Perfect study material curated for high-scoring CBSE board exams. Includes brief revisions, formulas, and verified step-by-step NCERT solutions.
1. The Fundamental Theorem of Arithmetic
Every composite number can be broken down (factorized) into a unique combination of prime numbers. This factorization is completely unique for every number, no matter what order you write the prime factors in.
Example: Consider the composite number 140.
If we find its prime factors: 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7.
2. Finding HCF and LCM Using Prime Factorization
Once you write out the prime factors of two or more numbers, you can easily calculate their HCF and LCM:
- HCF (Highest Common Factor): Take the smallest power of each common prime factor.
- LCM (Lowest Common Multiple): Take the highest power of every prime factor involved.
⭐ Golden Formula:
HCF(a, b) × LCM(a, b) = a × b
Note: This formula is only valid for a pair of two numbers!
3. Rational and Irrational Numbers
Rational Numbers: Numbers that can be written in the form $p/q$, where $q \neq 0$ and $p, q$ are integers.
Irrational Numbers: Numbers that cannot be expressed in $p/q$ form. Examples include $\sqrt{2}, \sqrt{3}, \sqrt{5}$, and $\pi$.
Repeated Board PYQ (CBSE 2020, 2023, 2024)
Question: Prove that $\sqrt{5}$ is an irrational number.
Detailed Contradiction Proof Steps:
1. Let us assume the opposite: assume $\sqrt{5}$ is a rational number.
2. Therefore, we can write $\sqrt{5} = a/b$, where $a$ and $b$ are integers and are co-prime (they have no common factors other than 1).
3. Squaring both sides gives: $5 = a^2 / b^2 \implies a^2 = 5b^2$.
4. This shows that $a^2$ is divisible by 5. According to theorems, if $a^2$ is divisible by 5, then $a$ must also be divisible by 5. So, let $a = 5c$.
5. Substituting $a = 5c$ into our equation: $(5c)^2 = 5b^2 \implies 25c^2 = 5b^2 \implies b^2 = 5c^2$.
6. This proves that $b^2$ is divisible by 5, which means $b$ is also divisible by 5.
7. Conclusion: Both $a$ and $b$ have 5 as a common factor. This completely contradicts our initial assumption that $a$ and $b$ are co-prime. Therefore, $\sqrt{5}$ must be irrational.