Real Numbers

The Fundamental Theorem of Arithmetic

Theorem Statement:

Every composite number can be uniquely expressed (factorised) as a product of prime numbers, apart from the order in which the prime factors occur.

$$\text{Composite Number } N = p_1^{a_1} \cdot p_2^{a_2} \cdot p_3^{a_3} \cdots p_k^{a_k}$$

Fundamental Relation between HCF and LCM:

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

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

[!CAUTION]

This relation holds strictly for two numbers only! For three numbers $a, b, c$: $\text{HCF}(a, b, c) \times \text{LCM}(a, b, c) \neq a \times b \times c$.

Revisiting Proofs of Irrationality (Proof by Contradiction)

Theorem: Prove that $\sqrt{5}$ is an irrational number.

Formal Proof:

1. Assume the contrary, that $\sqrt{5}$ is a rational number.
2. Therefore, there exist coprime positive integers $a$ and $b$ ($b \neq 0$, $\text{HCF}(a,b) = 1$) such that:

$$\sqrt{5} = \frac{a}{b} \implies a = b\sqrt{5}$$

1. Squaring both sides:

$$a^2 = 5b^2 \tag{1}$$

1. This shows that $5$ divides $a^2$. By prime property theorem, if a prime $p$ divides $a^2$, then $p$ divides $a$. Thus, $5$ divides $a$.
2. Let $a = 5c$ for some integer $c$. Substituting into equation (1):

$$(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2 \tag{2}$$

1. This shows that $5$ divides $b^2$, which implies $5$ divides $b$.
2. From steps 4 and 6, $5$ is a common factor of both $a$ and $b$.
3. But this contradicts our initial assumption that $a$ and $b$ are coprime ($\text{HCF}=1$).
4. Hence, our supposition was incorrect. We conclude that $\sqrt{5}$ is irrational. $\quad \blacksquare$

CBSE Board Practice Questions & Model Solutions

• Q1 (2 Marks): If $\text{HCF}(306, 657) = 9$, find $\text{LCM}(306, 657)$.
• Solution:

$$\text{LCM} = \frac{a \times b}{\text{HCF}} = \frac{306 \times 657}{9} = 34 \times 657 = 22,338$$

• Q2 (3 Marks): Explain why $7 \times 11 \times 13 + 13$ is a composite number.
• Solution:

$$7 \times 11 \times 13 + 13 = 13 \times (7 \times 11 + 1) = 13 \times 78 = 13 \times 2 \times 3 \times 13$$

Since the number can be expressed as a product of more than two prime factors, by the Fundamental Theorem of Arithmetic, it is composite.