1. Squaring the two given irrational numbers: (√2)² = 2 and (√3)² = 3.
2. Let p be a rational number between √2 and √3, so 2 < p² < 3.
3. One possible value: p² = 2.25  ⇒  p = 1.5, which rounds to 2.

1. Rational number between −2 and 6 = −2 + 62 = 42 = 2.
2. Check: −2 < 2 < 6. ✓

1. (a) → (ii) Any real number has a non-negative square (since squares are always ≥ 0).
2. (b) → (iv) Numbers of the form pq (q≠0) are rational numbers by definition.
3. (c) → (i) A number that neither terminates nor repeats is irrational.
4. (d) → (iii) A number with exactly two factors (1 and itself) is prime.

1. (a) → (iv) The least prime number is 2.
2. (b) → (i) The least composite number is 4.
3. (c) → (ii) The least whole number is 0.
4. (d) → (iii) The least natural number is 1.

1. (a) → (iii) Every natural number is also an integer.
2. (b) → (ii) The least natural number is 1 ("One").
3. (c) → (iv) There are infinitely many integers.
4. (d) → (i) The least whole number is 0 ("Zero").

1. The product of two negative integers is always positive: 7 × 12 = 84.

1. Performing long division of 1 by 11 gives 0.090909…
2. 111 = 0.0̄9̄, which is a non-terminating recurring (repeating) decimal expansion.

1. We get the maximum value of c−ba when c is the largest and a, b are the smallest values.
2. Take c = 25 (largest), a = 2 (smallest), b = 3 (smallest): 25−32 = 222 = 11.

1. Let x = 18.48̄ = 18.4848… …(i).
2. Multiply (i) by 100: 100x = 1848.4848… …(ii).
3. Subtract (i) from (ii): 99x = 1830  ⇒  x = 183099 = 61033.

1. A rational number lying between −2 and −3 is ½[(−2)+(−3)] = −52 = −2.5.
2. A rational number between −2 and −52: ½[(−2)+−52] = −94.
3. A rational number between −52 and −3: ½[−52+(−3)] = −114.
4. Ordering: −2 > −94 > −52 > −114 > −3.

1. Consider the squares 2² = 4 and 3² = 9. We need irrational numbers whose squares lie strictly between 4 and 9.
2. Since 4 < 5 < 6 < 9, we get 2 < √5 < √6 < 3.
3. √5 and √6 are irrational (5 and 6 are not perfect squares) and both lie between 2 and 3.

1. Taking a = 1.3, b = 1.4: ½(1.3 + 1.4) = ½(2.7) = 1.35.

1. Let x = 0.357̄ = 0.35777… So, 100x = 35.777… …(i) and 1000x = 357.777… …(ii).
2. Subtracting (i) from (ii): 900x = 322  ⇒  x = 322900 = 161450.

1. Multiply numerator and denominator by the conjugate (7−2√14): 5−3√147+2√14 × 7−2√147−2√14.
2. Numerator: (5−3√14)(7−2√14) = 35 − 10√14 − 21√14 + 6(14) = 35 + 84 − 31√14 = 119 − 31√14.
3. Denominator: 7² − (2√14)² = 49 − 56 = −7.
4. Result = 119 − 31√14−7 = −119 + 31√147.

1. By long division: 11 ÷ 24 = 0.45833…

1. By long division method: 33 ÷ 26 = 1.2692307692307…, which repeats from "692307".

1. Let x = 0.99999… …(a). Multiply both sides by 10: 10x = 9.9999… …(b).
2. Subtract (a) from (b): 9x = 9  ⇒  x = 1.
3. So 0.99999… = 1. This makes sense because 0.999… goes on forever, leaving no gap between it and 1 — so they represent the same number.

1. i(a) Let a = √2 and b = √3, two different irrational numbers. Then a+b = √2 + √3 is also irrational.
2. i(b) Let a = √2 and b = √3. Then a×b = √6 is also irrational.
3. ii Let a = 2√3 and b = 5√3, two different irrational numbers. Then ab = 2√35√3 = 25, which is rational.

1. Here a = 13, b = 12, n = 3. Using the formula, three rational numbers between a and b are: a + d, a + 2d, a + 3d, where d = b−an+1.
2. d = 1/2 − 1/34 = 1/64 = 124.
3. So the numbers are: 13 + 124 = 38,   13 + 2124 = 512,   13 + 3124 = 1124.

1. (i) 64100 = 0.64 — Terminating decimal.
2. (ii) 211 = 0.1̄8̄ = 0.181818… — Repeating decimal.
3. (iii) 7⅛ = 578 = 7.125 — Terminating decimal.
4. (iv) 513 = 0.3̅84615̅ — Repeating decimal.
5. (v) 7⅓ = 223 = 7.3̄ — Repeating decimal.
6. (vi) 169400 = 0.4225 — Terminating decimal.

1. Consider some rational numbers between 3 and 5 that are perfect squares of terminating decimals: 3.24, 3.61, 4.41, and 4.84.
2. Their square roots: √3.24 = 1.8, √3.61 = 1.9, √4 = 2, √4.41 = 2.1, √4.84 = 2.2.
3. So √3 < 1.8 < 1.9 < 2 < 2.1 < 2.2 < √5 (since √3 ≈ 1.732 and √5 ≈ 2.236).

