Real Numbers — RBSE Class 10 (Mathematics)

Every whole number you will ever meet — 12, 100, 2025 — is built from primes the way every word is built from letters. 12 = 2 × 2 × 3, and there is no other way. This chapter takes that simple, almost obvious fact and squeezes from it the HCF, the LCM, and a clean proof that √2 can never be written as a fraction. Small ideas, surprisingly powerful.

1. The Fundamental Theorem of Arithmetic

Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order of the factors.

For example: $3825=3\times 3\times 5\times 5\times 17=3^2\times 5^2\times 17$

This uniqueness is the bedrock. It is why there is exactly one HCF and one LCM for any pair of numbers, and it is the engine behind the irrationality proofs later in the chapter.

Writing a number in prime-factor form (the "factor tree") is the first skill to master — keep dividing by the smallest prime that fits until you reach 1.

2. HCF and LCM by prime factorisation

Once two numbers are in prime-power form:

• HCF (Highest Common Factor) = product of the smallest power of each common prime.
• LCM (Lowest Common Multiple) = product of the greatest power of every prime that appears.

Example — find HCF and LCM of 96 and 404. $96=2^5\times 3,404=2^2\times 101$

• HCF = $2^2=4$ (only common prime is 2; smallest power is $2^2$).
• LCM = $2^5\times 3\times 101=9696$.

The product rule (two numbers only)

$\text{HCF}(a,b)\times \text{LCM}(a,b)=a\times b$

A genuine time-saver: find the HCF (usually quick), then get the LCM by dividing the product by the HCF. Note: this rule holds for two numbers only — it does not extend to three numbers.

3. Revisiting irrational numbers

A rational number can be written as $\frac{p}{q}$ with integers p, q and $q\ne 0$. A number that cannot be is irrational (e.g. √2, √3, √5, π).

Proving √2 is irrational (proof by contradiction)

1. Assume the opposite: √2 is rational, so $\sqrt{2}=\frac{p}{q}$ where p, q are coprime (no common factor) and $q\ne 0$.
2. Then $2q^2=p^2$, so $p^2$ is even ⇒ p is even. Write $p=2k$.
3. Substitute: $2q^2=4k^2\Rightarrow q^2=2k^2$, so $q^2$ is even ⇒ q is even.
4. But then p and q share the factor 2 — contradicting "coprime".
5. Our assumption was wrong. Hence √2 is irrational. ∎

The same template proves √3, √5 and √p (for any prime p) irrational. A key supporting fact: if a prime p divides $a^2$, then p divides a.

Useful results

• rational + irrational = irrational (e.g. $5+\sqrt{2}$ is irrational).
• (non-zero rational) × irrational = irrational (e.g. $3\sqrt{2}$ is irrational).

4. Decimal expansions — terminating or not?

Look at a rational number $\frac{p}{q}$ in lowest terms. Factorise the denominator q:

$\frac{p}{q}$ has a terminating decimal iff $q=2^n{5}^m$ — i.e. the denominator's only prime factors are 2 and 5.

If q has any other prime factor (3, 7, 11, …), the decimal is non-terminating but recurring (it repeats).

Examples:

• $\frac{13}{3125}=\frac{13}{5^5}$ → denominator is $5^5$ → terminates.
• $\frac{17}{8}=\frac{17}{2^3}$ → terminates.
• $\frac{6}{15}=\frac{2}{5}$ → terminates (after simplifying!).
• $\frac{1}{7}$ → 7 is not 2 or 5 → non-terminating recurring (0.142857…).

Always reduce the fraction to lowest terms first, then inspect the denominator.

5. Closing thought

This whole chapter rests on one sentence: numbers factorise into primes in exactly one way. From it you got:

• a reliable method for HCF and LCM,
• a watertight way to prove irrationality, and
• a one-glance test for terminating decimals (only 2s and 5s downstairs).

In the RBSE board these are short, high-certainty marks — a √-proof and an HCF/LCM problem appear almost every year. The reasoning here (prime factorisation, proof by contradiction) is also the first taste of the logical rigour that all of higher mathematics is built on.