Real Numbers — Practice

Question 1

Express 156 as a product of its prime factors.

Accepted Answer

Step 1 — Divide by smallest primes: 156 ÷ 2 = 78, ÷ 2 = 39, ÷ 3 = 13 (prime).

Step 2 — 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13.

✦ Answer: 2² × 3 × 13.

Question 2

Without long division, state whether 13/3125 has a terminating decimal expansion.

Accepted Answer

Step 1 — 3125 = 5⁵, which is of the form 2ⁿ5ᵐ (n = 0, m = 5).

Step 2 — Hence the decimal terminates.

✦ Answer: Yes, it terminates.

Question 3

Find the HCF of 6 and 20 by prime factorisation.

Accepted Answer

Step 1 — 6 = 2 × 3; 20 = 2² × 5.

Step 2 — Common prime: 2; smallest power = 2¹.

✦ Answer: HCF = 2.

Question 4

Find the HCF and LCM of 96 and 404 by prime factorisation.

Accepted Answer

Step 1 — 96 = 2⁵ × 3; 404 = 2² × 101.

Step 2 — HCF = 2² = 4 (common prime 2, smallest power).

Step 3 — LCM = 2⁵ × 3 × 101 = 32 × 303 = 9696.

✦ Answer: HCF = 4, LCM = 9696.

Question 5

The HCF of two numbers is 9 and their LCM is 360. If one number is 45, find the other.

Accepted Answer

Step 1 — HCF × LCM = product of the two numbers.

Step 2 — 9 × 360 = 45 × x ⇒ 3240 = 45x.

Step 3 — x = 3240 / 45 = 72.

✦ Answer: The other number is 72.

Question 6

State whether 17/30 terminates, giving a reason.

Accepted Answer

Step 1 — 17/30 is already in lowest terms.

Step 2 — 30 = 2 × 3 × 5; it contains the prime 3 besides 2 and 5.

Step 3 — Since q ≠ 2ⁿ5ᵐ, the decimal does not terminate.

✦ Answer: Non-terminating recurring (because the denominator has a factor of 3).

Question 7

Prove that √5 is irrational.

Accepted Answer

Step 1 — Assume √5 is rational: √5 = p/q with p, q coprime and q ≠ 0.

Step 2 — Then 5q² = p², so 5 divides p² ⇒ 5 divides p. Write p = 5k.

Step 3 — Substitute: 5q² = 25k² ⇒ q² = 5k², so 5 divides q² ⇒ 5 divides q.

Step 4 — Now 5 divides both p and q, contradicting 'coprime'.

Step 5 — The assumption is false, so √5 is irrational. ∎

✦ Answer: proof by contradiction as above.

Question 8

Show that 7 × 11 × 13 + 13 is a composite number.

Accepted Answer

Step 1 — Take 13 common: 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1).

Step 2 — 7 × 11 + 1 = 77 + 1 = 78, so the number = 13 × 78.

Step 3 — It has factors other than 1 and itself (13 and 78), so it is composite.

✦ Answer: It equals 13 × 78, hence composite.

Question 9

Three bells ring at intervals of 9, 12 and 15 minutes. If they ring together at 8:00 a.m., at what time will they next ring together?

Accepted Answer

Step 1 — They ring together after the LCM of 9, 12, 15 minutes.

Step 2 — 9 = 3², 12 = 2²×3, 15 = 3×5. LCM = 2² × 3² × 5 = 180.

Step 3 — 180 minutes = 3 hours after 8:00 a.m.

✦ Answer: They next ring together at 11:00 a.m.

Question 10

Prove that 3 + 2√5 is irrational, given that √5 is irrational.

Accepted Answer

Step 1 — Assume 3 + 2√5 is rational, say equal to r (a rational number).

Step 2 — Then 2√5 = r − 3, so √5 = (r − 3)/2.

Step 3 — The right side is a difference/quotient of rationals, hence rational; so √5 would be rational.

Step 4 — This contradicts the given fact that √5 is irrational.

Step 5 — Therefore 3 + 2√5 is irrational. ∎

✦ Answer: contradiction — rearranging would make √5 rational.

Practice — Real Numbers

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