Number Systems — Practice

Question 1

Is √169 rational or irrational?

Accepted Answer

Step 1 — Check if the radicand is a perfect square.
  169 = 13². So √169 = 13.

Step 2 — Classify the result.
  13 is an integer, and every integer is rational (write it as 13/1).

✦ Answer: Rational.

Key insight: a square root is irrational only when the radicand is NOT a perfect square. Always check first.

Question 2

Is 0.101001000100001… rational or irrational?

Accepted Answer

Step 1 — Examine the decimal pattern.
  After each '1', the number of '0's increases by one: 0.1, 0.10, 0.100, 0.1000, …
  So the digits never settle into a repeating block.

Step 2 — Apply the decimal classification.
  Non-terminating AND non-repeating → irrational.

✦ Answer: Irrational.

Watch out: 'non-terminating' alone is NOT enough — 0.3̄ = 1/3 is non-terminating but it's rational because it repeats.

Question 3

Will 7/250 give a terminating or non-terminating decimal?

Accepted Answer

Step 1 — Ensure the fraction is in lowest terms.
  gcd(7, 250) = 1, so 7/250 is already reduced.

Step 2 — Prime-factorise the denominator.
  250 = 2 × 5³. Only the primes 2 and 5 appear.

Step 3 — Apply the terminating-decimal test.
  Denominator = 2^a · 5^b ⇒ terminating.

Step 4 — Verify by computation.
  7/250 = 28/1000 = 0.028.

✦ Answer: Terminating decimal. 7/250 = 0.028.

Question 4

Will 11/30 give a terminating or non-terminating decimal?

Accepted Answer

Step 1 — Check lowest terms.
  gcd(11, 30) = 1 → already reduced.

Step 2 — Prime-factorise the denominator.
  30 = 2 × 3 × 5. The factor 3 is NOT 2 or 5.

Step 3 — Apply the test.
  Denominator has a prime other than 2 or 5 ⇒ non-terminating (but repeating, since the number is rational).

Step 4 — Verify.
  11/30 = 0.3666… = 0.36̄.

✦ Answer: Non-terminating, repeating. 11/30 = 0.36̄.

Question 5

Simplify √48 − √27 + √12.

Accepted Answer

Step 1 — Factor each radicand into perfect-square × other.
  48 = 16 × 3 → √48 = √16 · √3 = 4√3.
  27 = 9 × 3 → √27 = √9 · √3 = 3√3.
  12 = 4 × 3 → √12 = √4 · √3 = 2√3.

Step 2 — Combine like surds.
  4√3 − 3√3 + 2√3 = (4 − 3 + 2)√3 = 3√3.

✦ Answer: 3√3.

Key strategy: pull out the largest perfect square from inside the radical, then combine like terms.

Question 6

Find (125)^(2/3).

Accepted Answer

Step 1 — Express the base as a power.
  125 = 5³.

Step 2 — Apply the power-of-a-power rule (a^p)^q = a^(pq).
  (5³)^(2/3) = 5^(3 · 2/3) = 5².

Step 3 — Evaluate.
  5² = 25.

✦ Answer: 25.

Alternative reading: a^(p/q) = (q-th root of a)^p = cube-root of 125, then square = 5² = 25.

Question 7

Simplify (16)^(−3/4).

Accepted Answer

Step 1 — Express the base as a power.
  16 = 2⁴.

Step 2 — Apply (a^p)^q = a^(pq).
  (2⁴)^(−3/4) = 2^(4 · −3/4) = 2^(−3).

Step 3 — Apply the negative-exponent rule a^(−n) = 1/a^n.
  2^(−3) = 1/2³ = 1/8.

✦ Answer: 1/8.

Sanity check: a negative fractional exponent always gives a fraction smaller than 1 (for base > 1). 1/8 fits.

Question 8

Insert two rational numbers between 3/5 and 4/5.

Accepted Answer

Step 1 — Convert to a common denominator with room to insert.
  3/5 = 9/15, 4/5 = 12/15. There's only ONE integer (10, 11) between 9 and 12 — not enough to pick two yet, so scale further.

Step 2 — Multiply both by something larger, say ×4.
  3/5 = 36/60, 4/5 = 48/60.
  Now any integer numerator between 36 and 48 works.

Step 3 — Pick two values.
  Pick 37/60 and 47/60 (or 38/60 = 19/30 and 46/60 = 23/30 — many answers possible).

Step 4 — Verify.
  3/5 = 36/60 < 37/60 < 47/60 < 48/60 = 4/5. ✓

✦ Answer: 37/60 and 47/60 (one valid set; infinitely many possible).

Alternative method — averaging: mean of 3/5 and 4/5 = 7/10. Mean of 3/5 and 7/10 = 13/20. So 13/20 and 7/10 also work.

Question 9

Express 0.6̄ as a fraction.

Accepted Answer

Step 1 — Let the decimal equal x and write it out.
  Let x = 0.6̄ = 0.6666…

Step 2 — Multiply by 10^n where n is the repeat-block length.
  Repeat block is '6' (length 1), so multiply by 10.
  10x = 6.6666…

Step 3 — Subtract the original from the multiplied form.
  10x − x = 6.6666… − 0.6666…
  9x = 6.

