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**Real Numbers**

- Real numbers constitute the union of all
**rational and irrational numbers**. - In general, all the arithmetic operations can be performed on these numbers and they can be
**represented on the number line**, also.

**Euclid’s Division Lemma**

- Given positive integers a and b, there exist unique integers q and r satisfying:
**a = bq + r**, 0 ≤ r < b. - This lemma is essentially equivalent to:
**dividend = (divisor x quotient) + remainder**

where q is called the**quotient**and r is called the**remainder.** **Lemma**is a proven statement used for proving another statement.

**Euclid's Division Algorithm**

- This is based on Euclid's Division Lemma. It is a method used to find the H.C.F of two numbers.
- Let there be two numbers a and b where a > b, the HCF is obtained by the following method:
**Step I:**We apply Euclid’s Division Lemma to find two integers q and r, such that a = bq + r, 0 ≤ r < b.**Step II:**If r = 0, then b is the required HCF.**Step III:**If r ≠ 0, then again obtain two integers using Euclid’s Division Lemma and continue till the remainder becomes zero. The divisor when the remainder becomes zero is the required HCF.

Note:It can be extended for all integers, except zero i.e., b ≠ 0.

**Example 1: Find HCF of 56 and 72.****Solution:**

Apply lemma to 56 and 72.

Take a bigger number and locate ‘b’ and ‘r’. 72 = 56 × 1 + 16

Since 16 ≠ 0, consider 56 as the new dividend and 16 as the new divisor. 56 = 16 × 3 + 8

Again, 8 ≠ 0, consider 16 as the new dividend and 8 as the new divisor. 16 = 8 × 2 + 0.

Since the remainder is zero, divisor 8 is HCF.

Try yourself:The largest number that divides 70 and 125, which leaves the remainders 5 and 8, is:

View Solution

**Fundamental Theorem of Arithmetic**

- Every composite number can be factorised as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.
- Example: 54=2×3×3×3
- Therefore, 54 is represented as a product of prime factors (One 2 and three 3s) ignoring the arrangement of the factors.

**Rational Number**

- The decimal expansion of every rational number is either terminating or non-terminating repeating.
- Let x be a rational number with a
**terminating decimal representation**. Then we can express x as p/q where p and q are coprime, and the prime factorisation of q is of the form 2^{n}5^{m}, where n, m are some non-negative integers. - Let x = p/q be a rational number such that prime factorisation of q is of the form 2
^{n}5^{m}, where n, m are some non-negative integers. Then x has a terminating decimal expansion. - The rational number p/q will have a
**non-terminating repeating**(recurring) decimal representation if, in standard form, the prime factorisation of q is not of the form 2^{n}5^{m}, where n, m are some non-negative integers.

- A number is irrational if and only if, its decimal representation is
**non-terminating and non-repeating**(non-recurring).**OR** - A number that cannot be expressed in the form of p/q, q ≠ 0 and p, q ∈ I, will be an irrational number. The set of irrational numbers is generally denoted by S.Try yourself:Find out the decimal expansion of 13/125.View Solution

- If p is a prime and p divides a
^{2}, then p divides a, where a is a positive integer.

**LCM & HCF of Two Numbers**

- For any two positive integers p and q, we have:

HCF (p,q) x LCM [p, q] = p x q

**Example 2: There are two positive numbers such that one exceeds the other by 20 and their LCM and HCF are 504 and 4 respectively. Find the numbers.****Solution **Let the numbers be 'x' and 'x + 20'.

We know that:

HCF (p,q) x LCM (p, q) = p x q

x(x + 20) = 4 x 504

x^{2} + 20x = 2016

x^{2} + (56 - 36)x - 2016 = 0

(x + 56)(x - 36) = 0

x = -56 & x = 36

x cannot be negative.

So two numbers are: x =** 36** & (x + 20) =** 56**

**LCM & HCF of Three Numbers**

- For any three positive integers a, b and c, we have:

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