Class 9 Maths Chapter 1 Number Systems

Exercise 1.1

Question 1.

Is zero a rational number? Can you write it in the form , where p and q are integers and q ≠ 0?
Solution:
Yes, zero is a rational number.
Zero can be written in any of the following forms :

Thus, 0 can be written as , where p = 0 and q is any non-zero integer.
Hence, 0 is a rational number.

 Question 2.
Find six rational numbers between 3 and 4.
Solution:
The rational numbers between 3 and 4.


 Question 3.
Find five rational numbers between  and .
Solution:
Since we want 5 rational numbers between  and , so we write

 Question 4.
State whether the following statements are true or false. Give reasons for your answers.

1. Every natural number is a whole number.
2. Every integer is a whole number.
3. Every rational number is a whole number.

Solution:
(1) True: Every natural number lies in the collection of whole numbers.
(2) False: -3 is not a whole number.
(3) False:   is not a whole number.

Exercise 1.2

Question 1.
State whether the following statements are true or false. Justify your answers.

1. Every irrational number is a real number.
2. Every point on the number line is of the form , where m is a natural number.
3. Every real number is an irrational number.

Solution:
(1) True (∵ Real numbers = Rational numbers + Irrational numbers.)
(2) False (∵ no negative number can be the square root of any natural number.)
(3) False (∵ rational numbers are also present in the set of real numbers.)

 Question 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.
e.g., 
Here, ‘4’ is a rational number.

 Question 3.
Show how  can be represented on the number line.
Solution:
We know that, = 


Draw of right angled triangle OQP, such that
OQ = 2 units
PQ= 1 unit
and   ∠OQP = 90°
Now, by using Pythagoras theorem, we have
OP2 =   OQ2 +PQ2
=    22 +12
= op=   = 
Now, take O as center OP = 45 as radius, draw an arc, which intersects the line at point R.
Hence, the point R represents .

 Question 4.
A classroom activity (constructing the ‘square root spiral’).
Solution:
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 figure).

Now, draw a line segment P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to 0P3. Continuing in this manner, you can get the line segment Pn-1Pn by drawing a line segment of unit length perpendicular to OPn-1. In this manner, you will have created the points P2, P3, ….,Pn…, and joined them to create a beautiful spiral depicting   ,………..

Ex 1.3

 Question 1.
Write the following in decimal form and say what kind of decimal expansion each has :
(1) 
(2) 
(3) 4
(4) 
(5) 
(6) 
Solution:
(1)   = 0.36, terminating.
(2) By long division, we have

(3)

(4)

∴  = 0.23076923 = 
(5)

∴  = 0.181818 = 

(6)

Question 2.
You know that  =  .Can you predict what the decimal expansions of  are, without actually doing the long  division? If so, how? [Hint: Study the remainders while finding the value of  carefully.]
Solution:
We have, = 

 Question 3.
Express the following in the form , where p and q are integers and q ≠ 0.

1. 
2. 0.
3.   

Solution:
(1) Let x =  = 0.666…………     …(i)
Multiplying Eq. (i) by 10, we get
10x = 6.666…                 …(ii)
On subtracting Eq. (i) from Eq. (ii), we get
(10x – x) = (6.666…) – (0.666…)
9x = 6 => x = 6/9           => x = 2/3
x = 0.  = 0.4777

(2) Let
Multiplying Eq. (i) by 10, we get
10x = 4.777…                       …(ii)
Multiplying Eq. (ii) by 10, we get
100x = 47.777                                               … (iii)
On subtracting Eq. (ii) from Eq. (iii), we get
(100x-10x)= (47.777…) – (4.777…)
90x =43      => x = 
x =  = 0.001001001…

(3) Let
Multiplying Eq. (i) by 1000, we get
1000x = 1.001001001…
On subtracting Eq. (i) from Eq. (ii), we get
(1000x – x) = (1.001001001…) – (0.001001001…)
999x = 1  => 

 Question 4.
Express 0.99999… in the form . Are you surprised by your answer? With your teacher and classmates discuss why the answer makes sense.
Solution:
Let x = 0.9999…                                                                              …(i)
Here, we have only one repeating digit. So, we multiply both sides of (i) by 10 to get
10x = 9.999….                                        …(ii)
Subtracting (i) from,(ii), we get
10x-x = (9.999…)-(0.9999…)
=>   9x = 9
=> x = 1
Hence, 0.9999…= 1
Since, 0.9999… goes on forever. So, there is no gap between 1 and 0.9999… and hence they are equal.

Question 5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion ? Perform the division to check your answer.
Solution:

Thus, = 
∴ The maximum number of digits in the quotient while computing  are 16.

Question 6.
Look at several examples of rational numbers in the form  (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:
Consider many rational numbers in the form  (q ≠ 0) , where p and q are
integers with no common factors other than 1 and having terminating decimal representations.



From the above, we find that the decimal expansion of the above numbers is terminating. Along with we see that the denominator of the above numbers are in the form 2m x 5n, where m and n are natural numbers. So, the decimal representation of rational numbers can be represented as a terminating decimal.

