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# Mathematics Class 10 – Chapter 1- Real Numbers – Notes

Number

A number is a mathematical object used to count, measure, and label.

Real number=R (Generally denoted by R):

All rational and irrational numbers are called real numbers.

Any real number can be plotted on the number line.

Q = Rational Numbers:

It is in the form p/q where, q ≠ 0, p, q ∈ I are rational numbers. (“∈” means “belongs to”)

All integers can be expressed as rational numbers, for example, 5 = 5/1.

And having a decimal expansion of rational numbers terminating or non-terminating recurring.

Q’ = Irrational Numbers:

Real numbers which cannot be expressed in the form p/q and whose decimal expansions are non-terminating and non-recurring.

Roots of primes like √2, √3, √5, etc. are irrational.

N = Natural Numbers:

Counting numbers are called natural numbers. N = {1, 2, 3, …}.

W = Whole Numbers:

Zero along with all-natural numbers are together called whole numbers {0, 1, 2, 3,…}.

Even Numbers:

Natural numbers of the form 2n are called even numbers. (2, 4, 6, …}

Odd Numbers:

Natural numbers of the form 2n -1 are called odd numbers. {1, 3, 5, …}

Important Points:

All Natural Numbers are whole numbers.
All Whole Numbers are Integers.
All Integers are Rational Numbers.
All Rational Numbers are Real Numbers.

Prime Numbers:

The natural numbers greater than 1 are divisible by 1 and the number itself only is called a prime number.
Prime numbers have two factors i.e., 1 and the number itself, For example, 2, 3, 5, 7 & 11, etc.

Composite Numbers:

The natural numbers which are divisible by 1, itself and any other number or numbers are called composite numbers. For example, 4, 6, 8, 9, 10, etc.
Note: 1 is neither prime nor a composite number.
Also, 1 is not a prime number as it has only one factor.

Euclid’s Division Lemma:

For a given positive integers a and b,
there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.

As we know “dividend = divisor × quotient + remainder”.
It also means that for a given pair of dividends and divisors, the quotient and remainder obtained are going to be unique.

Euclid’s Division Algorithm:

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.

Fundamental Theorem of Arithmetic: Every composite number can be expressed (factorized) as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur.

Method of Finding HCF:
H.C.F can be found using two methods –

(1) Prime factorisation and

(2) Euclid’s division algorithm.

Prime Factorisation:

Given two numbers, we express both of them as products of their respective prime factors. Then, we select the prime factors that are common to both the numbers
Example – To find the H.C.F of 20 and 24
20=2×2×5 and 24=2×2×2×3
The factor common to 20 and 24 is 2×2, which is 4, which in turn is the H.C.F of 20 and 24.

Euclid’s Division Algorithm:

It is the repeated use of Euclid’s division lemma to find the H.C.F of two numbers.

Important:
“For any two positive integers a and b,
a×b=H.C.F×L.C.M.”

Remarks:

1. Euclid’s division lemma and algorithm are so closely interlinked that people often
call former as the division algorithm also.
2. Although Euclid’s Division Algorithm is stated for only positive integers, it can be
extended for all integers except zero, i.e., b ≠ 0. However, we shall not discuss this
aspect here.

Example: Find the LCM and HCF of 6 and 20 by the prime factorization method.
Solution : We have : 6 = 2^1× 3^1 and 20 = 2 × 2 × 5 = 2^2×5^1
Note that
HCF(6, 20) = 2^1= Product of the smallest power of each common prime factor in the numbers.

LCM (6, 20) = 2^2× 3^1× 5^1= Product of the greatest power of each prime factor, involved in the numbers.

From the example above, you might have noticed that HCF(6, 20) × LCM(6, 20)= 6 × 20. In fact, we can verify that for any two positive integers a and b,
HCF (a, b) × LCM (a, b) = a × b. We can use this result to find the LCM of two positive integers if we have already found the HCF of the two positive integers.

Rational and Irrational Numbers:

If a number can be expressed in the form p/q where p and q are integers and q ≠ 0, then it is called a rational number.

If a number cannot be expressed in the form p/q where p and q are integers and q ≠ 0, then it is called an irrational number:

If p (a prime number) divides a^2 , then p divides a as well. For example, 3 divides 6^2, resulting in 36, implying that 3 divides 6.

The sum or difference of a rational and an irrational number is irrational

A non-zero rational and irrational number’s product and quotient are both irrational.

√p is irrational when p is a prime number. For example, 7 is a prime number and √7 is irrational.

In the method of contradiction, to check whether a statement is TRUE
(i) We assume that the given statement is TRUE.

(ii) We arrive at some result that contradicts our assumption, thereby proving the contrary.

Eg: Prove that √7 is irrational.
Assumption: √7 is rational.
Since it is rational √7 can be expressed as √7 = a/b, where a and b are co-prime Integers, b ≠ 0. On squaring, a^2/b^2=7 ⇒a^2=7b^2.

Hence, 7 divides a. Then, there exists a number c such that a=7c. Then, a^2=49c^2. Hence, 7b^2=49c^2 or b^2=7c^2.

Hence 7 divides b. Since 7 is a common factor for both a and b, it contradicts our assumption that a and b are coprime integers.

Hence, our initial assumption that √7 is rational is wrong. Therefore, √7 is irrational.

Terminating and nonterminating decimals:

Terminating decimals are decimals that end at a certain point. Example: 0.2, 2.56, and so on.
Non-terminating decimals are decimals where the digits after the decimal point don’t terminate. Example: 0.333333….., 0.13135235343…

Non-terminating decimals can be :
a) Recurring – a part of the decimal repeats indefinitely (0.142857142857….)
b) Non-recurring – no part of the decimal repeats indefinitely. Example: π=3.1415926535…

Check if a given rational number is terminating or not
If a/b is a rational number, then its decimal expansion would terminate if both of the following conditions are satisfied :

a) The H.C.F of a and b is 1.
b) b can be expressed as a prime factorization of 2 and 5 i.e b=2^m×5^n where either m or n, or both can = 0.

If the prime factorization of b contains any number other than 2 or 5, then the decimal expansion of that number will be recurring

Example:

1/40=0.025 is a terminating decimal, as the H.C.F of 1 and 40 is 1, and the denominator (40) can be expressed as 2^3×5^1.

3/7=0.428571 is a recurring decimal as the H.C.F of 3 and 7 is 1 and the denominator (7) is equal to 7^1

Example: Prove that √3 is irrational.

Solution: Let us assume that √3 is rational.

That is, we can find integers a and b (≠ 0) such that 3 =a/b.

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
So, b√3=a

Squaring both sides, we get 3b^2= a^2

Therefore, a^2 is divisible by 3, and we know that If p(a prime number) divides a^2 , then p divides a as well.

It means a is also divisible by 3.
So, we can write a = 3c for some integer c.

Substituting for a, we get
3b^2= 9c^2,
that is, b^2= 3c^2
This means that b^2 is divisible by 3, and so b is also divisible by 3.

Therefore, a and b have at least 3 as common factors. But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that √3 is irrational.

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