Class 9 - Maths - Number Systems

Question 1:

Is zero a rational number? Can you write it in the form p/q where p and q are integers and       q ≠ 0?

Answer:

Yes, zero is a rational number. It can be written in the form p/q.

For example: 0/1, 0/2, 0/5 etc. are rational numbers where p and q are integers and q ≠ 0.

Question 2:

Find six rational numbers between 3 and 4.

Answer:

To get six rational numbers between 3 and 4, the denominator must be 6 + 1 = 7

Here, 3 = (3 * 7)/7 = 21/7 and 4 = (4 * 7)/7 = 28/7

So, the six rational can be obtained by changing numerator from 22 to 27.

Therefore, six rational numbers are: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7

Question 3:

Find five rational numbers between 3/5 and 4/5.

Answer:

By converting these numbers into decimal, we have

3/5 = 0.6 and 4/5 = 0.8

Hence, the five rational numbers between 3/5 and 4/5 are: 0.61, 0.62, 0.63, 0.64, 0.65.

Question 4:

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

(i) Every natural number is a whole number.

(ii) Every integer is a whole number

(iii) Every rational number is a whole number.

Answer:

(i) True statement

The collection of all natural numbers and 0 is called whole numbers.

(ii) False statement

Integers such as –1, –2 are non-whole numbers.

(iii) False statement

Rational numbers like 1/2, 1/5, 2/7, etc. are not a whole number.

Exercise 1.2

Question 1:

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

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form √m where m is a natural number.

(iii) Every real number is an irrational number.

Answer:

(i) True statement, because all rational numbers and all irrational numbers form the group

(collection) of real numbers.

(ii) False statement, because no negative number can be the square root of any natural

number.

(iii) False statement, because rational numbers are also a part 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.

Answer: No, if we take a positive integer, say 4 its square root is 2, which is a rational number.

According to the Pythagoras theorem, in a right-angled triangle, the square of the hypotenuse

is equal to the sum of the squares of the other two sides.

In the figure:

OB2 = OA2 + AB2

=> OB2 = 12 + 12

=> OB2 = 2

=> OB = √2

Question 3:

Show how √5 can be represented on the number line.

Answer:

Let us take the horizontal line XOX’ as the x-axis. Mark O as its origin such that it represents 0.

Cut off OA = 1 unit, AB = 1 unit.

=> OB = 2 units

Draw a perpendicular BC Ʇ OX

Cut off BC = 1 unit.

Since OBC is a right triangle.

OB2 + OC2 = OC2

=> 22 + 12 = OC2

=> OC2 = 4 + 1

=> OC2 = 5

=> OC = √5

With O as centre and OC as radius, draw an arc intersecting OX at D.

Since OC = OD

So, OD represents √5 on XOX’.  Exercise 1.3

Question 1:

Write the following in decimal form and say what kind of decimal expansion each has:

(i) 36/100       (ii) 1/11       (iii) 4            (iv) 3/13          (v) 2/11             (vi) 329/400

Answer:

(i) 36/100 = 0.36

So, the decimal expansion of 36/100 is terminating.

(ii) 1/11 = 0.090909….. = 0.09

Thus, the decimal expansion of 1/11 is non-terminating repeating.

(iii) 4 = 4 + 1/8 = 33/8 = 4.125

Thus, the decimal expansion of 4 is terminating.

(iv) 3/13 = 0.230769230769…. = 0.230769

Thus, the decimal expansion of 3/13 is non-terminating repeating.

(v) 2/11 = 0.181818………. = 0.18

Thus, the decimal expansion of 2/11 is non-terminating repeating.

(vi) 329/400 = 0.8225

Thus, the decimal expansion of is terminating.

Question 2:

You know that 1/7 = 0.142857 can you predict what the decimal expansion of 1/7, 3/7, 4/7, 5/7, 6/7. Are, without actually doing the long division? if so how?

Answer:

Given, 1/7 = 0.142857

Now,

2/7 = 2 * (1/7) = 2 * 0.142857 = 0.285714

3/7 = 3 * (1/7) = 3 * 0.142857 = 0.428571

4/7 = 4 * (1/7) = 4 * 0.142857 = 0.571428

5/7 = 5 * (1/7) = 5 * 0.142857 = 0.714285

6/7 = 6 * (1/7) = 6 * 0.142857 = 0.857142

Thus, without actually doing the long division we can predict the decimal expansions of the

above given rational numbers.

Question 3:

Express the following in the form p/q where p and q are integers and q ≠ 0.

(i) 0.6                                 (ii) 0.47                        (iii) 0.001

Answer:

(i) Let x = 0.6

=> x = 0.66666…….                  ……….1

Multiply equation 1 by 10 on both sides, we get

10x = 6.6666….