1. Take LCM of denominators (6, 1) = 6, so −2 = −126.
2. Rational numbers between −56 and −126 include −66, −76, −86, −96, −106, −116.
3. Simplified: −1, −76, −43, −32, −53, −116.

1. If x and y are two rational numbers such that x < y, then x+y2 is a rational number between them.
2. Here 3 < 3+42 < 4  ⇒  3 < 72 < 4.

1. (i) Using (a+b)(a−b) = a²−b²: 5² − (√5)² = 25 − 5 = 20.
2. (ii) Using (a+b)² = a²+2ab+b²: (√5)² + 2√5√2 + (√2)² = 5 + 2√10 + 2 = 7 + 2√10.
3. (iii) = √2·√5 + √2·√7 + √3·√5 + √3·√7 = √10 + √14 + √15 + √21.
4. (iv) = 4√3 − 4√7 + √5·√3 − √5·√7 = 4√3 − 4√7 + √15 − √35.

1. 23.3408 = 23340810000 = 23340810⁴.
2. The denominator's prime factors are 2 and 5 (since 10 = 2×5): 23340810⁴ = 2334082⁴×5⁴.
3. 233408 = 2⁶ × 7 × 521, so 2334082⁴×5⁴ = 2²×7×5215⁴ after cancelling common factors.

1. By long division, 17 = 0.1̅42857̅ (repeating block: 142857).
2. We can predict the others by multiplying this repeating block and cyclically shifting it: 27 = 2 × 0.1̅42857̅ = 0.2̅85714̅.
3. 37 = 0.4̅28571̅, 47 = 0.5̅71428̅, 57 = 0.7̅14285̅, 67 = 0.8̅57142̅.
4. Each is a cyclic rotation of the same six digits (142857), just starting at a different point in the cycle.

1. An irrational number lying between a and b (in this pattern) can be taken as √(ab).
2. Irrational number between √5 and √7 = √(√5 × √7) = (35)¹⁴.

1. √3 ≈ 1.732 and √8 ≈ 2.828.
2. Two irrational (non-repeating, non-terminating) numbers between them: 1.9090090009… and 2.5151151115…
3. These satisfy √3 < 1.9090090009… < 2.5151151115… < √8.

1. Let y = −13, x = −12 (unlike denominators). Compute d = y−xn+1 with n=2: d = [−13−−12]/3 = 1/63 = 118.
2. Two rational numbers between them: x + d = −12 + 118 = −818 = −49, and x + 2d = −12 + 218 = −718.

1. 0.18696 can be written directly as 18696100000.
2. Simplifying by cancelling common factors: 18696100000 = 233712500.

1. Suppose, for contradiction, that √5 is rational, so √5 = pq where p, q are integers with no common factor other than 1.
2. Squaring both sides: 5q² = p². Since 5 divides p², and 5 is prime, 5 must also divide p.
3. Let p = 5m for some integer m. Substituting: 5q² = 25m²  ⇒  q² = 5m². So 5 divides q², and hence 5 divides q too.
4. This means 5 is a common factor of both p and q, contradicting our assumption that they share no common factor.

Steps of construction

1. Draw a line segment XY.
2. Draw OB = 1 cm, perpendicular to line XY at O.
3. From B, draw an arc of radius 3 cm cutting XY at A.
4. Join BA and OA.

Construction: OAB is a right-angled triangle with OB = 1 cm, BA = 3 cm

1. OAB is a right-angled triangle. By the Pythagorean theorem: AB² = OB² + OA².
2. 3² = 1² + OA²  ⇒  OA² = 9 − 1 = 8  ⇒  OA = √8 cm.

1. (i) x = 1115 − 25. Using LCM 15: x = 1115 − 615 = 515 = 13.
2. (ii) x = 56 + 13. Using LCM 6: x = 56 + 26 = 76.

1. (i) Brahmagupta used Fortunes (Dhana) for positive numbers (wealth) and Debts (ṛiṇa) for negative numbers. The set including positives, negatives, and zero is called Integers, denoted by Z.
2. (ii) Financial equation: 1200 + (−850) + (−450) + 600 = 1200 − 850 − 450 + 600 = ₹500 (fortune).
3. (iii)(a) (−12)×(−15) = +180 — product of two debts is a fortune (⊖×⊖ = ⊕).
4. (iii)(b) (−8)×7 = −56 — product of a debt and a fortune is a debt (⊖×⊕ = ⊖).
5. (iii)(c) 0−(−25) = 0+25 = 25 — subtracting a debt is the same as adding a fortune.

OR

1. (−3)×(−4) + (−6)×5 − 0×100 = 12 + (−30) − 0 = −18.

−18 plotted on the number line, 18 units to the left of zero

1. −18 is not a Natural Number (those start from 1) and not a Whole Number (those start from 0), but it is an Integer (Z includes all negative numbers, zero, and positive numbers).

