Class 9 Maths Chapter 1 Number System Notes.

Chapter 1 Number System
Exercise 1.1
i.e., Natural numbers = 1,2,3,4…
Whole numbers – Numbers starting from 0 to infinity (without fractions or decimals)
i.e., Whole numbers = 0,1,2,3…
Or, we can say that whole numbers have all the elements of natural numbers and zero.
Every natural number is a whole number; however, every whole number is not a natural number.
(ii) Every integer is a whole number.
Solution:
False
Integers- Integers are set of numbers that contain positive, negative and 0; excluding fractional and decimal
numbers.
i.e., integers= {…-4,-3,-2,-1,0,1,2,3,4…}
Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)
i.e., Whole numbers= 0,1,2,3….
Hence, we can say that integers include whole numbers as well as negative numbers.
Every whole number is an integer; however, every integer is not a whole number.
(iii) Every rational number is a whole number.
Solution:
False
Rational numbers- All numbers in the form p/q, where p and q are integers and q≠0.
i.e., Rational numbers = 0, 19/30 , 2, 9/-3, -12/7…
i.e., Whole numbers= 0,1,2,3….
Whole numbers- Numbers starting from 0 to infinity (without fractions or decimals)
Hence, we can say that integers include whole numbers as well as negative numbers.
All whole numbers are rational, however, all rational numbers are not whole numbers.
Exercise 1.2
1. State whether the following statements are true or false. Justify your answers.
(i) Every irrational number is a real number.
Solution:
True
Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are
integers and q ≠ 0.
i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….
Real numbers – The collection of both rational and irrational numbers are known as real numbers.
i.e., Real numbers = √2, √5, π, 0.102…
Every irrational number is a real number, however, every real number is not an irrational number.
(ii) Every point on the number line is of the form √m where m is a natural number.
Solution:
False
The statement is false since as per the rule, a negative number cannot be expressed as square roots.
E.g., √9 =3 is a natural number.
But √2 = 1.414 is not a natural number.
(iii) Every real number is an irrational number.
Solution:
False
Similarly, we know that there are negative numbers on the number line, but when we take the root of a negative
number it becomes a complex number and not a natural number.
E.g., √-7 = 7i, where i = √-1
The statement that every point on the number line is of the form √m, where m is a natural number is false.
The statement is false. Real numbers include both irrational and rational numbers. Therefore, every real number
cannot be an irrational number.
Real numbers – The collection of both rational and irrational numbers are known as real numbers.
i.e., Real numbers = √2, √5, , 0.102…
Irrational Numbers – A number is said to be irrational, if it cannot be written in the p/q, where p and q are
integers and q ≠ 0.
i.e., Irrational numbers = π, e, √3, 5+√2, 6.23146…. , 0.101001001000….
Every irrational number is a real number, however, every real number is not irrational.
Maths Chapter 1
Number System
2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a
number that is a rational number.
Solution:
No, the square roots of all positive integers are not irrational.
For example,
√4 = 2 is rational.
√9 = 3 is rational.
Hence, the square roots of positive integers 4 and 9 are not irrational. ( 2 and 3, respectively).
3. Show how √5 can be represented on the number line.
Solution:
Step 1: Let line AB be of 2 unit on a number line.
Step 2: At B, draw a perpendicular line BC of length 1 unit.
Step 3: Join CA
Step 4: Now, ABC is a right angled triangle. Applying Pythagoras theorem,
2
2
AB +BC = CA
2
2
2
22+1 = CA = 5
Thus, √5 is represented on the number line as shown in the figure.
⇒
CA = √5 . Thus, CA is a line of length √5 unit.
Step 4: Taking CA as a radius and A as a center draw an arc touching the number line. The point at which number
line get intersected by arc is at √5 distance from 0 because it is a radius of the circle whose center was A.
