#### Chapter 1 Real Numbers – Part 1 Class 10 Math

**Introduction to Real Numbers**

**Real Numbers**

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

**Euclid’s Division Lemma**

- Euclid’s Division Lemma states that given two integers
*a*and*b*, there exists a unique pair of integers*q*and*r*such that a=b×q+r and 0≤r<b. - This lemma is essentially equivalent to :
*dividend = divisor × 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**

- Euclid’s Division Algorithm is a method used to find the
**H.C.F**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×q+r and 0≤r<b. - 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.
- The above method is repeated until a remainder of zero is obtained. The divisor in that step is the H.C.F of the given set of numbers.

**The Fundamental Theorem of Arithmetic**

Prime Factorisation

- Prime Factorisation is the method of expressing a natural number as a product of prime numbers.
- Example: 36=2×2×3×3 is the prime factorisation of 36.

**Fundamental Theorem of Arithmetic**

- The Fundamental Theorem of Arithmetic states that the prime factorisation for a given number is unique if the arrangement of the prime factors is ignored.
- Example: 36=2×2×3×3 OR, 36=2×3×2×3
- Therefore, 36 is represented as a product of prime factors (Two 2s and two 3s) ignoring the arrangement of the factors.

**Method of Finding LCM**

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

- 36=2×2×3×3

56=2×2×2×7 - The common prime factors are 2×2
- The uncommon prime factors are 3×3 for 36 and 2×7 for 56.
- LCM of 36 and 56 = 2×2×3×3×2×7 which is 504

**Method of Finding HCF**

H.C.F can be found using two methods – Prime factorisation and 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.

Example: To find the HCF of 18 and 30

- Finding the HCF of 18 and 30
- The required HCF is
**6**.

- The required HCF is

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

- For any
**two**positive integers a and b,

a×b=H.C.F×L.C.M. - Example – For 36 and 56, the H.C.F is 4 and the L.C.M is 504

36×56=2016

4×504=2016

Thus, 36×56=4×504 - 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**

**Irrational Numbers**

Any number that cannot be expressed in the form of p/q (where p and q are integers and q≠0.) is an irrational number. Examples √2,π, e and so on.

Number theory: Interesting results

- If a number p (a prime number) divides a2, then p divides a. Example: 3 divides 62 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.

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, a2/b2=7

⇒a2=7b2.

Hence, 7 divides a. Then, there exists a number c such that a=7c. Then, a2=49c2. Hence, 7b2=49c2 or b2=7c2.

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.

**Revisiting Rational Numbers and Their Decimal Expansions**

**Rational Numbers**

Rational numbers are numbers that can be written in the form p/q, where p and q are integers and q≠0.

Examples -1/2, 4/5, 1,0,−3 and so on.

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**.**142857*142857**….*)

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 factorisation of 2 and 5 i.e b=2m×5n 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 23×51.

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 71