hand side.

(A∪B) ∪C

A∪B={12,13,14,15,16,17,18,20,22}

Joining the values is the set C with this A∪B.

(A∪B)∪C={11,12,13,14,15,16,17,18,20,22}

So the condition is proofed.

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Example for Foundations of Difference Set

1)A={2,5,7,8,9} B={2,3,5,6,7} and C={3,4,5,7,9}. Find the i)A-(B∪C) ii)B-(A∪C).

Solution

The given sets are A={2,5,7,8,9} B={2,3,5,6,7} and C={3,4,5,7,9}

i) A-(B∪C)

B∪C

B∪C={2,3,5,6,7,9}
A-( B∪C) condition is a difference of set. It means we select the values from the A set. But that values is not present in the B∪C set.
So A- B∪C={8}
ii)B-(A∪C)
A∪C={2,3,4,5,7,8,9}
B-(A∪C)={6}
These are the examples of foundations of set theory.