Number System Class 9 Notes Maths Chapter 1

Students can go through AP 9th Class Maths Notes Chapter 1 Number System to understand and remember the concepts easily.

Class 9 Maths Chapter 1 Notes Number System

→ Natural numbers : The numbers we use to count objects like starting from 1 are called natural numbers. These are denoted by the letter (N).
N = {1,2, 3, 4, l……}

• Smallest natural number = 1
• Largest natural number is (∝), because natural numbers are uncountable.
• Observe all natural numbers are positive numbers.

→ Whole numbers : After introducing zero (0) to natural numbers, the newer number set which starts from ‘0’ and followed by natural numbers is called whole numbers.

• Whole numbers set is denoted by ‘W’.
  ∴ W = {0, 1, 2, 3,}
• Smallest whole number is zero.
• Largest whole number is (∝), because whole numbers are uncountable.
• Except zero, all whole numbers are positive numbers.
• As shown below whole numbers are on right half-side of number line.

• On the left side of zero on the number line that is moving to left of zero on number line we have -1, -2, -3, ………. such negative numbers can be seen.
  Now, negative numbers including whole numbers (together) are called integers and is denoted by the letter (Z).
  ∴ Integers (Z) = {…… -3, -2, -1, 0, 1, 2, 3, …..}
• We observe number on left is one unit lesser than its immediate right one.
  For example -1 is (left of zero) one unit less than zero (which is immediate right).
• It is clear, we cannot say (recognise) the lowest or highest number of integers.

• Now, the question for the points on the number line that are in between any two integers.
• A part of above answer, another number system “Rational numbers” is introduced and defined as below.

→ Rational numbers : A number ‘Q’ is called , a rational number if it can be expressed as a ratio of two numbers (Integers) in which successor is a non-zero integer.
(Note : Those two are co-primes)
(Or)
A number ‘Q’ is called a rational number if it can be written in the form of \(\frac{p}{q}\), where ‘p’ is an integer and ’q’ is a non-zero integer (which means q ≠ 0) also p, q are co-prime numbers.
Rational numbers pe denoted by ‘Q’.

Examples of Rational numbers :

1. \(\frac{3}{4}\) (here 3, 4 are integers and 4 ≠ 0)
2. \(\frac{-1}{8}\) (here -1,8 are integers and 8 ≠ 0)
3. 6, because ‘6’ can be written as \(\frac{6}{1}\) (here 6, 1 are integers and 1 ≠ 0)
4. ‘0’, because ‘0’ can be written as \(\frac{0}{2}\) (here 0, 2 are integers and 2 ≠ 0)
5. \(\frac{3}{-5}\) (here 3, -5 are integers and -5 ≠ 0)

Thus all integers/whole numbers/natural numbers are rational numbers.
Note : We may notice \(\frac{1}{2}\), \(\frac{2}{4}\), \(\frac{3}{6}\), \(\frac{5}{10}\), \(\frac{10}{20}\), \(\frac{50}{100}\) etc., are equal. Hence, they represent same point on number line. We
choose \(\frac{1}{2}\) which is mid point of 1 and 2 on number line to represent all above equal ones.

→ Irrational numbers : A number which cannot be expressed in the form of \(\frac{p}{q}\), where p, q are integers and q ≠ 0 is called irrational number.
Irrational number can be described as
If the decimal expansion does not terminate (or) the decimal expansion does not end with a repeating sequence.’
Such numbers are called irrationals.
Ex :
i) 3.75 ……. is an irrational, because decimal expansion (75 …….) is not terminated.
ii) 4.1121231234 …… is an irrational, because it does not end with repeating sequence.
iii) π is an irrational, because
\(\frac{22}{7}\) = 3.14159 ……..

Decimal expansion not terminated nor end with a repeating sequence.
Hence π is an irrational.
Note : Square root of a prime number is always an irrational.
Ex: √2, √3, √5,√7,\(\sqrt{11}\) etc.
Note : Irrational numbers are denoted by (P) or (Q’).

→ Real numbers : All rational and irrational numbers together are called real numbers.
(Or)

• A real number is either a rational number or an irrational number.
• It means every real number can be represented by a unique point on the number line. And every point on the number line represents a unique real number.
• So, we call the number line as real number line.
• It means every point on number line represents either a rational number or an irrational number.

→ Real numbers and their decimal expansions:

• We can use decimal expansions to distinguish between rationals and irrationals.
• Visualising the representation of real numbers (either rational or irrational) on number line using a decimal expansion.

Remember,

1. If the remainder is zero it is terminating decimal, then it is a rational number.
2. If the remainder is a non-zero, then the following two cases arise (this is called non-terminating)
   i) Non-terminating recurring
   Ex : 5.676767……. = \(5 . \overline{67}\)
   These are also rational numbers.
   ii) Non-terminating, non-recurring
   Ex : 5.6712486 ……
   These are called irrational, numbers.

→ Definition :
i) The decimal expansion of a rational number is either terminating or nonterminating recurring.
ii) The decimal expansion of an irrational number is non-terminating nonrecurring.
Note : {√2 and π, Euler’s number e are called famous irrationals}
√2 = 1.414213562373 ……… (approximate value 1.414)
π = 3.14159265358979 ………….. (approximate value 3.141)
e = 2.718 (approximate value)

→ Number line representation of real numbers : Representing a terminating decimal expansion.
Ex : 3.467
It is clear that it lies in between 3 and 4.
Step 1 : Divide the gap in between 3 and 4 into 10 equal divisions, then they would be 3.1, 3.2, 3.3 …………. 4

It is clear that given 3.46 is in between 3.4 and 3.5. So by zoom in 3.4 and 3.5.

Step 2 : Divide the gap between 3.4 and 3.5 into 10 equal divisions. Then they will be 3.41, 3.42, 3.43, ……… 3.49, 3.5.

