In simple words, real numbers are the combination of both rational and irrational numbers in a number system. The class 10 Math Real Numbers talks about real numbers, donated as “R,”

which can be both positive (+) and negative (-). Real numbers include all natural numbers, fractions, whole numbers, and decimals. 

 

Some of the examples of real numbers are 6.99, 5/2, 23, -12, π, and so on. The set of real numbers is an amalgamation of various categories, including the natural numbers (N = {1,2,3,4,……}, integers  -infinity (-∞),……..-4, -3, and so on, whole numbers 0, 1, 2, 3, 4,5,6,…..…, rational  ½, 5/4 and 12/6, etc., and irrational numbers √2.

Properties of real numbers

There are majorly four properties of real numbers:

Real Number Properties	Examples
Commutative property	Addition:  5 + 3 = 3 + 5, 2 + 4 = 4 + 2 Multiplication: 5 × 3 = 3 × 5, 2 × 4 = 4 × 2
Associative property	Addition: 10 + (3 + 2) = (10 + 3) + 2. Multiplication: (2 × 3) 4 = 2 (3 × 4)
Distributive property	5(2 + 3) = 5 × 2 + 5 × 3.
Identity property	Addition: m + 0 = m (0 is the additive identity) Multiplication: m × 1 = 1 × m = m. (1 is the multiplicative identity)

 

Introduced by Euclid’s Division Lemma, in the Euclidean division algorithm, if there are two integers a and b, then the other two unique integers say q and r that satisfy the condition a = bq + r where 0 ≤ r < b. This algorithm is majorly used to calculate the Highest Common Factor (HCF) of two positive integers, a and b.

Let’s understand the Euclidean division algorithm through an example:

Example: Use the Euclidean division algorithm to find the HCF of 81 and 675.

In the given question, the larger integer is 675, and according to the Euclidean division algorithm, the equation will be a = bq + r where 0 ≤ r < b, we have

a = 675 and b = 81

⇒ 675 = 81 × 8 + 27

If applied again, we have,

81 = 27 × 3 + 0 

Since the remainder is zero, we cannot proceed further. Here the divisor is 27; therefore, the HCF of 65 and 81 will be 27.

 

According to the Fundamental Theorem of Arithmetic, if the integer is greater than 1 then the number will be either a prime number or will be expressed as a prime number. For example, the number is 35, which can be written as the prime number as 35 = 7 × 5. Here, 7 and 5 are the prime factors of 35.

 

Similarly, 114560 can be written as the product of its prime factors by using the prime factorization method. 

114560 = 27 × 5 × 179

Example Question: In a formula racing competition, the time taken by two racing cars A and B, to complete 1 round of the track is 30 minutes and 45 minutes, respectively. After how much time will the cars meet again at the starting point?

Solution: This time can be calculated by finding the L.C.M of the time taken by each.

30 = 2 × 3 × 5

45 = 3 × 3 × 5

The L.C.M is 90.

Thus, both cars will meet at the starting point after 90 minutes.

 

Any number which cannot be presented in the form of a/b, where both are integers and b≠ 0 is known as an irrational number. These are categorized under real numbers, which cannot be written in the form of a simple fraction. The irrational numbers are denoted by “P” and are the contradiction of rational numbers.

List of Irrational Numbers-

Pi, π	3.14159265358979…
Euler’s Number, e	2.71828182845904…
Golden ratio, φ	1.61803398874989….

 

Example of irrational number: 

Which of the following are Rational Numbers or Irrational Numbers?

2, -.45678…, 6.5, √ 3, √ 2

Solution: Rational Numbers – 2, 6.5 as these have terminating decimals.

Irrational Numbers – -.45678…, √ 3, √ 2 as these have a non-terminating non-repeating decimal expansion.

 

Contradictory to irrational numbers, any number which can be written or represented as p/q or in the form of a fraction is termed as a rational number denoted by “R.” 6,−8.1 are examples of rational numbers. 

 

Any real number which can be terminated or recurred is generally a rational number. For example- the number is 33.33333…….. Since it is represented as 100/3, it is a rational number that cannot be terminated (.333..). Such numbers are known as recurring decimal numbers.

1- Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r, for some integer q ≥ 0, and r = 0, 1, 2, 3, 4, 5, because 0≤r<6.

Now substituting the value of r, we get,

If r = 0, then a = 6q

Similarly, for r= 1, 2, 3, 4 and 5, the value of a is 6q+1, 6q+2, 6q+3, 6q+4 and 6q+5, respectively.

If a = 6q, 6q+2, 6q+4, then a is an even number and divisible by 2. A positive integer can be either even or odd. Therefore, any positive odd integer is of the form of 6q+1, 6q+3, and 6q+5, where q is some integer.

2- Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution: As we know that,

HCF×LCM=Product of the two given numbers

Therefore,

9 × LCM = 306 × 657

LCM = (306×657)/9 = 22338

Hence, LCM(306,657) = 22338

3- Check whether 6n can end with the digit 0 for any natural number n.

Solution: If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with the unit place as 0 or 5 is divisible by 5.

Prime factorization of 6n = (2×3)n

Therefore, the prime factorization of 6n doesn’t contain the prime number 5.

Hence, it is clear that for any natural number n, 6n is not divisible by 5, and thus it proves that 6n cannot end with the digit 0 for any natural number n.

4- Prove that √5 is irrational.

Solutions: Let us assume that √5 is a rational number.

i.e. √5 = x/y (where, x and y are co-primes)

y√5= x

Squaring both the sides, we get,

(y√5)2 = x2

⇒5y2 = x2……………………………….. (1)

Thus, x2 is divisible by 5, so x is also divisible by 5.

Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,

5y2 = (5k)2

⇒y2 = 5k2

which is divisible by 5. It means y is divisible by 5.

Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.

Hence, √5 is an irrational number.

5- Write down the decimal expansions of those rational numbers in Question 1 above, which have terminating decimal expansions.

Solutions:

(i) 13/3125

13/3125 = 0.00416

    1. What is the first chapter of class 10 maths?

The first chapter of class 10 maths is Real Numbers which talks about real numbers and their various types. 

   2. How many exercises are there in class 10 maths Chapter 1 Real Numbers? 

There are in total four exercises in class 10 maths Real Numbers.

    3. What are the real numbers in Class 10? 

Real Numbers is one of the most important chapters in class 10 maths which talks about: 

• Euclid’s Division Algorithm
• The Fundamental Theorem of Arithmetic
• Revisiting Rational & Irrational Numbers
• Decimal Expansions