Why are Natural Numbers called Natural Numbers?

A number system in maths or System of Numeration represents or expresses numbers on the number line in a consistent manner using digits that consist of 0,1,2,3,4,5,6,7,8,9. Furthermore, a number system is also used to represent a set of values to represent quantity. The Number system is used in countless numbers of tasks performed on day-to-day activities right away from counting stars in the nighty sky or solving complex mathematical equations, the Number System is used everywhere.

For Example: {1,2,3,4,5…………..∞} can be represented as a set of natural numbers in a number system that starts from 1 and goes up to infinity.

Types of Numbers

• Natural Numbers (N): These numbers consist of a set of all the positive numbers starting from 1 and goes up to Infinity.
  The set of Natural Numbers can be defined as N ={1, 2, 3, 4, 5, …………. ∞}.
• Whole Numbers (W): These numbers consist of a set of all Natural Numbers along with 0 (zero). 
  The set of Whole Numbers can be defined as W = {0, 1, 2, 3, 4, 5, …………… ∞}.
• Integers (Z): These numbers consist of a set of all the positive and negative Natural Numbers including 0 (zero). 
  The set of Integers can be defined as Z = {-∞ ……….. , -5, -4, -3, -2, -1 , 0, 1, 2, 3, 4, 5, ……………. ∞ }.
• Rational Numbers (Q): These are the numbers that can be written or expressed in the form of a/b or a ratio of two numbers. 
  Examples of Rational Numbers are 12/4, -3/5, 6/-12, 0/1, etc.
• Irrational Numbers (P): These are the numbers that cannot be expressed in the form of a/b or the form of the ratio of two numbers. 
  Examples of Irrational Numbers are √5, e = 2.718281, √7, etc.
• Imaginary Numbers (Im (Z)): These are the numbers that are products of real numbers along with an imaginary unit “i”.
  Examples of Imaginary Numbers are i², 2i, √5i, etc.
• Real Numbers (R): These are the numbers that consist of all the positive, negative, fractional, or decimal numbers without any imaginary unit or quantity.
  Examples of Real Numbers are 3/4, 2.3432, √3, -4, 10, etc.
• Complex Numbers (Re (Z)): These are the numbers that are written or expressed in the form of a+bi, where a and b are real numbers and i is an imaginary unit.
  Examples of Complex Numbers are 1+3i, 2+i, 4+√3i, etc.

Why Natural Numbers are called Natural Numbers?

All the numbers that can represent some actual life existing quantity are known as Natural Numbers. Natural numbers range from all the positive numbers starting from 1 that go up to infinity. Natural numbers do not contain 0 (zero) as zero in real life represents nothing or simply means that the quantity does not exist, which implies that these are the naturally occurring numbers.

For Example: You can put one or more apples in your shopping bag but you can’t put zero apples in it or hold zero apples in your hands as at that moment the apple simply does not exist for you.

From the above example, we can determine that Natural numbers have a natural reference to some actual existing quantity, as in the real-life counting of quantities start from 1 and not 0. 

Types of Number Systems

• Binary Number System: This kind of number system comprises only two digits that are 0 and 1. The base of the numbers in this number system is 2. In this system 0 and 1 are bits and eight of these bits together make a byte. A bit is the smallest memory unit in a computer, all the operations performed on the computer whether they are basic or complex they are accomplished in binary form that is in the form of 0 and 1.
  Example: 011101₂, 000111₂, 010101₂, etc.
• Octal Number System: This kind of number system comprises eight digits that range from 0-7. The base of the numbers in this number system is 8.
  Example: 47₈, 109₈, 343₈, etc.
• Decimal Number System: This kind of number system comprises digits ranging between 0-9. The base of the numbers in this number system is 10. Decimal numbers are used to represents actual life quantities as these Decimal numbers are the ones that we use in our day-to-day life. If a number has no base it simply means that the number is a base of 10.
  Example: 2893₁₀, 12₁₀, 456₁₀, 909₁₀, etc.
• Hexadecimal Number System: This kind of number system comprises decimal numbers ranging from 0-9 and six alphabetical characters that are A, B, C, D, E, F, these characters replace the numbers as A=10, B=11, C=12, D=13, E=14, and F=15, which together makes up a total of 16 digits. The base of the numbers in this number system is 16.
  Example: 134A₁₆, 3B4₁₆, 12A4F₁₆, etc.

Sample Problems based on Number Systems

Question 1. Divide a group of 456 students in the ratio of 12:16.

