Theorem (Euclid’s Division Lemma)

Chapter -1

Real Number

Theorem (Euclid’s Division Lemma): Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
To obtain the HCF of two positive integers, say c and d, with c > d, follow the steps below:
Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.
Step 2: If r = 0, d is the HCF of c and d. If r ≠ 0, apply the division lemma to d and r.
Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
This algorithm works because HCF (c, d) = HCF (d, r) where the symbol HCF (c, d) denotes the HCF of c and d, etc.

The Fundamental Theorem of Arithmetic
In your earlier classes, you have seen that any natural number can be written as a product of its prime factors. For instance, 2 = 2, 4 = 2 × 2, 253 = 11 × 23, and so on. Now, let us try and look at natural numbers from the other direction. That is, can any natural number be obtained by multiplying prime numbers? Let us see.
Take any collection of prime numbers, say 2, 3, 7, 11 and 23. If we multiply some or all of these numbers, allowing them to repeat as many times as we wish, we can produce a large collection of positive integers (In fact, infinitely many).
Let us list a few:

7 × 11 × 23 = 1771 3 × 7 × 11 × 23 = 5313
2 × 3 × 7 × 11 × 23 = 10626 23 × 3 × 73 = 8232
22 × 3 × 7 × 11 × 23 = 21252

and so on.
Theorem (Fundamental Theorem of Arithmetic): Every composite number can be expressed (factorised) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.
The prime factorisation of a natural number is unique, except for the order of its factors.
In general, given a composite number x, we factorise it as x = p₁p₂ ... pn, where p₁, p₂,..., p_n are primes and written in ascending order, i.e., p₁ ≤ p₂ ≤. . . ≤ p_n. If we combine the same primes, we will get powers of primes. For example,
                            32760 = 2 × 2 × 2 × 3 × 3 × 5 × 7 × 13 = 23 × 32 × 5 × 7 × 13