4. Classroom activity (Constructing the ‘square root spiral’) : Take a large sheet of paper and construct the
‘square root spiral’ in the following fashion. Start with a point O and draw a line segment OP1 of unit
length. Draw a line segment P1P2 perpendicular to OP of unit length (see Fig. 1.9). Now draw a line
2 3
2
1
segment P P perpendicular to OP . Then draw a line segment P P perpendicular to OP . Continuing in
Fig. 1.9 :
3 4
3
Maths Chapter 1
Number System
n-1
Constructing this manner, you can get theline segment P Pn by square root spiral drawing a line segment
of unit length perpendicular to OP . In thismanner, you will have created the points P , P ,….,Pn,… .,
n-1
and joined them to create a beautiful spiraldepicting √2, √3, √4, …
Solution:
2
3
Chapter 1
Number System
Step 1: Mark a pointOon the paper. Here, O will be the center of the square root spiral.
Step 2: From O, drawastraight line, OA, of 1cm horizontally.
Step 3: From A, drawaperpendicular line, AB, of 1 cm.
Step 4: Join OB. Here,OB will be of √2
Step 5: Now, from B,draw a perpendicular line of 1 cm and mark the end point C.
Step 6: Join OC. Here,OC will be of √3
Step 7: Repeat the stepsto draw √4, √5, √6….
Maths Chapter 1
Number System
Exercise 1.3
1. Write the following in decimal form and say what kind of decimal expansion each has :
(i) 36/100
Solution:
= 0.36 (Terminating)
(ii)1/11
Solution:
Solution:
= 4.125 (Terminating)
(iv) 3/13
Solution:
(v) 2/11
Solution:
(vi) 329/400
Solution:
= 0.8225 (Terminating)
2. You know that 1/7 = 0.142857. Can you predict what the decimal expansions of 2/7, 3/7, 4/7, 5/7, 6/7 are,
without actually doing the long division? If so, how?
[Hint: Study the remainders while finding the value of 1/7 carefully.]
Solution:
(i)
Solution:
3. Express the following in the form p/q, where p and q are integers and q 0.
Assume that x =0.666…
Then,10x = 6.666…
10x=6+ x
9x=6
x = 2/3
(ii)
Solution:
= (4/10)+(0.777/10)
Assume that x = 0.777…
Then, 10x = 7.777…
10x = 7 + x
x = 7/9
(4/10)+(0.777../10) = (4/10)+(7/90) ( x = 7/9 and x = 0.777…0.777…/10 = 7/(9×10) = 7/90 )
= (36/90)+(7/90) = 43/90
Solution:
Assume that x = 0.001001…
Then, 1000x = 1.001001…
1000x = 1 + x
999x = 1
x = 1/999
Solution:
Assume that x = 0.9999…..Eq (a)
Multiplying both sides by 10,
10x = 9.9999…. Eq. (b)
4. Express 0.99999…. in the form p/q . Are you surprised by your answer? With your teacher and
classmates discuss why the answer makes sense.
Eq.(b) – Eq.(a), we get
10x = 9.9999 –x = -0.9999…
_____________
9x = 9
x = 1
The difference between 1 and 0.999999 is 0.000001 which is negligible.
Hence, we can conclude that, 0.999 is too much near 1, therefore, 1 as the answer can be justified.
5. What can the maximum number of digits be in the repeating block of digits in the decimal expansion of
1/17 ? Perform the division to check your answer.
Solution:
1/17
Dividing 1 by 17:
There are 16 digits in the repeatingblock of the decimal expansion of 1/17.
6. Look at several examples of rational numbers in the form p/q (q ≠ 0), where p and q are integers with no
common factors other than 1 and having terminating decimal representations (expansions). Can you guess
what property q must satisfy?
Solution:
We observe that when q is 2, 4, 5, 8, 10… Then the decimal expansion is terminating. For example:
1/2 = 0. 5, denominator q = 2
1
7/8 = 0. 875, denominator q =2
4/5 = 0. 8, denominator q = 5
Solution:
1
3
We can observe that the terminating decimal may be obtained in the situation where prime factorization of the
denominator of the given fractions has the power of only 2 or only 5 or both.