Step 4 — Solve and reduce.
  x = 6/9 = 2/3.

Step 5 — Verify.
  2/3 = 0.6666… ✓

✦ Answer: 0.6̄ = 2/3.

Question 10

Express 1.32̄ as a fraction (the 32 repeats).

Accepted Answer

Step 1 — Let x = the decimal.
  x = 1.32̄ = 1.32323232…

Step 2 — Identify the repeat-block length.
  Repeat block = '32', length 2. Multiply by 10² = 100.
  100x = 132.32323232…

Step 3 — Subtract.
  100x − x = 132.32323… − 1.32323…
  99x = 131.

Step 4 — Solve.
  x = 131/99.

Step 5 — Check for further simplification.
  gcd(131, 99) = 1 (131 is prime).

✦ Answer: 1.32̄ = 131/99.

Note: had the decimal been 1.3̄2̄ vs 1.3̄2̄, the meaning would differ — always confirm exactly which digits repeat.

Question 11

Rationalise 1/(√5 − √2).

Accepted Answer

Step 1 — Identify the conjugate.
  The denominator is (√5 − √2). Its conjugate is (√5 + √2).

Step 2 — Multiply numerator and denominator by the conjugate.
  1/(√5 − √2) × (√5 + √2)/(√5 + √2).

Step 3 — Apply (a − b)(a + b) = a² − b² to the denominator.
  Denominator = (√5)² − (√2)² = 5 − 2 = 3.

Step 4 — Write the simplified fraction.
  = (√5 + √2)/3.

✦ Answer: (√5 + √2)/3.

Why multiply by the conjugate? Because the conjugate kills the radicals via the difference-of-squares identity.

Question 12

Rationalise 3/(√3 + √2) and write in simplest form.

Accepted Answer

Step 1 — Conjugate of (√3 + √2) is (√3 − √2).

Step 2 — Multiply numerator and denominator by the conjugate.
  3/(√3 + √2) × (√3 − √2)/(√3 − √2).

Step 3 — Simplify the denominator using (a + b)(a − b) = a² − b².
  Denominator = (√3)² − (√2)² = 3 − 2 = 1.

Step 4 — Multiply out the numerator.
  Numerator = 3(√3 − √2) = 3√3 − 3√2.

Step 5 — Combine (denominator is 1, so just keep the numerator).
  = 3√3 − 3√2.

✦ Answer: 3√3 − 3√2.

Observation: when the denominator simplifies to 1, the rationalised form is just the numerator. Always look for this short-circuit.

Question 13

Find the value of (3 + √2)(3 − √2).

Accepted Answer

Step 1 — Recognise the difference-of-squares pattern.
  (a + b)(a − b) = a² − b², where a = 3 and b = √2.

Step 2 — Apply the identity directly.
  = 3² − (√2)² = 9 − 2.

Step 3 — Evaluate.
  = 7.

✦ Answer: 7.

Why useful? The conjugate identity is THE workhorse of rationalisation. Memorise it.

Question 14

Expand (√5 + √3)².

Accepted Answer

Step 1 — Apply the identity (a + b)² = a² + 2ab + b².
  Here a = √5 and b = √3.

Step 2 — Substitute.
  (√5)² + 2·√5·√3 + (√3)².

Step 3 — Simplify each term.
  (√5)² = 5.
  2·√5·√3 = 2·√(5·3) = 2√15.
  (√3)² = 3.

Step 4 — Combine.
  5 + 2√15 + 3 = 8 + 2√15.

✦ Answer: 8 + 2√15.

Common error: students sometimes write (√5 + √3)² = 5 + 3 = 8 (wrong — they forget the cross-term 2√15).

Question 15

Locate √3 on the number line.

Accepted Answer

Step 1 — Recall the Pythagorean spiral idea.
  To locate √n, find a right-triangle whose hypotenuse is √n, then transfer that hypotenuse onto the number line via a compass arc.

Step 2 — Build a triangle with hypotenuse √3.
  Use 1² + (√2)² = 1 + 2 = 3. So a right triangle with legs 1 and √2 has hypotenuse √3.

Step 3 — Construct.
  (i) On the number line, mark O at 0 and A at 1.
  (ii) From the previous construction (or by drawing a unit perpendicular at A), locate the point at √2 — call it B (so that OB has length √2; B sits above the number line one unit up from A).
  (iii) At B, draw a unit perpendicular to OB, reaching a point C. Now OC is the hypotenuse of a right triangle with legs OB = √2 and BC = 1. By Pythagoras OC = √((√2)² + 1²) = √3.

Step 4 — Transfer the length.
  With O as centre and radius OC, draw an arc cutting the number line at a point P. The point P represents √3 on the number line.

✦ Answer: P (constructed as above) is √3.

Alternative spiral root method (the spiral of Theodorus): start with the unit-1-1-√2 triangle, then add unit perpendiculars to generate √3, √4, √5, … one after another.

