Let us recall the concepts in Real Numbers:
Fundamental Theorem of Arithmetic:
Fundamental Theorem of Arithmetic: 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.
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Important!
- If \(p\) is a prime number and \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a positive integer.
- If a prime number \(p\) divides the product \(ab\), then \(p\) must divide either \(a\) or \(b\).
- If a composite number \(n\) divides the product \(ab\) where and \(a\) and \(b\) are prime number, then \(n\) neither divide \(a\) nor \(b\).
Applications of the fundamental theorem of Arithmetic:
- \(HCF = \) Product of the smallest power of each common prime factor in the numbers.
- \(LCM = \) Product of the greatest power of each prime factor involved in the numbers.
- For any two integers \(a\) and \(b\), \(HCF(a, b)\ \times \ LCM(a, b)\ = \ a\times b\)
- \(HCF(p,q,r)\times LCM(p,q,r) \neq p\times q\times r\), where \(p\), \(q\), \(r\) are positive integers.
- \(LCM(p,q,r) = \frac{p\times q\times r\times HCF(p,q,r)}{HCF(p,q)\times HCF(q,r)\times HCF(p,r)}\)
- \(HCF(p,q,r) = \frac{p\times q\times r\times LCM(p,q,r,)}{LCM(p,q)\times LCM(r,q)\times LCM(p,r)}\)
Revisiting irrational number
A number \(x\) is said to be an irrational number if it cannot be expressed in the form of \(\frac{p}{q}\), where \(q\neq 0\).
Important!
In this, we are going to prove that \(\sqrt{p}\) is irrational where \(p\) is a prime. Now, we shall prove the theorem using the fundamental theorem of arithmetic.
Theorem:
Let \(p\) be a prime number. If \(p\) divides \(a^2\), then \(p\) divides \(a\), where \(a\) is a postive integer.
Proof: The prime factorisation of \(a\) can be written as:
\(a = p_1p_2.....p_n\)
where \(p_1p_2.......p_n\) are unique prime factors of \(a\).
Then, the prime factorisation of \(a^2\) can be written as:
\(a^2 = a\times a\)
\(a^2 = (p_1p_2.......p_n)\times (p_1p_2.......p_n)\)
\(a^2 = (p_1)^2(p_2)^2.......(p_n)^2\)
It is given that \(p\) divides \(A^2\). We know that from the fundamental theorem of arithemtic that \(p\) is one of the prime factors of \(a^2\). That is, \(p\) is one of the \((p_1, p_2, ......p_n)\).
Since \(a=p_1p_2......p_n\) and \(p\) is one of \((p_1, p_2, .....p_n)\), then \(p\) divides \(a\).
Hence, we proved.