7. Write three numbers whose decimal expansions are non-terminating non-recurring.
We know that all irrational numbers are non-terminating non-recurring. three numbers with decimal expansions
that are non-terminating non-recurring are:
1.
2.
3.
√3 = 1.732050807568
√26 =5.099019513592
√101 = 10.04987562112
8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.
Solution:
Three different irrational numbers are:
1. 0.73073007300073000073…
2. 0.75075007300075000075…
3. 0.76076007600076000076…
9. Classify the following numbers as rational or irrational according to their type:
(i)√23
Solution:
√23 = 4.79583152331…
(ii)√225
Solution:
Since the number is non-terminating and non-recurring therefore, it is an irrational number.
√225 = 15 = 15/1
Since the number can be represented in p/q form, it is a rational number.
(iii) 0.3796
Solution:
Since the number,0.3796, is terminating, it is a rational number.
(iv) 7.478478
Solution:
The number,7.478478, is non-terminating but recurring, it is a rational number.
(v) 1.101001000100001…
Solution:
Since the number,1.101001000100001…, is non-terminating non-repeating (non-recurring), it is an irrational
number.
Exercise 1.4
1.Visualise3.765onthe number line, using successive magnification.
Solution:

Exercise 1.5
1. Classify the following numbers as rational or irrational: (i) 2 –√5 Solution: We know that, √5 =
2.2360679… Here, 2.2360679…is non-terminating and non-recurring. Now, substituting the value
of √5 in 2 –√5, we get, 2-√5 = 2-2.2360679… = -0.2360679 Since the number, – 0.2360679…, is
non-terminating non-recurring, 2 –√5 is an irrational number. (ii) (3 +√23)- √23 Solution: (3 +√23)–√23 = 3+√23–√23 = 3 = 3/1 Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational. (iii)
2√7/7√7
Solution:
2√7/7√7 = ( 2/7)× (√7/√7)
We know that (√7/√7) = 1
Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7
Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.
(iv) 1/√2
Solution:
Multiplying and dividing numerator and denominator by √2 we get,
(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)
We know that, √2 = 1.4142…
Then, √2/2 = 1.4142/2 = 0.7071..
(v) 2
Solution:
We know that, the value of = 3.1415
Since the number, 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.
Hence, 2 = 2×3.1415.. = 6.2830…
Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.
2. Simplify each of the following expressions:
(i) (3+√3)(2+√2)
Solution:
(3+√3)(2+√2 )
Opening the brackets, we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)
= 6+3√2+2√3+√6
(ii) (3+√3)(3-√3 )
Solution:
2
(3+√3)(3-√3 ) = 3-(√3) = 9-3
= 6
(iii) (√5+√2) 2
Solution:
2
2
2
(√5+√2) = √5 +(2×√5×√2)+ √2
= 5+2×√10+2 = 7+2√10
(iv) (√5-√2)(√5+√2)
Solution:
2
2
2
(√5-√2)(√5+√2) = (√5-√2 ) = 5-2 = 3
Solution:
3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter, (say d). That is, π
=c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?
There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never
obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to
22/7 or 3.142857…
4. Represent (√9.3) on the number line.
Solution:
Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.
Step 2: Now, AC = 10.3 units. Let the centre of AC be O.
Step 3: Draw a semi-circle of radius OC with centre O.
Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.
Step 5: OBD, obtained, is a right angled triangle.
Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1
OB = OC – BC
⟹ (10.3/2)-1 = 8.3/2
Using Pythagoras theorem,
We get,
2
2
OD =BD +OB
2
2
2
⟹ (10.3/2) = BD +(8.3/2)
2 =
2
⟹ BD (10.3/2)-(8.3/2)
2
2
2
⟹ (BD) = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)
2
⟹ BD = 9.3
⟹ BD = √9.3
Thus, the length of BD is √9.3.
Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it
touches the line segment is at a distance of √9.3 from O as shown in the figure.
5. Rationalize the denominators of the following:
(i) 1/√7
Solution:
Multiply and divide 1/√7 by √7
(1×√7)/(√7×√7) = √7/7
(ii) 1/(√7-√6)
Solution:
Multiply and divide 1/(√7-√6) by (√7+√6)
2
2
[1/(√7-√6)]×(√7+√6)/(√7+√6) = (√7+√6)/(√7-√6)(√7+√6)
=(√7+√6)/√7-√6 [denominator is obtained by the property, (a+b)(a-b) = a-b ]
=(√7+√6)/(7-6)
=(√7+√6)/1
2 2
=√7+√6
(iii) 1/(√5+√2)
Solution:
Multiply and divide 1/(√5+√2) by (√5-√2)
2
2
[1/(√5+√2)]×(√5-√2)/(√5-√2) = (√5-√2)/(√5+√2)(√5-√2)
= (√5-√2)/(√5-√2 ) [denominator is obtained by the property, (a+b)(a-b) = a-b ]
2 2
= (√5-√2)/(5-2)
= (√5-√2)/3
(iv) 1/(√7-2)
Solution:
Multiply and divide 1/(√7-2) by (√7+2)
1/(√7-2)×(√7+2)/(√7+2) = (√7+2)/(√7-2)(√7+2)
2 2
= (√7+2)/(√7-2 ) [denominator is obtained by the property, (a+b)(a-b) = a-b ]
= (√7+2)/(7-4)
= (√7+2)/3
2 2
Exercise 1.6
1. Find:
(i)64
1/2
Solution:
1/2
64 = (8×8)
= (8 )
2 ½
1
1/2
= 8 [⸪2×1/2 = 2/2 =1]
= 8
(ii)32
1/5
Solution:
1/5
32 = (2 )
5 1/5
= (2 ) 5⅕
1
= 2 [⸪5×1/5 = 1]
= 2
(iii)125 1/3
Solution:
1/3
(125) = (5×5×5)
= (5 ) 3⅓
1
1/3
= 5 (3×1/3 = 3/3 = 1)
= 5
2. Find:
(i) 9 3/2
Solution:
93/2 = (3×3) 3/2
= (3 ) 2 3/2
3
= 3 [⸪2×3/2 = 3]
=27
(ii) 32
2/5
Solution:
2/5
32 = (2×2×2×2×2)
2/5
=(2 ) 5 2⁄5
2
=2 [⸪5×2/5= 2]
=4
(iii)16 3/4
Solution:
3/4
16 = (2×2×2×2)
= (2 ) 4 3⁄4
3
3/4
= 2 [⸪4×3/4 = 3]
= 8
(iv) 125-1/3 -1/3
125 = (5×5×5)
= (5 ) 3-1⁄3-1-1/3
= 5 [⸪3×-1/3 = -1]
= 1/5
3. Simplify:
2/3
(i) 2 ×2
1/5
Solution:
1/5
22/3×2 = 2
13/15
(2/3)+(1/5)
m n m+n
[⸪Since, a ×a =a ____ Laws of exponents]
= 2 [⸪2/3 + 1/5 = (2×5+3×1)/(3×5) = 13/15]
(ii) (1/3 )
3 7
Solution:
3 7
= 3 -21
1/2-3 7
(iii) 11 /11
Solution:
1/2
1/4
m n m x n
(1/3 ) = (3 ) [⸪Since,(a ) = a ____ Laws of exponents]
1/4
11 /11 = 11
1/4
(1/2)-(1/4)
= 11 [⸪(1/2) – (1/4) = (1×4-2×1)/(2×4) = 4-2)/8 = 2/8 = ¼ ]
1/2
(iv) 7 ×8
Solution:
1/2
1/2
1/2
m m
m
71/2×8 = (7×8) [⸪Since, (a ×b = (a×b) ____ Laws of exponent