1. (i) 4 fingers × 3 joints per finger = 12 joints. This basic pattern directly correlates to the historical base-12 (duodecimal) counting framework.
2. (ii) No, natural numbers are not closed under subtraction.
3. (iii) Let a = 3 and b = 5, where a, b ∈ N. Their difference is a − b = 3 − 5 = −2. The result −2 is an integer, but it is not a Natural Number.

OR

1. Ratio: Ingots : Bags = 152.
2. For 12 bags: Ingots = 152 × 12 = 15 × 6 = 90 copper ingots.

1. (i) The denominator is 20. Its prime factorization is 20 = 2² × 5. Since the prime factors consist only of 2 and 5, the decimal expansion will terminate.
2. (ii) 511 = 0.454545… = 0.4̅5̅.
3. (iii) When dividing by 7, the only possible non-zero remainders are 1, 2, 3, 4, 5, or 6. Because there are only 6 possible values, a remainder must inevitably repeat by the 7th step, forcing the sequence to loop indefinitely.

OR

1. 0.375 = 3751000. Dividing numerator and denominator by their GCD (125): 375÷125 = 3, 1000÷125 = 8. So 0.375 = 38.
2. Since HCF(3, 8) = 1, they are co-prime.

1. (i) √3 is an irrational number, since it cannot be written in the form pq (p, q integers, q≠0). Its decimal expansion is 1.73205080757…, which is non-terminating and non-repeating.
2. (ii) We should multiply √2 by √2 (or any non-zero multiple of √2): √2 × √2 = √4 = 2, which is rational 21.
3. (iii) First convert 3.2 to its simplest fractional form: 3.2 = 3210 = 165. Multiplying numerator and denominator by 2: 3210; by 3: 4815.

OR

1. We always get an irrational number. Property: Rational Number + Irrational Number = Irrational Number.
2. Example: 5 (rational) + √3 (irrational) = 5 + √3, which is completely irrational.

1. (i) Side of Tile A = √25 = 5 cm. Since 5 = 51, and both 5 and 1 are integers with denominator ≠ 0, it is a rational number.
2. (ii) Side of Tile B = √7 = 2.6457513… This decimal never ends and never repeats; it cannot be written as pq, so it is an irrational number.
3. (iii) Side of Tile A = 5 = 51, so p = 5 and q = 1.

OR

1. Yes. Both are real numbers, because the real number system includes all rational numbers and all irrational numbers — every measurable length is a real number.

1. Let √3 + 2 and √2 − 3 be two irrational numbers.
2. Their difference = (√3 + 2) − (√2 − 3) = √3 + 2 − √2 + 3 = √3 − √2 + 5.

1. The square of every rational number is non-negative, so a² ≥ 0, b² ≥ 0, c² ≥ 0.
2. Their sum is given to be zero: a² + b² + c² = 0.
3. A sum of non-negative numbers can only be zero when each term is zero: a² = 0, b² = 0, c² = 0.
4. Taking square roots: a = 0, b = 0, c = 0.

1. We know that 2√5 = √(2²×5) = √20, and 3√3 = √(3²×3) = √27.
2. Thus we have √20 < √21 < √22 < √23 < √24 < √25 < √26 < √27.
3. So any five irrational numbers between 2√5 and 3√3 are: √21, √22, √23, √24, and √26 (note: √25 = 5 is rational, so it's excluded).

Long division working for 103, 78, and 17, with remainder patterns shown

1. For 78, we find that the remainder becomes zero, so the decimal expansion is 78 = 0.875. We call this a terminating decimal expansion.
2. For 103 and 17, we notice that the remainders repeat after a certain stage, forcing the decimal expansion to go on forever — we have a repeating block of digits in the quotient.
3. 103 = 3.3̄ and 17 = 0.1̅42857̅. We say these expansions are non-terminating recurring.

1. Draw a number line and mark a point O, representing zero. Suppose point A represents 1 unit, so OA = 1.
2. Draw AB ⊥ OA at A such that AB = OA = 1 unit. By the Pythagorean theorem: OB² = OA² + AB² = 1² + 1² = 2, so OB = √2.
3. Draw an arc with centre O and radius OB, cutting the number line at A₂. Then OA₂ = √2.
4. Draw a right triangle OB₁B₂ such that B₁B₂ = 1 (where OB₁ = √2). Then OB₂² = (√2)² + 1² = 2 + 1 = 3, so OB₂ = √3.

Square-root spiral: successive right triangles construct √2, √3, …, √8 on the number line

1. Draw a right triangle OB₂B₃ such that B₂B₃ = 1 (where OB₂ = √3). Then OB₃² = (√3)² + 1² = 3 + 1 = 4, so OB₃ = 2... and continuing this pattern with the same unit-length perpendicular each time builds up through √5, √6, and √7.
2. Repeating the process once more from OB with length √7: OB² = (√7)² + 1² = 7 + 1 = 8, so OB = √8.
3. Each new hypotenuse OB_k, drawn with an arc onto the number line, gives the points representing √6, √7, and √8 in turn — this stepped construction is called the square root spiral.