Real Numbers Class 10: notes and important questions

Real Numbers — revision notes

1. The Fundamental Theorem of Arithmetic (FTA)

Every composite number can be written as a product of primes, and this factorisation is unique apart from the order of the factors. Write it in index form, e.g. \(3960 = 2^3 \times 3^2 \times 5 \times 11\).

• Find factors by repeated division by the smallest prime (2, 3, 5, 7, 11, …) or with a factor tree.
• Uniqueness lets you compare exponents: if \(2^x 3^y = 2^4 3^3\) then \(x = 4,\ y = 3\).
• A number ends in \(0\) only if its prime factorisation contains both \(2\) and \(5\); it ends in \(5\) only if it contains \(5\) (and no \(2\)).

2. HCF and LCM by prime factorisation

• HCF = product of the smallest power of each common prime factor.
• LCM = product of the greatest power of every prime factor involved.
• For two positive integers only: \(\text{HCF}(a,b) \times \text{LCM}(a,b) = a \times b\).
• HCF always divides LCM, so e.g. HCF \(14\), LCM \(204\) is impossible.

3. Word problems — which one?

• HCF: “greatest/largest”, splitting into equal groups, arranging in rows/stacks/trays, cutting into longest equal pieces. “Leaves remainder \(r\)” → take HCF of (numbers \(- r\)).
• LCM: “least/smallest”, events happening together again (bells, lights, laps). “Leaves remainder \(r\) in each case” → LCM \(+ r\).

4. Irrational numbers

Key result: if a prime \(p\) divides \(a^2\), then \(p\) divides \(a\). Use it to prove \(\sqrt2, \sqrt3, \sqrt5\) irrational by contradiction.

• rational \(\pm\) irrational = irrational; non-zero rational \(\times\) or \(\div\) irrational = irrational.
• irrational \(\pm\) irrational may be rational: \((2+\sqrt3) + (2-\sqrt3) = 4\).

Worked example 1

Find HCF and LCM of \(96\) and \(360\).
\(96 = 2^5 \times 3\), \(360 = 2^3 \times 3^2 \times 5\). HCF \(= 2^3 \times 3 = 24\); LCM \(= 2^5 \times 3^2 \times 5 = 1440\). Check: \(24 \times 1440 = 34560 = 96 \times 360\).

Worked example 2

Prove that \(\sqrt5\) is irrational.
Suppose \(\sqrt5 = a/b\), \(a, b\) coprime. Then \(a^2 = 5b^2\), so \(5 \mid a^2\) and hence \(5 \mid a\). Put \(a = 5c\): \(25c^2 = 5b^2 \Rightarrow b^2 = 5c^2\), so \(5 \mid b\). Then \(5\) divides both \(a\) and \(b\) — contradiction. Hence \(\sqrt5\) is irrational.

Worked example 3

Show that \(7 + 2\sqrt3\) is irrational (given \(\sqrt3\) irrational).
If \(7 + 2\sqrt3 = r\) (rational), then \(\sqrt3 = \dfrac{r-7}{2}\), which is rational — contradiction.

Common errors

• Using \(\text{HCF} \times \text{LCM} = \text{product}\) for three numbers.
• Taking the highest power for HCF (or the lowest for LCM).
• In irrationality proofs, forgetting to state that \(a\) and \(b\) are coprime, or not stating the contradiction clearly.
• Leaving a “prime factorisation” with a composite factor such as \(77\) or \(143\).

Board-exam tips

• Always show the prime factorisation — it carries a method mark even if the final answer slips.
• In word problems write a final sentence with units (“216 books per stack”, “9:00 a.m.”).
• Irrationality proofs are regular 2–3 mark items: learn the structure assume → square → prime divides → contradiction.