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, 0r < 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,..., pn are primes and written in ascending order, i.e., p1p2 ≤. . . ≤ pn. 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

Revisiting Irrational Numbers

Revisiting Irrational Numbers

In Class IX, you were introduced to irrational numbers and many of their properties. You studied about their existence and how the rationals and the irrationals together made up the real numbers. You even studied how to locate irrationals on the number line. However, we did not prove that they were irrationals. In this section, we will prove that 2, 3, 5  and, in general, p  is irrational, where p is a prime. One of the theorems, we use in our proof, is the Fundamental Theorem of Arithmetic.
Recall, a number‘s’ is called irrational if it cannot be written in the form  where p and q are integers and q0. Some examples of irrational numbers, with which you are already familiar, are:


Before we prove that 2 is irrational, we need the following theorem, whose proof is based on the Fundamental Theorem of Arithmetic.

Revisiting Rational Numbers and Their Decimal Expansions

In Class IX, you studied that rational numbers have either a terminating decimal expansion or a non-terminating repeating decimal expansion. In this section, we are going to consider a rational number, say

,  and explore exactly when the decimal expansion of   is terminating and when it is non-terminating and when it is non-terminating repeating (or recurring). We do so by considering several examples.

Let us consider the following rational numbers:

(i) 0.375       (ii) 0.104        (iii) 0.0875          (iv) 23.3408.



As one would expect, they can all be expressed as rational numbers whose denominators are powers of 10. Let us try and cancel the common factors between the numerator an denominator and see what we get:



Theorem: Let x be a rational number whose decimal expansion terminates.
Then x can be expressed in the form

,  where p and q are coprime, and the prime factorization of q is of the form 2n5m, where n, m are non-negative integers.
Theorem:   Let   be a rational number, such that the prime factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates.