CBSE • Class 9 • Mathematics • Chapter 1 (Number Systems) • Exercise 1.6

Exercise 1.6: Number Systems — NCERT Solutions

Representing irrational numbers like √x on the number line using constructions.

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What this exercise covers

√ on number linePythagoras constructionSpatial intuition

Step-by-step solutions — Exercise 1.6

2 solved questions • Each solution includes a Socratic hint, full working and a common-mistake callout • Last reviewed 2026-09-03

Exercise 1.6 Q1 • 3 marks

Represent 5\sqrt{5} on the number line using a geometric construction.
Hint (Socratic — try this first)▾
Can you build a right triangle whose hypotenuse squared equals 5 using two whole-number legs?
Step-by-step solution▾

Idea: Use the Pythagoras theorem, since 5=22+12\sqrt{5} = \sqrt{2^2 + 1^2}.

Construction steps:

  1. Draw a number line and mark point OO at 00 and point AA at 22 (so OA=2OA = 2 units).
  2. At AA, draw a line segment ABAB perpendicular to the number line with AB=1AB = 1 unit.
  3. Join OBOB. By Pythagoras: OB=OA2+AB2=22+12=4+1=5OB = \sqrt{OA^2 + AB^2} = \sqrt{2^2 + 1^2} = \sqrt{4+1} = \sqrt{5}
  4. With OO as centre and radius OB=5OB = \sqrt{5}, draw an arc cutting the number line at point PP.

Conclusion: Point PP represents 5\sqrt{5} on the number line.

Common mistake:
Students choose legs whose squares do not add up to 5 (e.g. legs 1 and 1, giving 2\sqrt2), or forget to transfer the hypotenuse onto the line with a compass arc.
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Exercise 1.6 Q2 • 3 marks

Using the spiral (square-root spiral) construction, explain how 3\sqrt{3} can be obtained starting from a segment of unit length.
Hint (Socratic — try this first)▾
If you already have a segment of length 2\sqrt{2}, what unit-length perpendicular gives you 3\sqrt{3} as the new hypotenuse?
Step-by-step solution▾

Building the spiral:

  1. Draw OA=1OA = 1 unit. At AA draw AB⊥OAAB \perp OA with AB=1AB = 1. Then OB=12+12=2.OB = \sqrt{1^2 + 1^2} = \sqrt{2}.
  2. At BB, draw BC⊥OBBC \perp OB with BC=1BC = 1 unit. By Pythagoras in triangle OBCOBC: OC=OB2+BC2=(2)2+12=2+1=3.OC = \sqrt{OB^2 + BC^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}.

Conclusion: The segment OCOC has length 3\sqrt{3}. Continuing this process (adding unit perpendiculars) generates 4,5,…\sqrt{4}, \sqrt{5}, \dots, forming the square-root spiral.

Common mistake:
Students use OAOA (the original leg) instead of the previous hypotenuse OBOB as the base for the next right triangle, breaking the spiral pattern.
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How to approach Exercise 1.6

  1. Re-read the chapter summary first. Open Number Systems and refresh the key concepts: Rational numbers, Irrational numbers, Decimal expansions, Laws of exponents.
  2. Try each problem yourself before opening the solution. Ten focused minutes per problem usually beats reading two finished solutions.
  3. Use Guru AI for the questions you get stuck on. The Socratic AI tutor walks you through it with hints instead of dictating the answer.
  4. Mark the questions you got wrong and revisit them after 24 hours — the spacing is what locks the method into long-term memory.

All exercises in Number Systems

  1. Exercise 1.1 — Rational vs irrational numbers — closure, density on the line, and locating simple surds.
  2. Exercise 1.2 — Decimal expansions of rationals — terminating and non-terminating recurring forms.
  3. Exercise 1.3 — Real numbers — operations and representability; approximating irrational numbers.
  4. Exercise 1.4 — Laws of exponents extended to rational and real bases and exponents.
  5. Exercise 1.5 — Rationalising denominators with surds and revisiting identities with roots.
  6. Exercise 1.6 — Representing irrational numbers like √x on the number line using constructions.

← Exercise 1.5

Rationalising denominators with surds and revisiting identities with roots.

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