## Introduction to Real Numbers

In mathematics, rational numbers and irrational numbers are together obtained from the set of real numbers. The set of real numbers is represented by the letter R. Therefore, it indicates that every real number is either a rational number or an irrational number. In either case, it contains a non–terminating decimal depiction. In the instance of rational numbers, the decimal depiction is repeating (including repeating zeroes) and if the decimal depiction is non–repeating, it is an irrational number.

## Topics and Sub Topics in Maths Real Numbers For Class 10

 Subject Maths Chapter Number 1 Chapter Name Real Numbers Topics 1.1  Introduction To Real Numbers 1.2  Euclid’s Division Lemma 1.3  The Fundamental Theorem of Arithmetic 1.4  Revisiting Irrational Numbers 1.5 Revisiting Rational Numbers and Their Decimal Expansions 1.6  Exercises 1.7  Summary

### Solved Examples

Here are a few examples of Class 10 real numbers solutions with practice problems:

Example: Numbers 135 and 225

Solution:

Step 1: Seeing that 225 is greater than 135, we will use Euclid’s division lemma, to

a =225

b=135

In order to find q and r, such that 225 = 135q+r, 0 is less than r

On dividing 225 by 135 we obtain the quotient as 1 and remainder as 90 i.e. 225 = 135 x 1 + 90

Step 2: Now, Remainder r which is 90 ≠ 0, we use Euclid’s division lemma to b =135 and r = 90 in order to determine the whole numbers q and r such that

135 = 90 x q + r, 0 r<90

On dividing 135 by 90 we obtain the quotient as 1 and remainder as 45 i.e. 135 = 90 x 1 + 45

Step 3: Remainder r = 45 ≠ 0 so we apply Euclid’s division lemma to

b =90

r = 45

In order to determine q and r such that 90 = 90 x q + r, 0 r<45

On dividing 90 by 45 we obtain the quotient as 2 and remainder as 0 i.e. 90 = 2 x 45 + 0

Step 4: Seeing that the remainder is zero, the divisor at this stage is HCF of (135, 225).

Seeing that the divisor at this stage is 45, thus, the HCF of 135 and 225 will be 45.

Example:

Find out the HCF of 867 and 255

Solution:

Using the Euclid’s division algorithm, we have

867 = 255 x 3 + 102

255 = 102 x 2 + 51

102= 51 x 2 + 0

Therefore, HCF of (867, 255) = 51

Example: Find out if 1009 is a prime or a composite number

Numbers are of two types – prime and composite. Prime numbers consist of only two factors namely 1 and the number itself while composite numbers consist of factors besides 1 and itself.

It can be observed that

7 x 11 x 13 + 13 = 13 x [7 x 11 + 1] = 13 x [77 + 1]

= 13 x 78

= 13 x 13 x 6

The provided expression consists of 6 and 13 as its factors. Thus, it is a composite number.

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)

= 5 x [1008 + 1)]

= 5 x 1009

1009 cannot be factorized any further. Thus, the given expression consists of 5 and 1009 as its factors. Therefore, it is a composite number.

### Conclusion

From the above explanations, we may conclude that to every real number there is a corresponding special point on the number line and contrarily, to each point on the number line there is a corresponding real number. Therefore, we observe that there is one–to–one correspondence between the real numbers and points on the number line ‘l’, which is why the number line is known as the ‘real number line’.

### Did you know

For every real number, there is a special corresponding point on the number line ‘l’ or we may also say that every point on the line ‘l’ corresponds to a real number (rational or irrational).