Rational and Irrational Numbers — Solutions

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1. Classify each number

   1. (a) √4: √4 = 2 = 2/1. Rational
   2. (b) √2: 2 is not a perfect square; 1.4142… non-terminating, non-recurring. Irrational
   3. (c) π: π = 3.14159… non-terminating, non-recurring. Irrational
   4. (d) 0.333…: Recurring decimal = 1/3. Rational
   5. (e) 1.5: Terminating decimal = 3/2. Rational
   6. (f) −7: −7 = −7/1. Rational
   7. (g) 0: 0 = 0/1. Rational
   8. (h) √9: √9 = 3 = 3/1. Rational
   9. (i) √5: 5 is not a perfect square; 2.2360… non-terminating, non-recurring. Irrational
   10. (j) 5/8: Already in form p/q. Rational

2. Order from smallest to largest

   1. (a) √3 ≈ 1.732, √5 ≈ 2.236, 2.1, 7/3 ≈ 2.333, π−1 ≈ 2.142. Order: √3, 2.1, √5, π−1, 7/3
   2. (b) −√2 ≈ −1.414, −1, −0.7, −1/2 = −0.5. Order: −√2, −1, −0.7, −1/2
   3. (c) 3.14, π ≈ 3.1416, 22/7 ≈ 3.1429, √10 ≈ 3.162, 3.2. Order: 3.14, π, 22/7, √10, 3.2

3. Convert recurring decimals to fractions

   1. (a) x=0.555…, 10x=5.555…, 9x=5, x=5/9
   2. (b) x=0.222…, 10x=2.222…, 9x=2, x=2/9
   3. (c) x=0.999…, 10x=9.999…, 9x=9, x=1
   4. (d) x=0.1818…, 100x=18.1818…, 99x=18, x=18/99=2/11
   5. (e) x=0.3636…, 100x=36.3636…, 99x=36, x=36/99=4/11
   6. (f) x=0.1666…, 10x=1.666…, 100x=16.666…, 90x=15, x=15/90=1/6
   7. (g) x=0.8333…, 10x=8.333…, 100x=83.333…, 90x=75, x=75/90=5/6
   8. (h) x=0.142857142857…, 999999x=142857, x=142857/999999=1/7

4. True or False

   1. True. Every integer n = n/1, which is in the form p/q.
   2. False. For example, 1/2 is rational but not an integer.
   3. False. √25 = 5, a whole number, so it is rational. The square root sign alone does not make a number irrational.
   4. False. 0.999… = 1 exactly (let x=0.999…, then 10x=9.999…, 9x=9, x=1).
   5. False. Counter-example: √2 + (−√2) = 0, which is rational.
   6. False. 22/7 is a rational approximation to π. The true value π = 3.14159… is irrational and cannot be written exactly as any fraction.

5. Number set membership

   1. 12: ℕ, ℤ, ℚ, ℝ
   2. −5: ℤ, ℚ, ℝ
   3. 3/4: ℚ, ℝ
   4. √2: Irrational, ℝ
   5. 0: ℤ, ℚ, ℝ (In QCAA, natural numbers start at 1, so 0 ∉ ℕ)
   6. −2.7: ℚ, ℝ (−2.7 = −27/10)
   7. √81: √81 = 9; ℕ, ℤ, ℚ, ℝ
   8. π/2: Irrational, ℝ (irrational divided by non-zero rational remains irrational)

6. Square garden

   1. Side = √50 = 5√2 m
   2. 5√2 is irrational because √2 is irrational, and multiplying by a non-zero rational (5) preserves irrationality.
   3. 5√2 ≈ 5 × 1.414 = 7.07 m. Yes, 7 m is a reasonable approximation (error < 1%).

7. Circumference of circular path

   1. The physicist rounded 5π to a finite decimal. 5π ≈ 15.70796…, not exactly 15.708.
   2. 5π is irrational: π is irrational, and 5 × (irrational) = irrational.
   3. A correct statement: “The circumference is exactly 5π metres, which is approximately 15.708 m (to 3 decimal places).”

8. Student claims

   1. Amir is incorrect. Recurring decimals (e.g. 0.333… = 1/3) go on forever but are rational. Only non-terminating, non-recurring decimals are irrational.
   2. √2 × √2 = 2 (rational)  √3 × √3 = 3 (rational)  √2 × √8 = √16 = 4 (rational)
      Conclusion: Bella is incorrect. The product of two irrational numbers is not always irrational — it can be rational.
   3. Carlos is correct. For example, between 1/3 and 1/2, we can find 5/12. In general, (p/q + r/s)/2 is a rational number between p/q and r/s.

9. Proof that √2 is irrational

   1. √2 = p/q ⇒ 2 = p²/q² ⇒ p² = 2q²
   2. 2q² is even, so p² is even. Since odd × odd = odd, p² even implies p must be even.
   3. Let p = 2k. Then (2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k². So q² is even, meaning q is even.
   4. Both p and q are even, so they share a common factor of 2. This contradicts the assumption that p/q is in simplest form. Therefore √2 cannot be written as p/q — it is irrational.

10. The Golden Ratio

    1. φ = (1 + √5)/2 is irrational because √5 is irrational, and (1 + irrational)/2 remains irrational.
    2. φ² = [(1+√5)/2]² = (1 + 2√5 + 5)/4 = (6 + 2√5)/4 = (3 + √5)/2. Meanwhile φ + 1 = (1+√5)/2 + 1 = (3+√5)/2. So φ² = φ + 1. ✓
    3. Let x = 0.999…. Then 10x = 9.999…. Subtracting: 9x = 9, so x = 1. Therefore 0.999… = 1 exactly. This shows a number can have two valid decimal representations.