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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
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)
Ans: We have
Multiply √7 both numerator and denominator
(ii)
Ans: We have,
Multiply √7 + √6 both numerator and denominator
(iii)
Ans: We have,
Multiply √5- √2 Both numerator and denominator
(iv)
Ans: We have,
(v)
Ans:
(vi)
Ans:
(vii)
Ans:
(viii)
Ans:
(ix)
Ans:
6. (i)
Ans:
(ii)
Ans:
(iii)
Ans:
| Exercise 1.6 |
1. Find:
(i) 64 ¹/²
Ans: We have,
(ii) 32 ¹/⁵
Ans: We have, 32¹/² = (2⁵)¹/⁵
(iii) 125 ¹/³
Ans: We have,
2. Find:
(i) 9³/²
Ans: We have,
Therefore, (9)³/² = 27
(ii) 32²/⁵
Ans: We have,
Therefore, (32)²/⁵ = 4
(iii) 16³/⁴
Ans: We have,
Therefore, (16)³/⁴ = 8
(iv) (125)–¹/³
Ans: We have,
Therefore,
3. Simplify
(i) 2²/³.2¹/⁵
Ans: We have, 2²/³.2¹/⁵
We know that aᵐ × aⁿ = aᵐ+ⁿ
Ans: We have,
Ans:
(iv) 7¹/².8¹/²
Ans: We have, 7¹/² × 8¹/²
We know that aᵐ×bᵐ = (a × b)ᵐ
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