=> 10x = 6 + 0.6666……

=> 10x = 6 + x                    [From equation 1]

=> 10x – x = 6

=> 9x = 6

=> x = 6/9

=> x = 2/3

(pi) Let x = 0.47

=> x = 0.477777…….                  ……….1

Multiply equation 1 by 10 on both sides, we get

10x = 4.7777….                  …….2

Multiply equation 2 by 10 on both sides, we get

=> 100x = 47 + 0.7777……

=> 100x = 43 + 4 + 0.7777……

=> 100x = 43 + 4.7777……

=> 100x = 43 + 10x                    [From equation 2]

=> 100x – 10x = 43

=> 90x = 43

=> x = 43/90

(iii) Let x = 0.001

=> x = 0.001001001…….                  ……….1

Multiply equation 1 by 1000 on both sides, we get

1000x = 1 + 0. 001001

=> 1000x = 1 + x                        [From equation 1]

=> 1000x – x = 1

=> 999x = 1

=> x = 1/999

Question 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.

Answer:

Let x = 0. 99999…….                  ……….1

Multiply equation 1 by 10 on both sides, we get

10x = 9. 99999….

=> 10x = 9 + 0. 99999…..

=> 10x = 9 + x                    [From equation 1]

=> 10x – x = 9

=> 9x = 9

=> x = 9/9

=> x = 1

The answer makes sense as 0.99999…. is very close to 1. That’s why we can say that

1. 99999… = 1

Question 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.

Answer:

Since, the number of entries in the repeating block of digits is less than the divisor. In 1/17 the

divisor is 17.

So, the maximum number of digits in the repeating block is 16.

To perform the long division, we have

1/17 = 0.588235294117647

Thus, there are 16 digits in the repeating block in the decimal expansion of 1/17.

Hence, our answer is verified.

Question 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?

Answer:

Let some examples are:

2/5 = 0.4,            1/10 = 0.1,             3/2 = 1.5,                   7/8 = 0.875

The denominator of all the rational numbers are in the form 2m * 5n where m and n are

integers.

Question 7:

Write three numbers whose decimal expansions are non-terminating non-recurring.

Answer:

Three numbers whose decimal expansions are non-terminating non-recurring are:

√2 = 1.414213562……….

√3 = 1.732050808……….

√5 = 2.236067978……….

Question 8:

Find three different irrational numbers between the rational numbers 5/7 and 9/11.

Answer:

5/7 = 0.714285714285……… = 0.714285

9/11 = 0.818181…………….. = 0.81

We know that there are infinite many irrational numbers between two rational numbers.

So, the three irrational numbers are:

1. 0.72722722272222…….., 2. 0. 73733733373333…….., 3. 0. 74744744474444……..

Question 9:

Classify the following numbers as rational or irrational:

(i) √23       (ii) √225       (iii) 0.3796        (iv) 7.478478…..            (v) 1.101001000100001...

Answer:

(i) Since 23 is not a perfect square.

So, √23 is an irrational number.

(ii) 225 = 15 * 15 = 152

So, 225 is a perfect square.

Thus, √225 is a rational number.

(iii) Since 0.3796 is a terminating decimal,

So, it is a rational number.

(iv) 7.478478 = 7.478

Since 7.478 is a non-terminating and recurring (repeating) decimal.

So, it is a rational number.

(v) Since 1.101001000100001... is a non-terminating and non-repeating decimal number.

So, it is an irrational number.

Exercise 1.4

Question 1:

Visualize 3.765 on the number line, using successive magnification.

Answer:

1. First of all, we observe that 3.765 lies between 3 and 4. Divide this portion in 10 equal parts.
2. In the next step, we locate 3.765 between 3.7 and 3.8
3. To get more accurate visualization of representation, we divide this portion of number line

into 10 equal parts and use a magnifying glass to visualize that 3.675 lies between 3.76 and

3.77.

1. Now to visualize 3.765 still more accurately, we divide the portion between 3.76 and and

3.77 into 10 equal parts and locate 3.765. Question 2:

Visualize 4.26 on the number line up to 4 decimal places.

Answer:

1. First of all, we observe that 4.2626(4.26) lies between 4 and 5. Divide this portion into 10

equal parts.

1. In the next step, we locate 4.2626 between 4.2 and 4.3
2. To get more accurate visualization of representation, we divide this portion of number line

into 10 equal parts and use a magnifying glass to visualize that 4.2626 lies between 4.262 and

4.263.

1. Now to visualize 4.2626 still more accurately, we divide the portion between 4.262 and

4.263 into 10 equal parts and locate 4.2626. Exercise 1.5

Question 1:

Classify the following numbers as rational or irrational:

(i) 2 – √5                (ii) (3 + √23) - √23            (iii) 2√7/7√7              (iv) 1/√2               (v) 2π

Answer:

(i) 2 – √5

Since it is a difference of a rational and irrational number,

So, 2 – √5 is an irrational number.

(ii) (3 + √23) - √23 = 3 + √23 - √23 = 3

Which is a rational number.

(iii) 2√7/7√7 = (2 * √7)/(7 * √7) = 2/7

Which is a rational number.

(iv) 1/√2

The quotient of rational and irrational is an irrational number.

