Estimating and Ordering Real Numbers — Solutions

Click any answer to watch the solution video.

1. Integer bounds for surds

   1. √5: 2²=4, 3²=9 → 2 < √5 < 3
   2. √14: 3²=9, 4²=16 → 3 < √14 < 4
   3. √30: 5²=25, 6²=36 → 5 < √30 < 6
   4. √60: 7²=49, 8²=64 → 7 < √60 < 8
   5. √90: 9²=81, 10²=100 → 9 < √90 < 10
   6. √110: 10²=100, 11²=121 → 10 < √110 < 11
   7. √3: 1²=1, 2²=4 → 1 < √3 < 2
   8. √55: 7²=49, 8²=64 → 7 < √55 < 8
   9. √200: 14²=196, 15²=225 → 14 < √200 < 15
   10. √0.5: 0.5²=0.25, 1²=1, and 0.7²=0.49<0.5 → 0 < √0.5 < 1 (between 0.7 and 0.8)

2. Estimate to 1 decimal place

   1. √7: 2²=4, 3²=9. Try 2.6: 6.76<7. Try 2.7: 7.29>7. 7 closer to 7.29 so √7 ≈ 2.6 (actually ≈2.646, rounds to 2.6)
   2. √20: 4²=16, 5²=25. Try 4.4: 19.36, 4.5: 20.25. √20 ≈ 4.5 (actually ≈4.472, rounds to 4.5)
   3. √50: 7²=49, 8²=64. 7.0²=49, 7.1²=50.41. √50 ≈ 7.1 (actually ≈7.071)
   4. √85: 9²=81, 10²=100. Try 9.2: 84.64, 9.3: 86.49. √85 ≈ 9.2 (actually ≈9.220)
   5. √130: 11²=121, 12²=144. Try 11.4: 129.96 ≈ 130. √130 ≈ 11.4 (actually ≈11.402)
   6. √3: 1²=1, 2²=4. Try 1.7: 2.89, 1.8: 3.24. 3 is closer to 3.24 so √3 ≈ 1.7 (actually ≈1.732)

3. Ascending order

   1. √6 ≈2.449, √7 ≈2.646, 5/2=2.5, 2.5, 2.6. Order: √6, 2.5=5/2, 2.6, √7 (tie: 5/2 = 2.5)
   2. √18=3√2≈4.243, 3√2≈4.243, 2√5≈4.472, 4, 4.1. Order: 4, 4.1, 3√2=√18, 2√5
   3. √9.9≈3.146, 3.14, π≈3.1416, 22/7≈3.1429, √10≈3.162. Order: 3.14, π, 22/7, √9.9, √10
   4. √50/5=5√2/5=√2≈1.414, √2≈1.414, 1.4, 7/5=1.4. Order: 1.4=7/5, √50/5=√2 (all equal or very close — 1.4 < √2 = √50/5)

4. Compare surd expressions

   1. 3√2 vs 2√3: (3√2)²=18, (2√3)²=12. 18>12, so 3√2 > 2√3
   2. 4√3 vs 3√5: (4√3)²=48, (3√5)²=45. 48>45, so 4√3 > 3√5
   3. √8 + √2 vs √18: √8+√2 = 2√2+√2 = 3√2. √18 = 3√2. So √8 + √2 = √18
   4. 5 − √3 vs 3: √3 > 0, so 5 − √3 < 5, but we need to compare with 3. 5 − √3 vs 3 ⇒ 2 vs √3. 2²=4, (√3)²=3, so 2 > √3, meaning 5 − √3 > 3
   5. 2 + √5 vs √13: (2+√5)² = 4 + 4√5 + 5 = 9 + 4√5. (√13)² = 13. Is 9 + 4√5 > 13? i.e. 4√5 > 4? √5 > 1 ✓. So 2 + √5 > √13

5. Problem solving

   1. √3 + √3 vs √6:
      1. 2√3 ≈ 2×1.7 = 3.4. √6 ≈ 2.4.
      2. (2√3)² = 12. (√6)² = 6. 12 ≠ 6, so 2√3 ≠ √6.
      3. √3 + √3 = 2√3
   2. Rope lengths: √20 = 2√5 ≈ 4.47, 4.4, 9/2 = 4.5. Ascending: 4.4, √20, 4.5 (i.e. 9/2). Total = 4.4 + √20 + 4.5 = 8.9 + 2√5 m
   3. Speed of sound:
      1. 380²=144400 > 142884, 370²=136900. Try 378: 378²=142884. So √142884 = 378 m/s exactly.
      2. 378² = 142884 ⇒ Yes, 142884 is a perfect square. √142884 = 378 m/s (a whole number).
   4. Pattern on number line: √2≈1.41, √3≈1.73, √5≈2.24, √7≈2.65, √11≈3.32. Gaps: 0.32, 0.51, 0.41, 0.67. The gaps are not uniform — they tend to get smaller as the numbers get larger (since the square root function grows more slowly for larger values). For example √100=10 and √101≈10.05; the gap is tiny.

6. Identifying a student error

   1. 2√3 ≈ 2 × 1.7 = 3.4; √6 ≈ 2.4
   2. (2√3)² = 12 and (√6)² = 6. Since 12 ≠ 6, we have 2√3 ≠ √6
   3. √3 + √3 = 2√3

7. Compare surd expressions without a calculator

   1. 3√2 vs 2√3: (3√2)² = 18; (2√3)² = 12. 18 > 12, so 3√2 > 2√3
   2. 4√3 vs 3√5: (4√3)² = 48; (3√5)² = 45. 48 > 45, so 4√3 > 3√5
   3. 5 − √3 vs 3: 5 − √3 > 3 iff 2 > √3 iff 4 > 3 ✓. So 5 − √3 > 3
   4. 2 + √5 vs √13: (2+√5)² = 9 + 4√5 ≈ 17.9; (√13)² = 13. So 2 + √5 > √13

8. Order from smallest to largest

   1. √5≈2.24, 2.2, 9/4=2.25, √6≈2.45, 2.4. Ascending: 2.2, √5, 9/4, 2.4, √6
   2. 3.14, π≈3.1416, 22/7≈3.1429, √10≈3.162, 3.2. Ascending: 3.14, π, 22/7, √10, 3.2
   3. √50/5 = √2 ≈ 1.414; 7/5 = 1.4. Ascending: 7/5 = 1.4, √2 = √50/5

9. Speed of sound

   1. 377² = 142 129 and 378² = 142 884, so √142 884 lies between 377 and 378 (or is exactly 378)
   2. 378 × 378 = 142 884 ✓. Exact speed = 378 m/s
   3. 378 m/s is rational — it is a whole number (an integer).

10. Surds on the number line — pattern investigation

    1. √2 ≈ 1.4, √3 ≈ 1.7, √5 ≈ 2.2, √7 ≈ 2.6, √11 ≈ 3.3
    2. Gaps: √3−√2≈0.3, √5−√3≈0.5, √7−√5≈0.4, √11−√7≈0.7. These vary because the surds skip non-consecutive integers. For consecutive integers √n and √(n+1), gaps get smaller as n increases.
    3. √(n+1) − √n = 1/(√(n+1) + √n). As n grows, the denominator grows, so the gap shrinks toward 0. The square root function grows more slowly for larger values.