Mathematics solution NCERT
Class 9 - Chapter 3: The World of Numbers
Q1: Convert the following rational numbers into decimal form by the process of long division.
(i) 3/50
Divide 3 by 50:
3 ÷ 50 = 0.06
Since the division terminates, it is a terminating decimal.
Answer: 3/50 = 0.06
(ii) 2/9
Divide 2 by 9:
2 ÷ 9 = 0.22222...
The digit 2 repeats forever.
Therefore, it is a non-terminating recurring decimal.
Answer: 2/9 = 0.22222...
Q2: Prove that √5 is an irrational number.
Proof:
Assume that √5 is rational.
Then it can be written in the form:
√5 = p/q
where p and q are integers having no common factor and q ≠ 0.
Squaring both sides:
5 = p²/q²
p² = 5q²
Therefore, p² is divisible by 5.
Hence p is also divisible by 5.
Let:
p = 5k
Substituting:
(5k)² = 5q²
25k² = 5q²
5k² = q²
This shows that q² is divisible by 5.
Hence q is also divisible by 5.
Thus both p and q are divisible by 5.
This contradicts the assumption that p and q have no common factor.
Therefore our assumption is false.
Hence, √5 is an irrational number.
Q3: Convert the following decimal numbers into the form p/q.
(i) 12.6
12.6 = 126/10
= 63/5
Answer: 63/5
(ii) 0.0120
0.0120 = 120/10000
= 3/250
Answer: 3/250
(iii) 3.052
3.052 = 3052/1000
= 763/250
Answer: 763/250
(iv) 1.235̅
Let
x = 1.235235235...
1000x = 1235.235235...
x = 1.235235...
Subtracting:
999x = 1234
x = 1234/999
Answer: 1234/999
(v) 0.23̅
x = 0.232323...
100x = 23.232323...
Subtract:
99x = 23
x = 23/99
Answer: 23/99
(vi) 2.05̅
x = 2.05555...
10x = 20.5555...
100x = 205.5555...
Subtract:
90x = 185
x = 185/90
= 37/18
Answer: 37/18
(vii) 2.125̅
x = 2.125125125...
1000x = 2125.125125...
Subtracting:
999x = 2123
x = 2123/999
Answer: 2123/999
(viii) 3.125̅
x = 3.125125125...
1000x = 3125.125125...
Subtract:
999x = 3122
x = 3122/999
Answer: 3122/999
(ix) 2.1625̅
x = 2.162516251625...
10000x = 21625.16251625...
Subtract:
9999x = 21623
x = 21623/9999
Answer: 21623/9999
Q4: Locate the following rational numbers on the number line.
(i) 0.532
0.532 lies between 0.53 and 0.54 on the number line.
0.53 -------- ● -------- 0.54
0.532
(ii) 1.15̅
1.15̅ = 1.151515...
It lies between 1.15 and 1.16.
1.15 -------- ● -------- 1.16
1.1515...
Q5: Find 6 rational numbers between 3 and 4.
Solution:
Write the numbers with denominator 7:
3 = 21/7
4 = 28/7
The six rational numbers between them are:
| Fraction | Decimal Form |
|---|---|
| 22/7 | 3.142857... |
| 23/7 | 3.285714... |
| 24/7 | 3.428571... |
| 25/7 | 3.571428... |
| 26/7 | 3.714285... |
| 27/7 | 3.857142... |
Answer:
22/7, 23/7, 24/7, 25/7, 26/7, 27/7
Q6: Find 5 rational numbers between 2/5 and 3/5.
Solution:
2/5 = 20/50
3/5 = 30/50
Any fractions between 20/50 and 30/50 will be rational numbers between 2/5 and 3/5.
The required five rational numbers are:
| Rational Numbers |
|---|
| 21/50 |
| 22/50 |
| 23/50 |
| 24/50 |
| 25/50 |
Answer:
21/50, 22/50, 23/50, 24/50, 25/50
Q7: Find 5 rational numbers between 1/6 and 2/5.
Solution:
LCM of 6 and 5 = 30
1/6 = 5/30
2/5 = 12/30
The five rational numbers between them are:
6/30, 7/30, 8/30, 9/30, 10/30
Simplifying where possible:
| Fraction |
|---|
| 1/5 |
| 7/30 |
| 4/15 |
| 3/10 |
| 1/3 |
Answer:
1/5, 7/30, 4/15, 3/10, 1/3
Q8: If x/3 + x/5 = 16/15, find the rational number x.
Solution:
x/3 + x/5 = 16/15
Taking x common:
x(1/3 + 1/5) = 16/15
x(5/15 + 3/15) = 16/15
x(8/15) = 16/15
x = (16/15) × (15/8)
x = 2
Answer: x = 2
Q9: Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.
Solution:
Given:
a + 1/b = 0
a = -1/b
Multiplying both sides by b:
ab = -1
Since:
ab = -1
and -1 is negative,
ab is always negative.
Answer: ab is negative.
Q10: A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.
Solution:
If the last non-zero digit occurs in the fourth decimal place, then the number can be written as:
a.bcdw
where w is a non-zero digit.
Such a decimal can always be written as:
p/10000
= p/10⁴
where p is an integer.
Since the fourth decimal place contains a non-zero digit, p cannot be divisible by 10.
Second Part:
It is not necessary that the denominator in lowest form contains 2⁴ or 5⁴.
Example:
0.2500 = 2500/10000
= 1/4
The denominator in lowest form is only:
4 = 2²
and does not contain 2⁴ or 5⁴.
Answer:
- The number can be written as p/10⁴.
- No, the denominator in lowest form need not contain 2⁴ or 5⁴.
Q11: Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.
Solution:
18/125
The denominator is:
125 = 5³
Since the denominator contains only the prime factor 5, the decimal expansion is terminating.
To determine the number of decimal places:
18/125 = (18 × 8)/(125 × 8)
= 144/1000
= 0.144
The decimal has 3 places.
Answer:
- Terminating decimal
- Number of decimal places = 3
Q12: A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
Solution:
Given denominator:
2³ × 5
= 8 × 5
= 40
To convert the denominator into a power of 10:
40 × 25 = 1000
= 10³
Therefore the decimal expansion will terminate after 3 decimal places.
Example:
1/40 = 0.025
which has 3 decimal places.
Answer: The decimal expansion will have 3 decimal places.
Q13: Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ − k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ − k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
Solution:
Given:
a = 7/12
b = 5/6
First express both fractions with the same denominator.
5/6 = 10/12
Thus,
a = 7/12
b = 10/12
Here,
k₂ - k₁ = 10 - 7 = 3
Since 3 is not greater than 6, multiply both fractions by 3:
a = 21/36
b = 30/36
Now,
k₂ - k₁ = 30 - 21 = 9
which is greater than 6.
Five rational numbers between a and b:
| Fraction |
|---|
| 22/36 |
| 23/36 |
| 24/36 |
| 25/36 |
| 26/36 |
Why k₂ − k₁ > n + 1?
To obtain n rational numbers between two fractions having the same denominator, there must be at least n integers available between their numerators.
If the difference between numerators is not sufficiently large, there will not be enough integers available to create n distinct fractions.
Therefore, the condition
k₂ − k₁ > n + 1
ensures that at least n distinct rational numbers can be formed between them.
Answer:
22/36, 23/36, 24/36, 25/36, 26/36
Q14: Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y and z must be simultaneously zero.
Proof:
Given:
x + y + z = 0
xy + yz + zx = 0
Squaring the first equation:
(x + y + z)² = 0
x² + y² + z² + 2(xy + yz + zx) = 0
Since
xy + yz + zx = 0
we get:
x² + y² + z² = 0
Now x², y² and z² are squares of rational numbers.
Each square is non-negative.
The sum of three non-negative numbers can be zero only when each one is zero.
Therefore,
x² = 0
y² = 0
z² = 0
Hence,
x = 0, y = 0, z = 0
Answer: The only possible values are
x = y = z = 0
Q15: Show that the rational number (a + b)/2 lies between the rational numbers a and b.
Proof:
Assume that
a < b
We must show that
a < (a+b)/2 < b
First inequality:
a < b
Adding a to both sides:
2a < a+b
Dividing by 2:
a < (a+b)/2
Second inequality:
a < b
Adding b to both sides:
a+b < 2b
Dividing by 2:
(a+b)/2 < b
Combining both results:
a < (a+b)/2 < b
Therefore,
(a+b)/2
lies between a and b.
| Example | Value |
|---|---|
| a | 2 |
| b | 8 |
| (a+b)/2 | 5 |
Clearly,
2 < 5 < 8
Hence proved.