Question 16

If a = 2 + √3 and b = 2 − √3, find a² + b² and a·b.

Accepted Answer

Step 1 — Notice a and b are conjugate pairs.
  This is going to be a (sum)(difference) situation, perfect for shortcuts.

Step 2 — Compute a + b and a − b.
  a + b = (2 + √3) + (2 − √3) = 4.
  a − b = (2 + √3) − (2 − √3) = 2√3.

Step 3 — Compute a·b using the conjugate identity.
  a·b = (2 + √3)(2 − √3) = 2² − (√3)² = 4 − 3 = 1.

Step 4 — Compute a² + b² using (a + b)² − 2ab.
  a² + b² = (a + b)² − 2(a·b) = 4² − 2(1) = 16 − 2 = 14.

✦ Answer: a·b = 1, a² + b² = 14.

Why this shortcut? Squaring 2 + √3 directly involves a cross-term. Using (a + b)² − 2ab keeps the arithmetic clean. Memorise this trick — appears in 1 of every 2 Class 9 papers.

Question 17

If x = 3 + 2√2, find x + 1/x.

Accepted Answer

Step 1 — Find 1/x by rationalising.
  1/x = 1/(3 + 2√2). Conjugate of the denominator: (3 − 2√2).

Step 2 — Multiply numerator and denominator by the conjugate.
  1/x = (3 − 2√2)/[(3 + 2√2)(3 − 2√2)].

Step 3 — Simplify the denominator using (a + b)(a − b) = a² − b².
  Denominator = 3² − (2√2)² = 9 − 4·2 = 9 − 8 = 1.

Step 4 — Simplified 1/x.
  1/x = (3 − 2√2)/1 = 3 − 2√2.

Step 5 — Add x and 1/x.
  x + 1/x = (3 + 2√2) + (3 − 2√2) = 6.

✦ Answer: x + 1/x = 6.

Follow-up favourite: find x² + 1/x². Use (x + 1/x)² = x² + 2 + 1/x², so x² + 1/x² = 36 − 2 = 34.

Question 18

If 2^x = 3^y = 6^z, prove that 1/x + 1/y = 1/z.

Accepted Answer

Step 1 — Let the common value be k.
  Set 2^x = 3^y = 6^z = k. (k > 0 since all bases are > 0.)

Step 2 — Solve each equation for the base.
  From 2^x = k → 2 = k^(1/x).
  From 3^y = k → 3 = k^(1/y).
  From 6^z = k → 6 = k^(1/z).

Step 3 — Use the relation 6 = 2 · 3.
  k^(1/z) = 6 = 2 · 3 = k^(1/x) · k^(1/y) = k^(1/x + 1/y).

Step 4 — Compare exponents (since k > 0, k ≠ 1 makes the bases match uniquely).
  1/z = 1/x + 1/y. ∎

✦ Answer: 1/x + 1/y = 1/z is proved.

Elegant pattern: any time you have a^x = b^y = (ab)^z (or similar), 1/x + 1/y = 1/z. Generalises to more bases.

Question 19

Prove that √2 + √3 is irrational.

Accepted Answer

Step 1 — Set up a proof by contradiction.
  Suppose √2 + √3 = r, where r ∈ ℚ (rational).

Step 2 — Isolate one of the surds.
  √3 = r − √2.

Step 3 — Square both sides.
  3 = r² − 2r√2 + 2.
  Rearrange: 2r√2 = r² + 2 − 3 = r² − 1.

Step 4 — Solve for √2.
  √2 = (r² − 1)/(2r).
  The RHS is rational (subtract, multiply, divide of rationals).
  But the LHS, √2, is known to be irrational.

Step 5 — Conclude.
  Contradiction: √2 ≠ rational. So our assumption (that √2 + √3 is rational) is false.
  Therefore √2 + √3 is IRRATIONAL. ∎

✦ Answer: Proof complete.

Generalises: the sum of two distinct surds √a and √b (where a, b are not perfect squares, ab not a perfect square) is always irrational. Same proof pattern.

Question 20

Find an irrational number between √2 and √3.

Accepted Answer

Step 1 — Estimate the interval numerically.
  √2 ≈ 1.41421…
  √3 ≈ 1.73205…
  So we need an irrational in (1.41421, 1.73205).

Step 2 — Construct a non-terminating, non-repeating decimal in this range.
  Recipe: pick a 'core' value between the bounds, then append a never-repeating tail.
  Try 1.5 as the core. Append 1.5010010001000010000010000001… (each block of 0s grows by one).

Step 3 — Verify the value lies in (√2, √3).
  Our number ≈ 1.501… which is > 1.414 and < 1.732. ✓

Step 4 — Verify it's irrational.
  The decimal expansion is non-terminating (it never ends) AND non-repeating (the gaps between 1s grow without bound). Therefore irrational.

✦ Answer: 1.501001000100001000001… (one example among infinitely many).

Many answers possible: any non-repeating decimal in (1.414, 1.732) works, e.g. √(2.5), (√2 + √3)/2.

Practice — Number Systems

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