Number System Class 9th Maths Chapter

2. (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

3. (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

4. (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 4
7
(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 , π
etc.

5. Type Of Rational Numbers
Terminating
Decimal
Non-Terminating
Decimal

6. (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.

7. (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.

8. 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=6
3 x 6
5
5
5 x 6 = 18
30
4 x 6
5 x 6
=
24
30
5 rational between 3 and 4 are:-
19 , 20 , 21 , 22 , 23
5
5 5
30 30 30 30 30

9. 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 .

10. 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

11. 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

12. The Greek genius
Archimedes was
the first o
compute digits in
the decimal
expansion of pi.

13. Question:-express the following in the form p/q
,where p and q are integers and q 0.
(i)0.6
Solution:-0.6 = x
Multiply both sides by 10
10x=6+.6
10x=6+x
10-x=6
9x=6
X=6
X=2
≠
9
3

14. Representing real numbers on number line
i) 5.37

15. Operations on real number
(i)Addition
(ii)Multiplication
(iii)Division
(iv)Subtraction

16. addition
Add :- 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

17. Multiplication
Multiply:6√5 and 2√5
Solution: 6√5 X 2√5
=6 X 2 X √5 X √5
=12 X 5
=60

18. Division
Divide: 8√15 by 2√3
Solution:- 8√15 ÷ 2√3
=8 √3 X √5
= 4√5
2√3

19. 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.

20. (4)With O as centre and
radius BD draw an arc to
cut the line AB at E.
(5)Therefore,BD=BE=√9.3.

21. Simplification
Simplify:(3 + √3) (2 +√2)
= 3(2 +√2) + √3 (2+√2)
= 6 + 3√2 + 2√3 + √6

22. Rationalisation
Rationalise:(i) 1
= 1 X
√7
√7
√7
√7
= √7
7

23. (ii) 1
√7 - √6
= 1
√7 - √6
X √7 + √6
√7 + √6
= √7 + √6
(√7) – (√6)
22
= √7 + √6
7-6
= √7 + √6

24. (i) a.a= a
(ii) (a)=a
(iii) a
(iv) a b=(ab)
pq p+q
P q pq
a = a
p
q
P - q
P p p

25. Find :
(i) 64
=8
1
2
2 x 1
2
=81
=8
(ii) 34
2
5
=2
5 x 2
5
=2
2
=4

26. simplify
(i) 11
=11
=11
=11
11
1
2
1
4
1 1
2 4
-
2-1
4
1
4
(ii)7 . 8
1
2
1
2
=(7x8)
1
2
=56
1
2