Solution: 

Given Number of students = 456

Given ratio = 12:26

Let S be the number of students i.e, S = 456

Let a = 12 and b = 16 then the sum of the ratio R will be : R = a+b

R = 12 + 26, R = 38 

Now to find the ratio in form of a/b or a:b 

For a:

a = (S/R)*a

a = (456/38)*12

a = (12)*12

a = 144

For b:

b = S – a

b = 456 – 144

b = 312

Answer: Obtained value of a and b are 144 and 312 therefore the ratio of the students will be = 144 : 312

Question 2. John earns Rs.56000 per month. He spends 1/4 of his income on food; 3/10 of the remainder on house rent and 5/21 of the remainder on the education of children. How much money is still left with him?

Solution:

Given, total salary of john = Rs.56000

Given expenses on food = 1/4

Given expenses on house rent = 3/10

Given expenses on education = 5/21

Salary of John that he spends on food = 56000 x 1/4

 = 14000

The total portion of the salary of John that he spends on food is Rs.14000, therefore the remaining 

salary of John will be: Total salary – food expenses 

56000 – 14000 = 42000

Now, the salary of john that he spends on house rents after spending on food = 42000 x 3/10

= 12600

Total remaining salary of John after spending house rent is house rent – remaining salary

42000 – 12600 = 29400

Now, the salary of john that he spends on the education of his children from his remaining salary after

 spending on food and house rent = 29400 x 5/21

 = 7000 

Total remaining salary of john after spending on education of his children = education – remaining salary

29400 – 7000 = 22400

So, the total remaining salary of john after spending on food, house rent and education of children is :

Answer: 22400

Question 3. Simplify the following numbers:

a. √-1

b. -1

c. √-16

d.  √-20

Solution:

We Know that : i = √-1

i^{2 =} -1

i^{3 = -i}

i^{4 =}  1

a. Simplify √-1.  

we know that : i = √-1

so, √-1 can be simplifies as i

b. Simplify -1.  

we know that : i² = -1

so, -1 can be simplifies as i²

c. Simplify  √-16.

√-16 can be written as : √(16 x -1) or √16 x √-1

that is equivalent to : 4 x √-1

we know that i = √-1

therefore √-16 can be simplified as 4i

d. Simplify √-20.

√-20 can be written as √(20 x -1) or √20 x √-1

that is equivalent to √20 x i

therefore √-20 can be simplified as i√20.

Question 4. Solve the following Quadratic equation  3x² + 6 = 6x.

Solution:

Given quadratic equation = 3x² + 6 = 6x     ……. eq(1)

converting the quadratic equation in the form of ax² + bx + c = 0

3x² – 6x + 6 = 0  

here, a = 3, b = -6, and c = 6.

using the formulae (-b ± √(b² – 4ac))/2a .

Putting the value of a,b and c in the formulae

(-(-6) ± √((-6)² – 4 x 3 x 6))/2 x 3 

we get,

= (6 ± √(36 – 72))/6

= (6 ± √-36)/6

= (6 ± 6i)/6

On dividing the equation by 6 we get, (1 ± i) i.e, 1 + i and 1 – i.

putting 1 + i and 1 – i in eq(1)

(1 + i):                                                                        (1 – i):

3(1 + i)² + 6 = 6(1 + i)                                                  3(1 – i)² + 6 = 6(1 – i)

3(1² + i^{2 +} 2i) + 6 = 6 + 6i                                            3(1² + i² – 2i) + 6 = 6 – 6i

3(1 + (-1) + 2i) + 6 = 6 + 6i                                         3(1 + (-1) – 2i) + 6 = 6 – 6i

3 – 3 + 6i + 6 = 6 + 6i                                                  3 – 3 – 6i + 6 = 6 – 6i

6i + 6 = 6i + 6 (hence proved)                                     -6i + 6 = -6i + 6 (hence proved)

Question 5. Rationalize 3/(4 – √6).

Solution:

Given Rational number = 3/(4 – √6).

We can see that denominator contains an irrational number so we need to multiply the additive inverse of the denominator to the numerator and denominator,

3/(4 – √6) x (4 + √6)/(4 + √6)

Now, the equation will be: 

3(4 + √6)/(4 – √6) x (4 + √6)

=  (12 + 3√6)/(16 – 6).

=  (12 + 3√6)/10

So the rationalized form of 3/(4 – √6) will be (12 + 3√6)/10.