CBSE Grade 10 Maths Chapter 1 - Real Numbers

Recap - Different Types of Numbers

 N=Natural Numbers

• Non-negative counting numbers excluding zero are known as natural numbers.
• 1, 2, 3, 4 ......etc are all natural numbers.
• 1 is the smallest Natural Number.

W=Whole numbers 

• All non-negative counting numbers including zero are known as whole numbers.
• 0, 1, 2, 3, 4.....etc are all Whole Numbers.
• 0 is the smallest Whole Number.

I=Integers

• All negative and non-negative numbers including zero altogether known as integers.
• ............– 3, – 2, – 1, 0, 1, 2, 3, 4, ………….. are integers.

Q=Rational Numbers

• Rational numbers are numbers that can be written in the form p/q, where p and q are integers and $q\ne 0$. 

         Example -1/2$,\; 4/5,1,0,-3$  and so on.R

• Rational numbers can be written in decimal form also which could be either terminating or non-terminating.  Example - 5/2 = 2.5 (terminating) and
  
  
  (non-terminating).

Q’=Irrational Numbers

• Any number that cannot be expressed in the form of p/q (where p and q are integers and $q\ne 0$) is an irrational number.

         Example - √2$,\pi ,$ e and so on.

Note :

• If p is a prime number and p divides a² , then p is one of the prime factors of a² which divides a, where a is a positive integer.

• If p is a positive number and not a perfect square, then √n is definitely an irrational number.

• If p is a prime number, then √p is also an irrational number.

R=Real Numbers

• Real numbers constitute the union of all rational and irrational numbers.
• Any real number can be plotted on the number line.
  

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....

Note -

• 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 which are divisible by 1 and the number itself are called prime numbers.
• Prime numbers have two factors i.e., 1 and the number itself . Example-  2, 3, 5, 7 & 11 etc.
• 1 is not a prime number as it has only one factor.

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.

Euclid’s Division Lemma

Algorithm vs Lemma

• An algorithm gives us some definite steps to solve a particular type of problem in a well-defined manner.
• A lemma is a statement which is already proved and is used for proving other statements.

Euclid’s Division Lemma

• Euclid’s Division Lemma states that for any given two integers a and b, there exists a unique pair of integers q and r such that $a=b\times q+rand0\le r<b$.
• This lemma is essentially equivalent to : 

  Dividend = Divisor $\times$ Quotient + Remainder

• In other words, for a given pair of dividend and divisor, the quotient and remainder obtained are going to be unique.

Euclid’s Division Algorithm

• This concept is based on Euclid’s division lemma. 
  
• Euclid’s Division Algorithm is a method used to find the H.C.F (Highest common factor) of two numbers, say a and b where a> b.
• We apply Euclid’s Division Lemma to find two integers q and r such that

$a=b\times q+rand0\le r<b$.

Steps to calculate HCF of two positive integers’ a and b where a > b

Step 1: Apply Euclid’s division lemma to find q and r where $a=b\times q+rand0\le r<b$. 

Step 2: If r = 0, the H.C.F is b, else, we apply Euclid’s division Lemma to b (the divisor) and r (the remainder) to get another pair of quotient and remainder. 

Step 3: Continue with this process until we get the remainder as zero. Now the divisor at this stage will be HCF(a, b)

• Example: To find the HCF of 18 and 30
  

The required HCF is 6.

The Fundamental Theorem of Arithmetic

The Fundamental Theorem of Arithmetic states that - 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.

Prime Factorisation

• Prime Factorisation is the method of expressing a natural number as a product of prime numbers.

        Example: $36=2\times 2\times 3\times 3$  is the prime factorisation of 36.

• According to The Fundamental Theorem of Arithmetic, the prime factorisation for a given number is unique if the arrangement of the prime factors is ignored.

        Example: $36=2\times 2\times 3\times 3$ OR, $36=2\times 3\times 2\times 3$

 

Therefore, 36 is represented as a product of prime factors (Two 2s and two 3s) ignoring the arrangement of the factors.

Methods to find Prime Factors 

There are two methods to find the prime factors of a number-

1. Prime factorisation using a factor tree

2. Repeated Division  Method

The prime factors of 36 are 2 and 3.

We can write 36 as a product of prime factors: 2 × 2 × 3 × 3

Note : To find prime factors using the repetitive division, it is advisable to start with a small prime factor and continue the process with bigger prime factors.

 Different Methods to find HCF

Example - Find the HCF of 60 and 72.

Solution: First, we have to find the prime factorisation of 60 and 72.

Then encircle the common factors.

HCF of 60 and 72 is 2 × 2 × 3 = 12.

Example - Find the LCM of 60 and 72.

Solution:

First, we have to find the prime factorisation of 60 and 72.

Note : To find the LCM, we will count the common factors one time and multiply them with the other remaining factors.