So, 1/√2 is an irrational number.

(v) 2π

2π = 2 * π = Product of a rational and an irrational (which is an irrational number)

So, 2π is an irrational number.

Question 2:

Simplify each of the following expressions:

(i) (3 + √3)(2 + √2)          (ii) (3 + √3)(3 - √3)          (iii) (√5 + √2)2            (iv) (√5 - √2)( √5 + √2)

Answer:

(i) (3 + √3)(2 + √2) = 3(2 + √2) + √3(2 + √2)

= 3 * 2 + 3 * √2 + √3 * 2 + √3 * √2

= 6 + 3√2 + 2√3 + √6

(ii) (3 + √3)(3 - √3) = 3(3 - √3) + √3(3 - √3)

= 3 * 3 – 3 * √3 + √3 * 3 - √3 * √3

= 9 - 3√3 + 3√3 – 3

= 6

(iii) (√5 + √2)2 = (√5 + √2) (√5 + √2)

= √5(√5 + √2) + √2(√5 + √2)

= √5 * √5 + √5 * √2 + √2 * √5 + √2 * √2

= 5 + √10 + √10 + 2

= 7 + 2√10

(iv) (√5 - √2)(√5 + √2) = √5( √5 + √2) - √2(√5 + √2)

= √5 * √5 + √5 * √2 - √2 * √5 - √2 * √2

= 5 + √10 - √10 – 2

= 3

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 = c/d.

This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Answer:

When we measure the length of a line with a scale or with any other device, we only get an

approximate rational value, i.e. c and d both are irrational.

=> c/d is irrational and hence π is irrational.

Thus, there is no contradiction in saying that π is irrational.

Question 4:

Represent √9.3 on the number line.

Answer:

To represent √9.3 on the number line, draw AB = 9.3 units. Now produce AB to C such that BC

= 1. Draw the perpendicular bisector of AC which intersects AC to O. Taking O as center and

OA as radius, draw a semi-circle which intersects D to the perpendicular at B. Now taking O as

center and OD as radius, draw an arc which intersects AC produced at E.

Hence, OE = √9.3 Question 5:

Rationalize the denominators of the following:

(i) 1/√7                        (ii) 1/(√7 - √6)                      (iii) 1/(√5 + √2)                   (iv) 1/(√7 - 2)

Answer:

(i) 1/√7 = (1/√7) * (√7/√7)

= (1 * √7)/( √7 * √7)

= √7/7

(ii) 1/(√7 - √6) = {1/(√7 - √6) * {((√7 + √6))/( (√7 + √6))}

= (√7 + √6)/{ (√7 - √6)  (√7 + √6)}

= (√7 + √6)/{( √7)2 – (√6)2 }

= (√7 + √6)/(7 - 6)

= (√7 + √6)

(iii) 1/(√5 + √2) = {1/(√5 + √2) * {((√5 - √2))/( (√5 - √2))}

= (√5 - √2)/{ (√5 + √2)  (√5 - √2)}

= (√5 - √2)/{( √5)2 – (√2)2 }

= (√5 - √2)/(5 - 2)

= (√5 - √2)/3

(iv) 1/(√7 - 2) = {1/(√7 - 2) * {((√7 + 2))/( (√7 + 2))}

= (√7 + 2)/{ (√7 - 2)  (√7 + 2)}

= (√7 + 2)/{( √7)2 – 22 }

= (√7 + 2)/(7 - 4)

= (√7 + 2)/3

Exercise 1.6

Question 1:

Find:             (i) 641/2                           (ii) 321/5                            (iii) 1251/3

Answer:

(i) 641/2 = (82)1/2 = 82 * 1/2 = 81 = 8

(ii) 321/5 = (25)1/5 = 25 * 1/5 = 21 = 2

(iii) 1251/3 = (53)1/3 = 53 * 1/3 = 51 = 5

Question 2:

Find:         (i) 93/2                    (ii) 322/5                          (iii) 163/4                          (iv) 125-1/3

Answer:

(i) 93/2 = (32)3/2 = 32 * 3/2 = 33 = 27

(ii) 322/5 = (25)2/5 = 25 * 2/5 = 22 = 4

(iii) 163/4 = (24)3/4 = 24 * 3/4 = 23 = 8

(iv) 125-1/3 = (53)-1/3 = 53 * (-1/3) = 5-1 = 1/5

Question 3:

Simplify: (i) 22/3 * 21/5               (ii) (1/33)7                  (iii) 111/2/111/4               (iv) 71/2 * 81/2

Answer:

(i) 22/3 * 21/5 = 2(2/3 + 1/5) = 2(10 + 3)/15 = 213/15

(ii) (1/33)7 = (3-3)7 = 3-3*7 = 3-21

(iii) 111/2/111/4 = 111/2 * 11-1/4 = 111/2 – 1/4 = 11(2 - 1)/4 = 111/4

(iv) 71/2 * 81/2 = (7 * 8)1/2 = 561/2

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