Notes of chapter: Real Numbers are presented below. Indepth notes along with worksheets and NCERT Solutions for Class 10.
(1) Real numbers-
Real numbers are collection of rational and irrational numbers.
(2) Euclid’s Division Lemma-
It is just equivalent to division equation
Dividend = Divisor × Quotient + Remainder
But a formal statement is given by Euclid which is known as Euclid’s Division Lemma.
Given positive integers a and b, there exist whole numbers q and r satisfying
a = bq + r, 0 ≤ r < b.
Here, q and r are quotient and remainder.
a = dividend and
b = divisor
Eg:- Find out the integers q and r for positive integers 4, 19.
Euclid’s Division Lemma
a = bq + r, 0 ≤ r < b.
19 = 4 × 4 + 3
(4 goes into 19 up to 4) (remainder is 3)
4 = 3 × 1 + 1
3 = 1 × 3 + 0
(3) Euclid’s Division Algorithm
This theorem is based on the Euclid’s Division Lemma.
According to this algorithm-
Certain steps should be follow to get HCF of two positive integers c and d with c>d.
Steps to follow-
Step 1- Apply Euclid’s Division Lemma , to c and d to find out 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 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.
Eg:- Use Euclid’s algorithm to find the HCF of 455 and 42.
Step 1- Apply Euclid’s Division Lemma, to c and d
455 = 42 × 10 + 35
Step 2- If r ≠ 0, apply lemma to d and r.
Here, r ≠ 0, apply Euclid’s Division Lemma
42 = 35 × 1 + 7
Again apply Euclid’s Division Lemma
35 = 7 5 + 0
Remainder is zero.
Divisor is 7.
Therefore, 7 is HCF of 455 and 42.
Utility of Euclid’s Division Algorithm –
(i) Euclid’s Algorithm is very useful for calculating HCF of very large numbers.
(ii) It is one of the earliest examples of an algorithm that a computer had been programmed to carry out.
(4)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.
The prime factorisation of a natural number is unique, except for the order of its factors.
Given a composite number x, we factorise it as x = p1p2… pn,
Where p1 , p2… pn are prime numbers written in ascending order.
120 = 2 × 2 × 2 × 3 × 5
= 23 3 5 (order is ascending)
Therefore, factorisation is unique.
(5) Revisiting Irrational Numbers-
Eg:- √2, √3, π, 0.10110111011110….
Let p be a prime number. If p divides a2, then p divides a, where a is positive integer.
Let the prime factorisation of a = p1p2………..pn,
Where p1,p2………..pn are primes, not necessarily distinct.
a2 = (p1p2………..pn) (p1p2………..pn)
= p12 p22………..pn2.
p divides a2 (Given)
Therefore, according to fundamental theorem of arithmetic p is one of the prime factors of a2.
By using uniqueness part of the fundamental theorem of arithmetic
Only prime factors of a2 are p1, p2,………..pn,
p is one of p1, p2,………..pn.
Now, since a = p1p2………..pn,
Therefore, p divides a.
Theorem 2- ∠2 is irrational.
Let us assume that ∠2 is rational.
Suppose a and b have a common factor other than 1.
Let, a and b are coprime numbers.
So, b√2 = a
2b2 = a2 (squaring both sides)…(1)
Therefore, 2 divides a2.
So, 2 divides a. (Theorem 1-If p divides a2, then p divides a)
a = 2c…(2)
Put value of a in equation 1
2b2 = (2c)2
2b2 = 4c2
b2 = 2 c2
Therefore, 2 divides b2.
So, 2 divides b. (Theorem 1- If p divides a2, then p divides a)
Therefore, a and b have at least 2 as a common factor.
But, a and b have no common factor other than 1 because both are coprime numbers. Therefore, it contradicts.
Hence, √2 is irrational.
Eg: – Show that 5 – √3 is irrational.
Let us assume that 5 – √3 is rational.
We can find coprime a and b, where b ≠ 0,
But, this contradicts the fact that √3 is irrational number.
Therefore, our assumption that 5 -√3 is rational is incorrect.
So, we conclude that 5 – √3 is irrational.
Revisiting Rational Numbers and their Decimal Expansion-
Clearly remainder is not zero and denominator is not of the form 2n5m. But after some time decimal expansion is repeating. Therefore it is non – terminating recurring decimal expansion.