RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.3 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 1 Real Numbers Exercise 1.3

Question 1.
Prove that \(\sqrt{5}\) is irrational.
Solution:
Let us suppose that \(\sqrt{5}\) is a rational number. So, we can find two integers r and s where s ≠ 0 and
\(\sqrt{5}\) = \(\frac{r}{s}\)
Now again let us suppose that r and s have some common factors other than 1. Then on dividing r and s by that common factor, we can obtain \(\sqrt{5}\) = \(\frac{a}{b}\). Here a and b are coprime, i.e.
by \(\sqrt{5}\) = a
squaring both sides
⇒ (b\(\sqrt{5}\))² = a²
⇒ b²(\(\sqrt{5}\))² = a²
⇒ 5b² = a² ……..(i)
So 5 divides a².
According to theorem 1.3 if a prime number p divides a², then p will divide a also, where a is a positive integer.
⇒ 5 divides a also. …..(ii)
so a = 5c where c in any integer
Putting the value of a in (i)
5b² = (5c)²
5b² = 25c²
b² = 5c²
or 5c² = b²
⇒ 5 divides b²
According to theorem 1.3, 5divides b also. ……(iii)
From (ii) and (iii), there is at least one common factor 5 of a and b. But it is contradicts the fact that a and b are coprime or these have no common factor other than
1. Therefore our hypothesis that \(\sqrt{5}\) is a rational number is wrong, i.e., \(\sqrt{5}\) in an irrational number.

Question 2.
Prove that 3 + 2\(\sqrt{5}\) is irrational.
Solution:
Suppose that 3 + 2\(\sqrt{5}\) is a rational number. §0 we can find coprime numbers a and b wheijc a and b are integers such that b ≠ 0 and
3 + 2\(\sqrt{5}\) = \(\frac{a}{b}\)
or 3 – \(\frac{a}{b}\) = 2\(\sqrt{5}\)
or \(\frac{3-\frac{a}{b}}{2}\) = \(\sqrt{5}\)
or \(\sqrt{5}\) = \(\frac{3}{2}-\frac{a}{2 b}\)
or \(\sqrt{5}\) = \(\frac{3 b-a}{2 b}\) …….(i)
Since a and b is both are integers, therefore
\(\frac{3 b-a}{2 b}\) = \(\frac{3 \times \text { integer }-\text { integer }}{2 \times \text { integer }}\)
= a rational number
Hence from (i) \(\sqrt{5}\) is a rational number. But this fact contradicts the statement that \(\sqrt{5}\) is an irrational number, therefore this hypothesis is wrong. Therefore the given number 3 + 2\(\sqrt{5}\) is an irrational number.

Question 3.
Prove that the following are irrationals :
(i) \(\frac{1}{\sqrt{2}}\)
(ii) 7\(\sqrt{5}\)
(iii) 6 + \(\sqrt{2}\)
Solution:
(i) \(\frac{1}{\sqrt{2}}\)
Contrary to the statement given in question let us assume that \(\frac{1}{\sqrt{2}}\) is a rational number. So we can find coprime integers a and b (b ≠ 0), i.e.,

Since the quotient of two integers is a rational number, therefore \(\frac{2 a}{b}\) = a rational number.
From (i) \(\sqrt{2}\) also is a rational number. But this statement is wrong, i.e., our hypothesis is incorrect. Therefore \(\frac{1}{\sqrt{2}}\) is an irrational number.

(ii) 7\(\sqrt{5}\)
Suppose that the given number 7\(\sqrt{5}\) is a rational number. Therefore we can find two coprime integers a and b (b ≠ 0) such that
7\(\sqrt{5}\) = \(\frac{a}{b}\)
or 7b\(\sqrt{5}\) = a
or \(\sqrt{5}\) = \(\frac{a}{7b}\) ….(i)
Since in (i) a, 7 and b all are integers and the quotient of two integers is also a rational number, i.e., •
\(\frac{a}{7b}\) = a rational number therefore from (i)
\(\sqrt{5}\) = a rational number which is a contradictory statement of the statement \(\sqrt{5}\) is one irrational number.’ Hence our assumption in incorrect. Therefore 7\(\sqrt{5}\) is an irrational number.

(iii) 6 + \(\sqrt{2}\)
Suppose that 6 + \(\sqrt{2}\) in a rational number. So we can find coprime numbers a and b (b ≠ 0) such that

As a and b are integers, so \(\frac{a-6 b}{b}\) will be a rational number because the quotient of integers is also a rational number, i.e.,
\(\frac{a-6 b}{b}\) = a rational number
∴ From (i)
\(\sqrt{2}\) = a rational number
But this is a contradictory statement of the statement that \(\sqrt{2}\) is an irrational number’.
Hence our assumption is incorrect, i.e., 6 + \(\sqrt{2}\) is an irrational number.