Chapter 1: Number Systems
Exercise Questions & Answers

Answer:
Yes, zero is a rational number. It can be written as 0/q where q is any non-zero integer. For example, 0/1, 0/2, 0/(-3), etc. all represent zero.

Answer:
We can find rational numbers between 3 and 4 by expressing them with a common denominator.
Let's write 3 = 21/7 and 4 = 28/7.
Six rational numbers between them are: 22/7, 23/7, 24/7, 25/7, 26/7, 27/7.
Alternatively: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 are also rational numbers between 3 and 4.

Answer:
We can find rational numbers between 3/5 and 4/5 by expressing them with a larger denominator.
Multiply numerator and denominator by 6: 3/5 = 18/30 and 4/5 = 24/30.
Five rational numbers between them are: 19/30, 20/30, 21/30, 22/30, 23/30.
Simplifying: 19/30, 2/3, 7/10, 11/15, 23/30.

Answer:
(i) True - Natural numbers (1, 2, 3, ...) are included in whole numbers (0, 1, 2, 3, ...).
(ii) False - Integers include negative numbers (-1, -2, -3, ...) which are not whole numbers.
(iii) False - Rational numbers include fractions like 1/2, 3/4, etc. which are not whole numbers.

Answer:
(i) True - Real numbers consist of both rational and irrational numbers, so every irrational number is a real number.
(ii) False - Not every point on the number line is of the form √m. For example, negative numbers and rational numbers like 1/2 cannot be expressed as √m where m is a natural number.
(iii) False - Real numbers include both rational and irrational numbers. Rational numbers like 2, 3/4, etc. are real but not irrational.

Answer:
No, the square roots of all positive integers are not irrational. For example, √4 = 2, √9 = 3, √16 = 4, etc. are rational numbers because they are integers.

Answer:
To represent √5 on the number line:
1. Draw a number line and mark point O at 0.
2. Mark point A at 2 units from O.
3. Draw a perpendicular at A of length 1 unit and mark the endpoint as B.
4. Join OB. By Pythagoras theorem, OB = √(2² + 1²) = √5.
5. With O as center and OB as radius, draw an arc intersecting the number line at point C.
6. OC represents √5 on the number line.

Answer:
(i) 36/100 = 0.36 → Terminating decimal
(ii) 1/11 = 0.090909... = 0.09̅ → Non-terminating recurring decimal
(iii) 4 1/8 = 33/8 = 4.125 → Terminating decimal
(iv) 3/13 = 0.230769230769... = 0.230769̅ → Non-terminating recurring decimal
(v) 2/11 = 0.181818... = 0.18̅ → Non-terminating recurring decimal
(vi) 329/400 = 0.8225 → Terminating decimal

Answer:
Yes, we can predict the decimal expansions by multiplying the repeating block of 1/7 by the numerator.
Since 1/7 = 0.142857̅, the repeating block is 142857.
2/7 = 2 × 0.142857̅ = 0.285714̅
3/7 = 3 × 0.142857̅ = 0.428571̅
4/7 = 4 × 0.142857̅ = 0.571428̅
5/7 = 5 × 0.142857̅ = 0.714285̅
6/7 = 6 × 0.142857̅ = 0.857142̅
Each fraction has the same digits in the repeating block but in a different cyclic order.

Answer:
(i) Let x = 0.666...
Then 10x = 6.666...
Subtracting: 10x - x = 6.666... - 0.666...
9x = 6 → x = 6/9 = 2/3
So, 0.6̅ = 2/3

(ii) Let x = 0.4777...
Then 10x = 4.777... and 100x = 47.777...
Subtracting: 100x - 10x = 47.777... - 4.777...
90x = 43 → x = 43/90
So, 0.47̅ = 43/90

(iii) Let x = 0.001001001...
Then 1000x = 1.001001001...
Subtracting: 1000x - x = 1.001001... - 0.001001...
999x = 1 → x = 1/999
So, 0.001̅ = 1/999

Answer:
Let x = 0.99999...
Then 10x = 9.99999...
Subtracting: 10x - x = 9.99999... - 0.99999...
9x = 9 → x = 1
So, 0.99999... = 1/1 = 1
This might seem surprising, but it makes sense because the difference between 1 and 0.999... is infinitely small (approaches zero). In mathematics, 0.999... is exactly equal to 1.

Answer:
For a rational number p/q, the maximum number of digits in the repeating block is (q-1).
For 1/17, the maximum number of digits in the repeating block is 16.
Performing the division: 1/17 = 0.0588235294117647̅ (16 digits in the repeating block).

