(i)Natural Numbers(N):-The numbers

1,2,3,…,etc are natural numbers.

(ii)Whole Numbers(W):-All natural numbers

along with 0(zero) form the whole numbers.0,1,2,3,…

(iii)Integers(Z):-All positive number and

negative numbers along with 0(zero) form

the integers. ……-3,-2,-1,0,1,2,3,…etc

(iv)Real Numbers(R):-The set of all rational

numbers and irrational numbers together

form real numbers.

Real Numbers

(ii)Irrational

Numbers

(i)Rational

Numbers

(i)Rational Numbers(Q):-A number of the

form p , where p & q are integers and q ≠0

is

q

called a rational number. For e.g. : 2 , -1 , 3 3 47

(ii)Irrational Numbers:- The numbers which

cannot be written in p form, where p and q q

are integers and q≠0. For e.g. : √2 , √3 , √5 , πetc.

Type Of Rational Numbers

Terminating

Decimal

Non-Terminating

Decimal

(i)Terminating Decimal Expansion:- When the

decimal expansion terminates or ends after a

finite number of steps then it is called

terminating decimal expansion.The terminating

decimal expansion is a rational number.

(ii) Non terminating repeating decimal

expansion:- the decimal representation of a number is said to be non terminating

repeating decimal if it repeats and doesn’t

come to an end.Note:- terminating decimal and non terminating

repeating decimal denotes rational numbers.

(iii) Non terminating non repeating decimal

expansion:- the decimal representation of a number is said to be non terminating non

repeating when it neither terminates nor

repeats.Note:- non terminating non repeating decimal

denotes irrational number.

There are infinitely many rational numbers

between two given rational numbers. Question:-Find five rational numbers between 3 and 4 .

Solution: n=5 n+1=63 x 6

55

5 x 6= 18

30

4 x 65 x 6

=2430

5 rational between 3 and 4 are:-

19 , 20 , 21 , 22 , 23

5

5 5

30 30 30 30 30

The Pythagoreans in Greece, followers of

the famous mathematician and

philosopher Pythagoras , were the first

to discover the numbers which were not

rational, around 400 BC. These numbers

are called irrational numbers, because

they cannot be written in the form of

a ratio of integers. There are many

myths surrounding the discovery of

irrational numbers by the Pythagorean,

Hippacus of Croton .

Representation of √2 on number line

1)Draw a no. line.

2)Mark the origin 0 as ‘O’ .

3)Name the pt. 1 as A. AO=1 unit.

4)Draw the AX at point A .

5)On AX cut AB=OA.

6)Join OB.

r

7) Triangle OAB is a right triangle by pythagoras theorum

OB=√2.8) With a compass OB as radius draw an arc . 9) It cuts no. line at a point , mark it D. OD=OB=√2.

Note:- We can also make √3 ,√5 and various other irrational numbers on number line by 2 methods :- i)Spiral method

ii)Direct method

The Greek genius

Archimedes was

the first o

compute digits in

the decimal

expansion of pi.

Question:-express the following in the form p/q ,where p and q are integers and q 0.

(i)0.6Solution:-0.6 = xMultiply both sides by 10

10x=6+.610x=6+x10-x=69x=6X=6

X=2

≠

9

3

Representing real numbers on number line

i) 5.37

Operations on real number

(i)Addition

(ii)Multiplication

(iii)Division

(iv)Subtraction

additionAdd :- 2√2 + 5√3 AND √2 - 3√3

SOLUTION :- (2√2 + 5√3 ) + (√2 - 3√3)

= (2√2 + √2 ) + (5√3 - 3√3)

= (2+1) √2 +(5+3)√3

=3√2 + 2√3

MultiplicationMultiply:6√5 and 2√5Solution: 6√5 X 2√5

=6 X 2 X √5 X √5=12 X 5

=60

DivisionDivide: 8√15 by 2√3

Solution:- 8√15 ÷ 2√3=8 √3 X √5

= 4√52√3

Representation of √9.3 on number line

(1)Draw AB of length 9.3 cm and extend to C such

that BC=1 cm.

(2)Draw the perpendicular bisector of AC at O.

With O as center and radius OA draw a semi

circle.

(3)From B draw a perpendicular to intersect the

semicircle at D.

(4)With O as centre and

radius BD draw an arc to

cut the line AB at E.

(5)Therefore,BD=BE=√9.3.

Simplification

Simplify:(3 + √3) (2 +√2)= 3(2 +√2) + √3 (2+√2)= 6 + 3√2 + 2√3 + √6

Rationalisation

Rationalise:(i) 1

= 1 X

√7

√7

√7

√7

= √77

(ii) 1

√7 - √6

= 1√7 - √6

X √7 + √6√7 + √6

= √7 + √6

(√7) – (√6) 22

= √7 + √6

7-6

= √7 + √6

(i) a.a= a (ii) (a)=a (iii) a

(iv) a b=(ab)

pq p+q

P q pq

a = ap

q

P - q

P p p

Find :

(i) 64

=8

12

2 x 12

=81

=8

(ii) 3425

=25 x 2

5

=22

=4

simplify

(i) 11

=11

=11

=11

11

12

14

1 1

2 4-

2-14

14

(ii)7 . 81

2

1

2

=(7x8)1

2

=561

2