- 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.
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:
Since 5 and 8 are the remainders of 70 and 125, respectively.
Thus after subtracting these remainders from the numbers, we have the numbers:
(70 − 5) = 65 & (125 − 8) = 117 which is divisible by the required number.
Now required number = H.C.F of (65, 117)
⇒ 117 = 65 * 1 + 52
⇒ 65 = 52 * 1 + 13
⇒ 52 = 13 * 4 + 0
⇒ HCF(65, 117) = 13
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.
- 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 2n 5m, 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 2n 5m, 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 2n 5m, 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).
- 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.
Since the denominator, 125 = 53 is of the form 2m5n where m = 0 and n = 3.
So 13/125 has a terminating decimal expansion.
- If p is a prime and p divides a2, 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
x2 + 20x = 2016
x2 + (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: