SEBA Class 9 Mathematics Chapter 1 Number Systems

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SEBA Class 9 Mathematics Chapter 1 Number Systems

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Number systems

Chapter – 1

Exercise 1.1

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

Ans: Yes, zero is a rational number and it can be written in the form of p/q as 0 = 0/1 here p = 0 and q = 1

Notice that q can be any number you wish. [∵ 0 = 0/2 = 0/3 = 0/4 etc.]

2. Find six rational numbers between 3 and 4.

Ans: Let a = 3, and b = 4

Rational number lying between 3 and 4 is a + b/2 i.e. 3 + 4/2 = 7/2

Now rational number between 3 and

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Then required six rational numbers between 3 and 4 are 7/2, 13/4, 15/4, 27/8, 29/8, 31/8

ALITER

We want to find all six at one go. Since we want 6 numbers, so write.

3 = 21/6 + 1 = 21/7 and 4 = 28/6 + 1 = 28/7

Now six rational numbers between 3 and 4 are 22/7, 23/7, 24/7, 25/7, 26/7, 27/7

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

Ans: To find five rational numbers between 3/5 and 4/5, we take 3/5 and 4/5 as a rational number with denominator 30. i.e. 3/5 = 3/5 × 6/6 = 18/30 and 4/5 = 4/5 × 6/6 = 24/30

Then, required six rational numbers between 3/5 and 4/5 are 19/30, 20/30, 21/30 and 23/30.

4. Are the following statements true or false? Give reasons for your answer.

(i) Every natural number is a whole number.

Ans: Yes, because each natural number is a whole number.

(ii) Every integer is a whole number.

Ans: False, because all negative natural numbers (-1, -2, -3, …) are not whole numbers.

(iii) Every rational number is a whole number.

Ans: False, because the rational numbers of the type 2/5, 2/5, 3/5 etc are not whole numbers.

Exercise 1.2

1. State whether the following statements are true or false. Jus-tify your answer.

(i) Every irrational number is a real number.

Ans: True, because the set of every rational and every irrational number is called a real number.

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

Ans: False, no negative number can be the square root of any natural number.

(iii) Every real number is an irrational number.

Ans: False, because the real number is the set of every irrational and rational number. For example, 2 is real but not irrational.

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.

Ans: No, the square roots of all positive integers are not always irrational.

For Example: √4= 2 is a rational number.

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

Ans: They represent √5 on the number line, taking 0 at zero on the number line. Again taking OA = 1 unit in the positive direction of the number line.

Then OB = √1² + 1² = √2

Construct OD of unit length perpendicular to OB.

Then OD= √(√2)² +1² = √3

Construct DE of unit length perpendicular to OD.

Then OE = √(√3)² + 1² = √4 = 2

Construct EF of unit length perpendicular to OE. Then

OF = √2² + 1² = √5

Using a compass, with centre) and radius OF, draw an arc which intersects the number line in the point R. Then R corresponds to √5.

4. Classroom activity (Constructing the ‘Square root spiral’).

Take a large sheet of paper and construct the ‘square spiral’ in the following fashion. Start with a point O and draw a line segment P₁ P₂ per-perpendicular to OP₁ of unit length. Now draw a line segment P₂P₃ perpendicular-lar to OP₂. Then draw a line segment P₃P₄ perpendicular to OP₃.

Continuing in this manner, you can get the line segment Pn – 1 Pn by drawing a line segment of unit length perpendicular to OPn – 1. In this man-ner, you will have created the points P₂, P₃,…, Pn, …, and join-ing them to a create beautiful spiral depicting.

√2,√3, √4,…

Ans: For self-practice.

Exercise 1.3

1. Write the following in decimal form and say what kind of deci- mal expansion each has:

(i) 36/100

Ans: 36/100 = 0.36 terminating decimal.

(ii) 1/11

Ans:

Remainders: 1, 1, 1, 1………,

Divisor: 11

The decimal expansion is non-terminating repeating.

(iii) 4 + 1/8

Ans: 4 + 1/8 = 33/8

∴ 4 + 1/8 = 4.125

The decimal representation is terminating.

(iv) 3/13

Ans:

The decimal expansion is non-terminating repeating.

(v) 2/11

Ans:

Remainders: 9, 2, 9, 2….

Divisor: 0.1818……..

We write; 2/11 = 0.1818……

(vi) 329/400

Ans:

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 divi-sion? If so, how?

Ans:

3. Express the following in the form p/q, where p and q are inte-gers and q ≠ 0.

Ans:

Now, adding 6 both side in equation (1)

x + 6 = 0.66666….+ 6

or x + 6 = 6.6666… (2)

Again, multiplying by 10 both sides in equation (1)

10 x = 0.66666…x10

or 10 x = 6.6666… (3)

Now, form equations (2) and (3)

x + 6 = 10 x or 9 x = 6 ∴ x = 6/9

or x = 0.47777

Now, multiply 10th side in equation (1)

10 x = 0.47777…x10

or 10 x = 4.77777

Again, multiply 100 both side in equation (1)

100 = 0.477777 … X 100

or 100 x = 47.777 …. (3)

Subtract equation (2) from equation (3)

90 x = 43.0000

Which is in the form of p/q

or, x = 0.001001001 …….(1)

Now, add 1 both side in equation (1)

x + 1 = 0.001001001…+1

or x + 1 = 1.001001001 ….. (2)

Again, multiply 1000 both side in equation (1)

1000 × x = 1000 x 0.001001001 …..

or 1000x = 1.001001001 …. (3)

From equations (2) and (3) we get

x + 1 = 1000x or 999x = 1 ∴ x = 1/999

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.

Ans: Let x = 0.9999….(1)

Add 9 both side in equation (1)

x + 9 = 9 + 0.99999…

or x + 9 = 9.99999.…(2)

Again, multiply 10 both side in equation (1)

10 x = 9.9999….(3)

From equations (2) and (3) we get 10 x = x + 9

or 9 x = 9 or x = 1 Therefore, 0.9999 … = 1

It is because there is an infinite 9 coming after the point; which is very-very close to 1.

5. 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?

Ans: Some examples of rational number having terminating decimal representations are:

(i) 1/2 = 0.5

(ii) 7/4 = 1.75

(iii) 7/4 = 0.875

(iv) 2/5 = 0.4

It is clear that the prime factorisation of q has only power of 2 or power of 5 or both.

6. Write three numbers whose decimal expansions are non termi- nating non-recurring.

Ans: We know that the decimal expansions of an irrational number is

non-terminating non-recurring. Three examples of such numbers are as

√2 = 1.4142135…

√3 = 1.722050807…

π = 3.1415926535…

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

Ans: To find an irrational number between 5/7 and 9/11 is non terminat-ing non recurring lying between them.

Therefore, the required three different irrational number which are ly-ing between 5/11 and 9/11 are:

0.720720072000 ….

0.730730073000… and 0.740740074000…

8. Classify the following number as rational or irrational:

(i) √23

Ans: We have √23 = 4.795831523…

It is non terminating non recurring.

So, √23 is an irrational number.

(ii) √225

Ans: We have √225 = 15 or √225 = 15/1 which is a rational number.

(iii) 0.3796

Ans:

So, 0.3796 is a rational number.

(iv) 7.478478…

Ans:

which is non terminating recurring.

Therefore, 7.478478…. is an irrational number.

(v) 1.101001000100001

Ans: We have 1.101001000100001….

which is non terminating non recurring.

Therefore, 1.101001000100001…. is an irrational number.

Exercise 1.4

1. Visualise 3.765 on the number line, using successive magnification.

Ans:

Ans:

Exercise 1.5

1. Classify the following numbers as rational irrational:

(i) 2 – √5

Ans: We have 2 -√5

оr 2-2.236067977…

or -0.236067977….

which is non terminating non recurring. So, it is an irrational number.

(ii) (3 + √23)-√23

Ans: We have 3 + √23 – √23

or 3 + √23 – √23 or 3

which is a rational number.

(iii) (2√7/7√7

Ans: We have, 2√7/7√7 or which is a rational number.

(iv) 1/√2

Ans:

Therefore the quotient of this number is irrational.

(ν) 2 π

Ans: We have, 2 π = 2 x 3.1415926535…..

= 6.2831853070….

which is non terminating non recurring.

Therefore, it is an irrational number.

2. Simplify each of the following expressions:

(i) (3 + √3) (2 + √2)

Ans: We have, (3 + √3) (2 +√2)

or 6 + 3√2 + 2√3 + √6

(ii) (3 + √3) (3 – √3)

Ans: We have, (3 + √3) (3 – √3)

we know that (a + b) (a – b) = a² – b²

∴ (3 + √3) (3 – √3) = (3)² – (√3)² = 9 – 3 = 6

So, (3 – √3) (3 – √3) = 6

(iii) (√5 + √2)²

Ans: We have, (√5 + √2)²

we know that (a + b)² = a² + b² + 2 ab

∴ (5 + √2)² = (√5)² + (√2)² + 2√5.√2 = 5 + 2 + 2√10

or (√5 + √2)² = 7 + 2√10

(iv) (√5 – √2) (√5 + √2)

Ans: We have, (√5 – √2) (√5 + √2) = (√5)² – (√2)²

[∵ (a + b) (a – b) = a² – b² = 5-2]

∴ (√5 – √2) (√5 + √2) = 3

(v) (3√5 – 4√3)²

Ans: (3√5 – 4√3)²

= (3√5)² – 2.3√5.4√3 + (4√3)²

= 9 x 5-24 √15 + 1 6 x 3

= 45 – 24 √15 + 48

= 93 – 24 √15

(vi) (√7 – 6) (√3 – √7)

Ans: (√7 – 6) (√3 – √7)

= √(√3 – √7) – 6(√3 – √7)

= √21 – √49 – 6√3 + 6√7

= √21 – 6√3 – 7 + 6√7

(vii) (2 + √6) (4 + √6)

Ans: (2 + √6) (4 + √6)

= 2(4 + √6) + √6 (4 + √6)

= 8 + 2√6 + 4√6 + √36

= 8 + 6 + 6√6 = 14 + 6√6

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?

Ans: Circumference = 2 π r (Irrational number) of the circle (c)

r → Radius of the circle

Diameter of circle (d) = 2 r (Rational number)

с/d = 2 πr/2 r Irrational number/Rational number = Irrational number

As we know, an irrational number divided by a rational number results in an Irrational number.

4. Represent √9.3 on the number line.

Ans: To represent √9.3 on the number line. We mark a point B on the number line so that AB = 9.3 units. Again mark a point C so that BC = 1 unit. Now take the mid point of AC and mark that point as 0. Draw a semicircle with centre 0 and radius OC. Draw the line perpendicular to AC passing through B and in-intersecting the semicircle at D. Then, BD = √9.3.

Again taking BD as a radius and draw an arc which intersecting the number line atE. Then E represents √9.3 on the number line.

Take B at zero on the number line, then point E represents √9.3.

5. Rationalise the denominators of the following:

(i)