NCERT SOLUTIONS FOR CLASS 10 MATHS CHAPTER 1 REAL NUMBERS EX 1.3

NCERT-SOLUTIONS-CLASS-10 MATH CH-1 REAL-NUMBERS-EX-1.3 has detailed explanations of concepts as well as answers just for the excellence of the students. We, at cbseinsights.com aspire to increase the intellectual levels of the students for various competitive exams……

Find the Theorem for reference :-

Question 1. Prove that √5 is irrational.
Solution:

Let us assume, that √5 is rational number.
i.e. √5 = x / y (where, x and y are co-primes)

Rearranging (Cross-multiplication) the equation √5 = x / y
we get y√5 = x …………………………. Eq. (1)
Squaring both the sides, we get,
(y√5)² = x²
⇒5y² = x² or x² = 5y² ………………… Eq. (2)

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

By theorem , (Let p be a prime number. If p divides a², then p divides a, where a is a positive integer.)

Let us say, x = 5k, for some value of k

Squaring both the sides, we get

x² = 25 k² and substituting the value of x in Eq. (2), we
get,
5y² = (5k)²
⇒ y² = 5k² 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.

Question 2. Show that 3 + 2√5 is irrational.
Solution:
Let us assume 3 + 2√5 is rational.
Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y
Rearranging, we get,
Since x and y are integers, thus, is a rational number.
Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.
So, we conclude that 3 + 2√5 is irrational.

NCERT-SOLUTIONS-CLASS-10 MATH CH-1 REAL-NUMBERS-EX-1.3

Question 3. Prove that the following are irrational.
(i) 1/√2
(ii) 7√5
(iii) 6 + √2
Solution:

(i) 1/√2
Let us assume 1/√2 is rational.

Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y
Rearranging, we get,
√2 = y/x
Since x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.
Hence, we can conclude that 1/√2 is irrational.

(ii) 7√5
Let us assume 7√5 is a rational number.
Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y
Rearranging, we get,
√5 = x/7y
Since x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5
is irrational.
Hence, we can conclude that 7√5 is irrational.

(iii) 6 +√2
Let us assume 6 +√2 is a rational number.
Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x / y⋅
Rearranging, we get,
√2 = (x / y) – 6
Since x and y are integers, thus (x / y) – 6 is a rational number, and therefore, √2 is rational.
This contradicts the fact that √2 is an irrational number.
Hence, we can conclude that 6 +√2 is irrational.