Class 10 Maths Notes Chapter 1 — Real Numbers#

Key terms#

• Prime number — a natural number greater than 1 whose only factors are 1 and itself (2, 3, 5, 7, …).
• Composite number — a natural number greater than 1 with a factor other than 1 and itself (4, 6, 9, …).
• 1 is neither prime nor composite.
• HCF — the largest number that divides each of the given numbers exactly.
• LCM — the smallest number that is a multiple of each of the given numbers.
• Coprime numbers — two numbers whose HCF is 1 (for example, 8 and 15).
• Rational number — can be written as $\frac{p}{q}$, where $p$, $q$ are integers and $q \neq 0$.
• Irrational number — cannot be written as $\frac{p}{q}$. Examples: $\sqrt{2}$, $\sqrt{3}$, $\sqrt{15}$, $\pi$, $0.10110111\ldots$
• Rational and irrational numbers together make up the real numbers.

Fundamental Theorem of Arithmetic#

Theorem 1.1: Every composite number can be expressed (factorised) as a product of primes, and this factorisation is unique, apart from the order in which the prime factors occur.

• With the primes in ascending order, every composite number $x$ has exactly one form: $$x = p_1^{a_1} \times p_2^{a_2} \times \cdots \times p_k^{a_k}$$
• Example: $32760 = 2^3 \times 3^2 \times 5 \times 7 \times 13$
• Ends in 0? A number ends in the digit 0 only if its prime factorisation contains both 2 and 5. By uniqueness, no other primes can appear.
  • $4^n = 2^{2n}$ has no 5, so $4^n$ never ends in 0.

HCF and LCM by prime factorisation#

For any two positive integers $a$ and $b$:

Example: $96 = 2^5 \times 3$ and $404 = 2^2 \times 101$
$\text{HCF}(96, 404) = 2^2 = 4$
$\text{LCM}(96, 404) = \frac{96 \times 404}{4} = 9696$

This relation holds for two numbers only. For three numbers, $\text{HCF}(p, q, r) \times \text{LCM}(p, q, r) \neq p \times q \times r$. For example, with 6, 72 and 120, HCF = 6 and LCM = 360, but $6 \times 360 \neq 6 \times 72 \times 120$.

Irrational numbers#

Theorem 1.2: Let $p$ be a prime number. If $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.

(The proof of Theorem 1.2 is not from the examination point of view.)

Theorem 1.3: $\sqrt{2}$ is irrational.

Proof (by contradiction): Assume $\sqrt{2} = \frac{a}{b}$, where $a$, $b$ are coprime integers and $b \neq 0$.

Squaring both sides:

So 2 divides $a^2$, and by Theorem 1.2, 2 divides $a$. Let $a = 2c$.

Substituting $a = 2c$ in equation (1):

So 2 divides $b^2$, and therefore 2 divides $b$.

Now 2 divides both $a$ and $b$, which contradicts that they are coprime. ∴ $\sqrt{2}$ is irrational.

The same proof works for $\sqrt{3}$, $\sqrt{5}$ and $\sqrt{p}$ for any prime $p$ (replace 2 by $p$).

Rational and irrational together#

• rational ± irrational = irrational
• non-zero rational × irrational = irrational
• non-zero rational ÷ irrational (or irrational ÷ non-zero rational) = irrational
• irrational with irrational = either: $\sqrt{2} \times \sqrt{2} = 2$ (rational), $\sqrt{2} \times \sqrt{3} = \sqrt{6}$ (irrational)

To prove $5 - \sqrt{3}$ is irrational: assume $5 - \sqrt{3} = \frac{a}{b}$, so:

The right side is rational, which would make $\sqrt{3}$ rational. This is a contradiction, so $5 - \sqrt{3}$ is irrational. Use the same method for $3\sqrt{2}$, $\frac{1}{\sqrt{2}}$ and $6 + \sqrt{2}$.

Common mistakes#

• Using HCF × LCM = product for three numbers. It works for two only.
• Calling 1 a prime. It is neither prime nor composite.
• Leaving out “coprime” in an irrationality proof. The contradiction depends on it.
• Saying $\sqrt{n}$ is irrational for every $n$. $\sqrt{4} = 2$ is rational.
• Saying the product of two irrationals is irrational. $\sqrt{2} \times \sqrt{8} = 4$.
• Checking only for 5 in “ends in 0” questions. Both 2 and 5 are needed.