Fundamental Theorem of Arithmetic (FTA)
The Fundamental Theorem of Arithmetic states that every composite number can be expressed as a product of prime numbers, and this factorisation is unique, except for the order of the primes.
In simple terms, every number has a unique prime identity. This theorem forms the backbone of number theory and is extensively used in problems related to HCF, LCM, divisibility, irrational numbers, and terminating decimals.
If a number can be written as: $$N = p_1^{a} \times p_2^{b} \times p_3^{c} \dots$$ where \(p_1, p_2, p_3\) are prime numbers, then this representation is unique.
Why This Theorem is Important for Board Exams
- Used to prove irrationality (e.g., √2, √3)
- Foundation of Euclid’s Division Lemma based questions
- Direct application in HCF and LCM problems
- Important for determining terminating/non-terminating decimals
- Frequently asked in case-study and assertion-reason questions
Visual Understanding of Prime Factorisation
Important Properties
- Every number greater than 1 is either prime or composite
- Prime factorisation is unique
- Used to express numbers in exponential form
- Helps in simplifying fractions and roots
Example 1: Check whether \(4^n\) can end with digit 0
Concept Used: A number ends with 0 if it is divisible by 10 = 2 × 5
Solution:
$$\begin{aligned} 4^n &= (2^2)^n \\\\ &= 2^{2n} \end{aligned}$$The prime factorisation contains only 2 and no 5. Hence, it is not divisible by 10.
Conclusion: \(4^n\) never ends with digit 0.
Example 2: Find HCF and LCM of 6 and 20
$$\begin{aligned} 6 &= 2^1 \times 3^1 \\\\ 20 &= 2^2 \times 5^1 \end{aligned}$$HCF: Product of smallest powers of common primes
$$HCF = 2^1 = 2$$LCM: Product of greatest powers of all primes
$$LCM = 2^2 \times 3 \times 5 = 60$$Prime Divisibility Theorem
Theorem: Let \(p\) be a prime number. If \(p \mid a^2\), then \(p \mid a\), where \(a\) is a positive integer.
This theorem is a direct application of the Fundamental Theorem of Arithmetic and plays a crucial role in proofs involving irrational numbers and divisibility properties.
If a prime divides the square of a number, it must already be present in the number itself.
Proof
Let the prime factorisation of \(a\) be:
$$a = p_1^{\alpha_1} \cdot p_2^{\alpha_2} \cdots p_n^{\alpha_n}$$Then,
$$a^2 = p_1^{2\alpha_1} \cdot p_2^{2\alpha_2} \cdots p_n^{2\alpha_n}$$Since \(p \mid a^2\), by uniqueness of prime factorisation, \(p\) must be one of the primes \(p_1, p_2, \dots, p_n\).
Therefore, \(p\) is already a factor of \(a\).
Hence proved: \(p \mid a\)
Proof that \(\sqrt{2}\) is Irrational
This is one of the most important proofs in Class X and is frequently asked in board exams.
Proof by Contradiction
Assume \(\sqrt{2}\) is rational.
$$\sqrt{2} = \frac{a}{b}, \quad b \neq 0$$where \(a\) and \(b\) are coprime integers.
Squaring both sides:
$$2 = \frac{a^2}{b^2} \Rightarrow a^2 = 2b^2 \quad (1)$$This implies \(2 \mid a^2\). Using the previous theorem:
$$\Rightarrow 2 \mid a$$Let \(a = 2c\)
Substitute in (1):
$$\begin{aligned} a^2 &= 4c^2 \\ 2b^2 &= 4c^2 \\ b^2 &= 2c^2 \end{aligned}$$Thus, \(2 \mid b\)
Hence, both \(a\) and \(b\) are divisible by 2, which contradicts the assumption that they are coprime.
Therefore, \(\sqrt{2}\) is irrational.
Operations on Rational and Irrational Numbers
- Rational + Irrational = Irrational
- Rational − Irrational = Irrational
- Non-zero Rational × Irrational = Irrational
- Non-zero Rational ÷ Irrational = Irrational
Product of two irrational numbers may or may not be irrational.
Example: \(\sqrt{2} \times \sqrt{2} = 2\) (rational)
Board Exam Insights
- Proof of \(\sqrt{2}\) irrational → very high probability question
- Theorem \(p \mid a^2 \Rightarrow p \mid a\) → used in proof-based questions
- Concept of coprime numbers → appears in MCQs and case studies
- Steps of contradiction proof must be written clearly and sequentially
Summary
- Every composite number has a unique prime factorisation.
- If a prime divides the square of a number, it must divide the number itself.
- \(\sqrt{2}\) is irrational, proven using contradiction and prime factorisation.
- Rational and irrational numbers follow strict operational rules important for exams.
Deep Conceptual Understanding: Prime Divisibility & Irrationality
The theorem \(p \mid a^2 \Rightarrow p \mid a\) is not just a standalone result—it is a special case of a more general principle in number theory:
If a prime number \(p\) divides \(a^n\) (where \(n\) is a natural number), then \(p \mid a\).
This result is heavily used in proving irrationality of square roots and higher roots.
Alternative Proof (Using Euclid’s Lemma)
Euclid’s Lemma states:
Since \(a^2 = a \cdot a\), if \(p \mid a^2\), then:
$$p \mid a \cdot a$$By Euclid’s Lemma:
$$p \mid a \quad \text{or} \quad p \mid a$$Hence, \(p \mid a\)
Generalisation: Irrationality of \(\sqrt{p}\)
The proof used for \(\sqrt{2}\) can be extended:
The same contradiction method applies:
- Assume \(\sqrt{p} = \frac{a}{b}\)
- Then \(a^2 = p b^2\)
- So \(p \mid a^2 \Rightarrow p \mid a\)
- This leads to contradiction
Hence, \(\sqrt{p}\) is irrational for every prime \(p\).
Common Exam Traps & Mistakes
- Assuming \(p \mid a\) directly from \(p \mid a^2\) without justification
- Forgetting to mention that \(a\) and \(b\) are coprime
- Skipping contradiction step in irrational proofs
- Writing incomplete prime factorisation
- Not using proper mathematical statements like “Assume”, “Therefore”, “Contradiction”
Advanced Examples
Example 1: Prove that \(\sqrt{3}\) is irrational
Following the same steps as \(\sqrt{2}\), we assume:
$$\sqrt{3} = \frac{a}{b}$$Then:
$$a^2 = 3b^2$$Using theorem:
$$3 \mid a^2 \Rightarrow 3 \mid a$$Let \(a = 3k\), substitute:
$$b^2 = 3k^2 \Rightarrow 3 \mid b$$Contradiction arises. Hence, \(\sqrt{3}\) is irrational.
Example 2: Can \(\sqrt{4}\) be irrational?
No. Because:
$$\sqrt{4} = 2$$Since 4 is not prime, the theorem does not apply. This highlights that the condition “\(p\) is prime” is essential.
Concept Flow Visualization
Must-Remember Results for Board Exams
- If \(p\) is prime and divides \(a^2\), then \(p \mid a\)
- \(\sqrt{2}, \sqrt{3}, \sqrt{5}\) are irrational
- If a number has prime factorisation of the form \(2^n \times 5^m\), its decimal expansion is terminating
- Otherwise, the decimal expansion is non-terminating recurring
Quick Revision Box
- FTA ensures uniqueness of prime factorisation
- Prime divisibility theorem is key for proofs
- Irrational numbers cannot be written as \(a/b\)
- Contradiction method is standard proof technique
Interactive Proof Visualizer: Irrationality of √2
Click each step to progressively reveal the logical flow of the proof. This helps in mastering step-wise presentation, which is crucial for board exams.
$$\sqrt{2} = \frac{a}{b}, \quad a,b \text{ coprime}$$
$$2 = \frac{a^2}{b^2} \Rightarrow a^2 = 2b^2$$
⇒ 2 divides \(a\)
Substitute back
⇒ 2 divides \(b\)
Assertion–Reason Questions (CBSE Pattern)
Directions: Choose the correct option:
- (A) Both Assertion and Reason are true, and Reason is the correct explanation
- (B) Both are true, but Reason is not the correct explanation
- (C) Assertion is true, Reason is false
- (D) Assertion is false, Reason is true
Q1. Assertion: If a prime number divides \(a^2\), then it divides \(a\).
Reason: Every composite number has a unique prime factorisation.
Answer: (A)
Q2. Assertion: √2 is irrational.
Reason: √2 can be written as a terminating decimal.
Answer: (C)
Q3. Assertion: Product of two irrational numbers is always irrational.
Reason: √2 × √2 = 2
Answer: (D)
Q4. Assertion: If a number has only factors 2 and 5, its decimal expansion terminates.
Reason: Prime factorisation determines decimal nature.
Answer: (A)
Case Study Based Questions
Case Study: A student claims that a number is rational if it can be expressed as \(\frac{a}{b}\), where \(b \neq 0\). He tries to check whether √5 is rational.
Q1. Which method is most appropriate to check irrationality?
(A) Prime factorisation
(B) Contradiction method
(C) Division method
(D) Estimation
Answer: (B)
Q2. If √5 = a/b, what is the next step?
(A) Multiply by 5
(B) Square both sides
(C) Add 1
(D) Factorise
Answer: (B)
Q3. Why does contradiction arise?
(A) a and b become equal
(B) a and b both divisible by 5
(C) Fraction becomes improper
(D) No contradiction
Answer: (B)
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