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Learn Exercise 1.3 with Free Lessons & Tips

Question 1:

Prove that $\sqrt{5 }$ is irrational.

Solution 1:

Let $\sqrt{5}$ is a rational number. Therefore. we can find two integers a, b (b $eq$ 0)
such that $\sqrt{5}$ = $a/b$
Let a and b have a common factor other than 1. Then we can divide
them by the common factor. and assume that a and b are co-prime.
a = $\sqrt{5}b$
$a^{2}$ = $5b^{2}$
Therefore. $a^{2}$ is divisible by 5 and it can be said that a is divisible by 5. Let a
= 5k. where k is an integer
$(5k)^{2}$ = $5b^{2}$
$b^{2}$ = $5k^{2}$ This mans that $b^{2}$ is divisible by 5 and hence, b is divisible
by 5.
This implies that a and b have 5 as a common factor. And this is a
contradiction to the fact that a and b are co-prime.
Hence, cannot be expressed as $p/q$ and it can be said that $\sqrt{5}$ is irrational.

Comments

Krishna Verma

Question 2:

Prove that 3+2 $\sqrt{5}$ is irrational.

Solution 2:

To prove : 3+2√5 is rational number.

Ans: 3 is rational number.

2 is rational number.

2√5 is irrational number (product of rational and irrational is an irrational number)

3+2√5 is an irrational number (sum of rational and irrational number is irrational)

Let P = 3+2√5 is rational number ( we are assuming )

P= 3+2√5

P-3 =2√5

(P-3)÷2=√5

L.H.S. we assume P is a rational number .

P-3 is also rational number substraction of rational number always rational number .

(P-3)÷2 rational number (P-3) divide by rational number 2 hence (P-3)÷2 is rational number .

L.H.S. is rational While R.H.S. √5 is irrational L.H.S ≠R.H.S.

Our assumption is wrong so we can say 3+2√5 is an irrational number.

Comments

Nitin Pathak

Question 3:

Prove that the following are irrationals : (i) $\frac{1}{\sqrt{2}}$ (ii) $7\sqrt{5}$ (iii) $6+\sqrt{2 }$

Solution 3:

(i) Assume that $\frac{1}{\sqrt{2}}$ is Rational.

Hence, $\frac{1}{\sqrt{2}} =\frac{a}{b}, \sqrt{2}=\frac{b}{a}$

$\frac{b}{a}$ is rational as a and b are integers.

Hence $\sqrt{2}$ is rational which contradicts the fact that $\sqrt{2}$ is irrational.

Therefore, our assumption was incorrect and $\frac{1}{\sqrt{2}}$ is irrational.

(ii) Assume that 7 $\sqrt{5}$ is rational.

$7\sqrt{5} = \frac{a}{b}$ where a and b are some integers.

therefore, $\sqrt{5}= \frac{a}{7b}$

a and 7b are ratonal and so is $\sqrt{5}$

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence our assumption was incorrect and $7\sqrt{5}$ is rational.

(iii) Assume $6+\sqrt{2}$ as rational

$6+\sqrt{2}= \frac{a}{b} , \sqrt{2}= \frac{a}{b} - 6$

Since a and b are integers, $\frac{a}{b}-6$ is also rational and $\sqrt{2}$ should be rational as well.

This contradicts the fact that $\sqrt{2}$ is irrational and hence our assumption was incorrect.

$6+\sqrt{2}$ is irrational.

Comments

Tasneem

All Exercises - Chapter 1 - Real Numbers

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

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Learn Exercise 1.3 with Free Lessons & Tips

All Exercises - Chapter 1 - Real Numbers

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