Observe 3.467 is more than 3.46 and less than 3.47. So by zooming in 3.46 and 3.47, we divide the gap between 3.46 and 3.47 into 10 equal divisions we get 3.461, 3.462, 3.463, ……. ,3.467, …

Step 3 : By zooming in 3.46 to 3.47, we get

Then, plot the point as above which will represent given 3.467 on number line. Thus by zooming in, we can spot the point which represents the given.

→ Square roots of real numbers :
Finding square root of a given positive real number (x) geometrically.
Example: Find \(\sqrt{2.7}\) geometrically, (or)
Finding √x for x is a positive real number.

Procedure:
Step 1 : Draw a line segment with a length of (x + 1).

Step 2 : Mark a point at a distance of 1 unit from either end point (P or Q) and name it ‘R’.

Step 3 : Find the mid point of line segment \(\overline{\mathrm{PQ}}\) and name it as ‘O’.

Step 4 : Draw a semi circle or cirlce from the point ‘O’ as centre.

Step 5 : Draw a perpendicular at ‘R’ to the line segment PQ which crosses the semi circle at ‘S’.

Result : Then \(\overline{\mathrm{RS}}\) is square root of ‘x’.
Proof : PQ = x + 1; PO = OQ = \(\frac{x+1}{2}\),
OR = OQ – QR = \(\frac{x+1}{2}\) – 1 = \(\frac{x-1}{2}\)
OP = OS = OQ (radius of semi circles) From Pythagorus theorem,
in ΔORS, ∠R = 90°, \(\overline{\mathrm{OS}}\) is hypotenuse.
∴ OS² = OR² + RS²
⇒ RS² = OS² – OR²
= (\(\frac{x+1}{2}\))² – (\(\frac{x-1}{2}\))²
= \(\frac{x^2+2 x+1}{4}\) – \(\frac{\left(x^2-2 x+1\right)}{4}\)
= \(\frac{x^2+2 x+1-x^2+2 x-1}{4}\)
= \(\frac{4 x}{4}\) = x
∴ RS = √x
Hence proved.

Thus by following above 5 steps we can find square root of a positive real numbers geometrically and then we can place it on number line (If wanted).

Observe the pattern for better understanding :
If a² = b, then ‘b’ is a positive real number, then √b = a.
If a³ = c, then \(\sqrt[3]{c}\) = a
If a⁴ = d, then \(\sqrt[4]{d}\) = a
If a⁵ = e, then \(\sqrt[5]{e}\) = a
Thus we can conclude if
aⁿ = p, then \(\sqrt[n]{p}\) = a where ‘p’ is a positive real number, ‘n’ is a positive integer.
We call the symbol ‘√’ as radical sign. In the case of ‘n’ = 2
We write ‘√’, which means \(\sqrt[2]{ }[latex] if and call it square root.

→ Identities related to square roots : Let ‘a’ and ‘b’ are positive real numbers, then
i) [latex]\sqrt{\mathrm{ab}}\) = (√a)(√b) Ex : \(\sqrt{15}\) = √5.√3

ii) \(\sqrt{\frac{a}{b}}\) = \(\frac{\sqrt{a}}{\sqrt{b}}\) Ex: \(\sqrt{\frac{9}{4}}\) = \(\frac{\sqrt{9}}{\sqrt{4}}\) = \(\frac{1}{2}\) = 1.5

iii) (√a + √b)(√a – √b) = (√a)² – (√b)²
= a – b

iv) (a + √b)(a – √b) = (a)² – (√b)²
= a² – b

v) (√a + √b)(√c – √d)
= \(\sqrt{\mathrm{ac}}\) + \(\sqrt{\mathrm{ad}}\) + \(\sqrt{\mathrm{bc}}\) + \(\sqrt{\mathrm{bd}}\)

vi) (√a + √b)²
= (√a)² + 2.√a.√b + (√b)²
= a + 2\(\sqrt{\mathrm{ab}}\) + b
To understand in an easier way we adopt rationalising denominator method for the fractions in which denominator is an irrational number.
In these cases we use its conjugate of a + √b is a – √b and vice-versa.

→ Recalling laws of exponents of natural numbers:
i) a^m.aⁿ = a^{m + n}
ii) \(\) = a^{m – n} (m < n)
iii) (a^m)ⁿ = a^mn
iv) (a^m) (b^m) = (ab)^m
Here a, b, m, n are natural numbers.
Now, can we extend the above for a, b, m, n which will be real numbers.
Case (I) : If ‘m’ is a negative number, then we write a^-m = \(\frac{1}{a^{\mathrm{m}}}\).
Before that let us look into the following case (for m = n)
\(\frac{a^m}{a^{\mathrm{n}}}=\frac{a \times a \times a \ldots \ldots(\mathrm{m} \text { times })}{a \times a \times a \ldots \ldots(\mathrm{m} \text { times })}\) (∵ m = n)
\(\frac{a^m}{a^m}\) = a^{m – m} = a⁰ = 1 (Using above second identity) ∴ a⁰ = 1

Consider the identity \(\frac{a^m}{a^n}\) = a^{m – n}
Let m = 0, then
\(\frac{a^m}{a^n}\) = a^{m – n}
\(\frac{a^0}{a^n}\) = a^{0 – n}
\(\frac{1}{a^n}\) = a^-n [∵ a⁰ = 1 already deduced]
Thus we can write and\(\frac{1}{a^n}\) = a^-n → (1)
and a⁰ = 1 → (2)

Now, consider m = am = 3/4 then
a^m = a^3/4 = \(\left(a^3\right)^{\frac{1}{4}}=\sqrt[4]{a^3}\)
So, by using sense of above identities and deduced identities (1), (2) we can simplify expressions in the form of above.