Answer:
For a rational number p/q to have a terminating decimal expansion, the prime factorization of q must be of the form 2^m × 5ⁿ, where m and n are non-negative integers.
Examples:
1/2 = 0.5 (q=2=2¹)
3/4 = 0.75 (q=4=2²)
7/8 = 0.875 (q=8=2³)
1/5 = 0.2 (q=5=5¹)
3/10 = 0.3 (q=10=2¹×5¹)

Answer:
Three numbers with non-terminating non-recurring decimal expansions (irrational numbers) are:
(i) √2 = 1.414213562...
(ii) π = 3.141592653...
(iii) 0.101001000100001...

Answer:
First, 5/7 ≈ 0.714285... and 9/11 ≈ 0.818181...
Three irrational numbers between them are:
(i) 0.720720072000720000...
(ii) 0.730730073000730000...
(iii) 0.750750075000750000...
These are non-terminating non-recurring decimals, hence irrational.

Answer:
(i) √23 → Irrational (23 is not a perfect square)
(ii) √225 = 15 → Rational (perfect square)
(iii) 0.3796 → Rational (terminating decimal)
(iv) 7.478478... = 7.478̅ → Rational (non-terminating recurring decimal)
(v) 1.101001000100001... → Irrational (non-terminating non-recurring decimal)

Answer:
(i) 2 - √5 → Irrational (difference of rational and irrational)
(ii) (3 + √23) - √23 = 3 → Rational (simplifies to integer)
(iii) 2√7/7√7 = 2/7 → Rational (simplifies to rational number)
(iv) 1/√2 → Irrational (reciprocal of irrational is irrational)
(v) 2π → Irrational (product of non-zero rational and irrational)

Answer:
(i) (3 + √3)(2 + √2) = 3×2 + 3×√2 + √3×2 + √3×√2 = 6 + 3√2 + 2√3 + √6
(ii) (3 + √3)(3 - √3) = 3² - (√3)² = 9 - 3 = 6
(iii) (√5 + √2)² = (√5)² + 2×√5×√2 + (√2)² = 5 + 2√10 + 2 = 7 + 2√10
(iv) (√5 - √2)(√5 + √2) = (√5)² - (√2)² = 5 - 2 = 3

Answer:
There is no contradiction. When we measure the circumference (c) and diameter (d) of a circle with physical instruments, we only get approximate rational values. The actual values of c and d are such that their ratio is exactly π, which is irrational. The irrationality of π means that it cannot be expressed as a ratio of two integers, but it can be expressed as the ratio of two real numbers (c and d).

Answer:
To represent √9.3 on the number line:
1. Draw a line segment AB of length 9.3 units.
2. Extend AB to point C such that BC = 1 unit.
3. Find the midpoint O of AC.
4. Draw a semicircle with center O and radius OA.
5. Draw a perpendicular to AC at B, intersecting the semicircle at D.
6. BD = √9.3
7. With B as center and BD as radius, draw an arc intersecting the number line at E.
8. BE represents √9.3 on the number line.

Answer:
(i) 1/√7 = (1×√7)/(√7×√7) = √7/7
(ii) 1/(√7 - √6) = [1×(√7 + √6)] / [(√7 - √6)(√7 + √6)] = (√7 + √6)/(7 - 6) = √7 + √6
(iii) 1/(√5 + √2) = [1×(√5 - √2)] / [(√5 + √2)(√5 - √2)] = (√5 - √2)/(5 - 2) = (√5 - √2)/3
(iv) 1/(√7 - 2) = [1×(√7 + 2)] / [(√7 - 2)(√7 + 2)] = (√7 + 2)/(7 - 4) = (√7 + 2)/3

Answer:
(i) 64^1/2 = √64 = 8
(ii) 32^1/5 = ⁵√32 = 2 (since 2⁵ = 32)
(iii) 125^1/3 = ³√125 = 5 (since 5³ = 125)

Answer:
(i) 9^3/2 = (9^1/2)³ = 3³ = 27
(ii) 32^2/5 = (32^1/5)² = 2² = 4
(iii) 16^3/4 = (16^1/4)³ = 2³ = 8
(iv) 125^-1/3 = 1/(125^1/3) = 1/5

Answer:
(i) 2^2/3 · 2^1/5 = 2^{(2/3 + 1/5)} = 2^{(10/15 + 3/15)} = 2^13/15
(ii) (1/3³)⁷ = 1/3²¹ = 3^-21
(iii) 11^1/2/11^1/4 = 11^{(1/2 - 1/4)} = 11^1/4
(iv) 7^1/2 · 8^1/2 = (7·8)^1/2 = 56^1/2 = √56 = 2√14