2. Repeated Division Method

If we have to find the LCM of so many numbers then we use this method.

Example - Find the LCM of 105, 216 and 314.

Solution:

Use the repeated division method on all the numbers together and divide until we get 1 in the last row.

 
LCM of 105,216 and 314 is 2 × 2 × 2 × 3 × 3 × 3 × 5 × 7 × 157 = 1186920

Example: Find the Least Common Multiple  (L.C.M) of 36 and 56,

Step 1 : Find the prime factors of 36 and 56

$36=2\times 2\times 3\times 3$
$56=2\times 2\times 2\times 7$

 

Step 2 : Find common prime factors .

 

                The common prime factors of 36 and 56 are $2\times 2$

$\mathrm{Step\; 3\; :\; Find\; uncommon\; factors\; of\; 36\; and\; 56}$

                The uncommon prime factors are  $3\times 3$ for 36 and $2\times 7$ for 56.
 

Step 4 : LCM of 36 and 56 = Product of all common prime factors and uncommon prime factors

 

                   LCM of 36 and 56 = $2\times 2\times 3\times 3\times 2\times 7$  which is 504

Summary :

We can factorize each composite number as a product of some prime numbers and of course, this prime factorization of a natural number is unique as the order of the prime factors doesn’t matter.

• HCF of given numbers is the highest common factor among all which is also known as GCD i.e. greatest common divisor.

• LCM of given numbers is their least common multiple.

• If we have two positive integers  ‘m’ and ‘n’ then the property of their  HCF and LCM will be:

  HCF (m, n) × LCM (m, n) = m × n.

•  In words we can say that
  

  Product of Two Numbers = HCF X LCM of the Two Numbers

Example – For 36 and 56, the H.C.F is 4 and the L.C.M is 504

• $36\times 56=2016$
  $4\times 504=2016$
  Thus, $36\times 56=4\times 504$
• The above relationship, however, doesn’t hold true for 3 or more numbers

Applications of HCF & LCM in Real-World Problems

L.C.M can be used to find the points of common occurrence. For example,ringing of bells that ring with different frequencies, the time at which two persons running at different speeds meet, and so on.

Revisiting Irrational Numbers

 Rational Number and their Decimal Expansions

• Let y be a real number whose decimal expansion terminates into a rational number which we can express in the form of a/b, where a and b are co-prime, and the prime factorization of the denominator b has the powers of 2 or 5 or both like 2ⁿ5^m, where n, m are non-negative integers.

• Let y be a rational number in the form of y = a/b, so that the prime factorization of the denominator b is of the form 2ⁿ5^m, where n, m are non-negative integers then y has a terminating decimal expansion.

• Let y = a/b where a and b are co-prime, be a rational number, if the prime factorization of the denominator b is not in the form of 2ⁿ5^m, where n, m are non-negative integers then y has a non-terminating repeating (recurring) decimal expansion.

Number theory:

• If a number p (a prime number) divides $a^2$, then p divides a. Example: 3 divides $6^2$ i.e 36, which implies that 3 divides 6.
• The sum or difference of a rational and an irrational number is irrational
• The product and quotient of a non-zero rational and irrational number are irrational.
• √p is irrational when ‘p’ is a prime. For example, 7 is a prime number and √7 is irrational. The above statement can be proved by the method of “Proof by contradiction”.

Proof by Contradiction

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 which contradicts our assumption, thereby proving the contrary.

Example - Prove that √7 is irrational.

Assumption - √7  is rational.

Since it is rational √7 can be expressed as 

$\surd 7\; =\; a/b$, where a and b are co-prime integers, b $\ne$ 0. 

On squaring,

$a^2$$/b2=7$

$\Rightarrow a^2=7b^2$. 

Hence, 7 divides a. 

Then, there exists a number c such that a=7c.

Then, $a^2=49c^2$. 

$7b^2=49c^2$ 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 co-prime integers.

Thus, our initial assumption that √7  is rational is wrong.

 Therefore, √7 is irrational.

Terminating and non-terminating 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: $\pi =\mathrm{3.1415926535\dots }$

$\mathrm{Some\; more\; Related\; videos}$

$\mathrm{1.\; Prove\; by\; using\; Euclids\; Division\; Lemma\; that\; "\; Any\; positive\; integer\; is\; of\; the\; form\; 3q,3q+1\; or\; 3q+2}$

$\mathrm{2.\; Prove\; that\; \surd 12\; -\; 6\; is\; an\; irrational\; number}$

$\mathrm{3.Every\; positive\; integer\; is\; of\; the\; form\; 2q}$

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 factorisation of 2 and 5 i.e $b=2^m\times 5^n$ where either m or n, or both can = 0.
If the prime factorisation 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 $2^3\times 5^1.$

$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 $7^1$