Question 7.
Write three numbers whose decimal expansions are non-terminating non-recurring.
Solution:
Three numbers whose decimal representations are non-terminating and non-repeating are
 , and  or we can say 0.100100010001…, 0.20200200020002… and 0.003000300003…

 Question 8.
Find three different irrational numbers between the rational numbers and .
Solution:
To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,

Thus, = 0.8181… = 
The required numbers are
0.73073007300073000073…
0.7650765007650007650000…
0.80800800080000…

Question 9.
Classify the following numbers as rational or irrational:
(1) 
(2) 
(3) 0.3796
(4) 478478…
(5) 1.101001000100001…
Solution:
(1)   is an irrational number as 23 is not a perfect square.
(2)  

Thus, 15 is a rational number.
(3) 0.3796 is a rational number as it is terminating decimal.
(4) 7.478478… is non-terminating but repeating, so, it is a rational number.
(5) 1.101001000100001… is non-terminating and non-repeating so, it is an irrational number.

 

Exercise 1.4

 Question 1.
Visualize 3.765 on the number line, using successive magnification.
Solution:
We know that 3.765 lies between 3 and 4. So, let us divide the part of the number line between 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7 and 3.8 [Fig. (i)]. Now, we imagine dividing this again into ten equal parts. The first mark will represent 3.71, the next 3.72 and so on. To see this clearly, we magnify this as shown in
[Fig. (ii)].

Again 3.765 lies between 3.76 and 3.77 [Fig. (ii)]. So, let us focus on this portion of the number line [Fig. (iii)] and imagine to divide it again into ten equal parts [Fig. (iii)]. Here, we can visualize that 3.761 is the first mark and 3.765 is the 5th mark in these subdivisions. We call this process of visualization of representation of numbers on the number line through a magnifying glass as the process of successive magnification. So, we get seen that it is possible by sufficient successive magnifications of visualizing the position (or representation) of a real number with a terminating decimal expansion on the number line.

 Question 2.
Visualise  on the number line, upto 4 decimal places.
Solution:
We adopt a process by successive magnification and successively decrease the lengths of the portion of the number line in which  is located. Since  is located between 4 and 5 and is divided into 10 equal parts [Fig. (i)]. In further, we locate  between 4.2 and 4.3 [Fig. (ii)].

To get more accurate visualization of the representation, we divide this portion into 10 equal parts and use a magnifying glass to visualize that  lies between 4.26 and 4.27. To visualise  more clearly we divide again between 4.26_and 4.27 into 10 equal parts and visualise the representation of  between 4.262 and 4.263 [Fig. (iii)].

Now, for a much better visualization between 4.262 and 4.263 is again divided into 10 equal parts
[Fig. (iv)]. Notice that  is located closer to 4.263 then to 4.262 at 4.2627.

Remark: We can adopt the process endlessly in this manner and simultaneously imagining the decrease in the length of the number line in which 4.26 is located.

Exercise 1.5

 Question 1.
Classify the following numbers as rational or irrational:
(1) 2-
(2) (3+ )-
(3) 
(4) 
(5) 2π
Solution:
(1) Irrational ∵ 2 is a rational number and  is an irrational number.
∴ 2 –  is an irrational number.
(∵ The difference of a rational number and an irrational number is irrational)

(2) 3 +  –  = 3 (rational)

(3)  = (rational)

(4)  (irrational)  ∵ 1 ≠ 0 is a rational number and 2 ≠ 0 is an irrational a/2 number.
∴   = is an irrational number.
(∵ The quotient of a non-zero rational number with an irrational number is irrational).
(5) 2π (irrational) ∵ 2 is a rational number and π is an irrational number.
∴  2π is an irrational number, (∵ The product of a non-zero rational number with an irrational number is an irrational).

Question 2.
Simplify each of the following expressions :
(1) (3 +  ) (2 + a/2)
(2) (3 + ) (3- )
(3) ( +  )2
(4) ( –  ) ( +  )
Solution:

 Question 3.
Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is
π = . This seems to contradict the fact that it is irrational. How will you resolve this contradiction?
Solution:
Actually  =  which is an approximate value of π.

 Question 4.
Represent  on the number line.
Solution:
Mark the distance 9.3 units from a fixed point A on a given line to obtain a point B such that AB = 9.3 units. From B, mark a distance of 1 unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with center O and radius OC. Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D

Then BD = . To represent  on the number line. Let us treat the line BC as the number line, with B as zero, C as 1, and so on. Draw an arc with center B and radius BD, which intersects the number line at point E. Then, the point E represent .

 Question 5.
Rationalise the denominators of the following:
(i) 
(ii) 
(iii) 
(iv) 
Solution:

Exercise 1.6

Question 1.
Find:
(i) 
(ii) 
(iii) 
Solution:


 Question 2.
Find:
(i) 
(ii)  
(iii) 
(iv) 
Solution:
.\

 Question 3.
Simplify
(i) .
(ii) 7
(iii) 
(iv) .
Solution: