Proof of Theorems : Real Numbers | Mathematics (Maths) Class 10 PDF Download

Clear all your Class 10 doubts with EduRev Join for Free

Join for Free

Theorem 1.1 - Fundamental Theorem of Arithmetic

Statement: 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.

Given: Let N be a composite number. It can be expressed as a product of prime numbers.
To Prove: The prime factorization of N is unique, apart from the order of factors
Proof:

•  Consider any composite number N, which can be written as a product of primes. 
•  For example, take 32760: 
• 32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13
•  Written in exponent form: 2³ × 3² × 5 × 7 × 13
•  Suppose there exists another way to factorize N into primes. 
•  By the Fundamental Theorem of Arithmetic, the set of prime factors obtained will always be the same, only differing in order. 
•  Thus, the prime factorization of a number is unique (except for the arrangement of factors). 

Conclusion: The Fundamental Theorem of Arithmetic ensures that every composite number has a unique prime factorization, apart from the order of the factors

Theorem 1.2

Statement: If a prime number p divides a², then p also divides a, where a is a positive integer.

Given: Let p be a prime number and a be a positive integer such that p divides a².
To Prove: If p divides a², then p must also divide a.
Proof:

• Let a be any positive integer, and let its prime factorization be:
  a = p₁ × p₂ × ... × p_n
• Squaring both sides, we get:
  a² = (p₁ × p₂ × ... × p_n)² = p_1² × p₂² × ... × p_n².
• If a prime number p divides a², then p must be one of the prime factors of a².
• Since each prime factor in a² appears as a square, p must already be a factor of itself.
• This follows directly from the Fundamental Theorem of Arithmetic, which states that prime factorization is unique.
• Therefore, if p divides a², then p must also divide a.

Conclusion:

The Theorem 1.2 proves that if a prime number divides the square of a number, then it must also divide the number itself. This theorem is widely used in number theory and proofs of irrationality, such as proving that √2 is irrational.

Theorem 1.3 

Statement: √2 is an irrational number.

To Prove: √2 cannot be expressed as a fraction pq, where p and q are integers and q ≠ 0.
Proof :

• Assume √2 is rational.
• Then, we can write √2 as pq, where p and q are coprime integers (i.e., they have no common factor other than 1).
• Squaring both sides:
  2 = p²q² ⇒ p² = 2q².
• This means p² is divisible by 2, so p must also be divisible by 2 (from Theorem 1.2).
• Let p = 2k for some integer k.
• Substituting in p² = 2q², we get:
  (2k)² = 2q² ⇒ 4k² = 2q² ⇒ q² = 2k².
• This means q² is also divisible by 2, so q must also be divisible by 2.
• But this contradicts our assumption that p and q are coprime (as both are divisible by 2).
• Therefore, our assumption that √2 is rational must be false and√2 is rational.

Conclusion: Thus, √2 is irrational.