exercise-1.3-class-10-maths

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Question 1:

Prove that \(\sqrt 5\) is irrational.

Solution 1:

Let \(\sqrt 5\) is a rational number. Therefore, we can find two integers \(a, b (b ≠ 0)\)

such that  \(\sqrt 5=\dfrac{a}{b}\)

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.

\(a =\sqrt{5} b\)

\(a^2 = 5 b^2\)

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

\(5k^2 = 5 b^2\)

\(b^2 = 5 k^2\).

This means 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 \(\dfrac{p}{q}\) or it can be said that \(\sqrt 5\) is irrational.

Question 2:

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

Solution 2:

Let is \(3 +2\sqrt{5}\) rational.

Therefore, we can find two integers \(a, b (b ≠ 0)\) such that

\(3 +2\sqrt{5}=\dfrac{a}{b}\)

\(\sqrt{5}=\dfrac{1}{2}(\dfrac{a}{b}-3)\)

Since \(a\) and \(b\) are integers,

\(\dfrac{1}{2}(\dfrac{a}{b}-3)\)

will also be rational and  therefore, \(\sqrt 5\) is rational.

This contradicts the fact that \(\sqrt 5\) is irrational. Hence, our assumption

that \(3 + 2 \sqrt 5\) is rational is false. 

Therefore, \(3 + 2 \sqrt 5\) is irrational.

 

Question 3:

Prove that the following are irrationals:

(i) \(\dfrac{1}{\sqrt{2}}\)

(ii) \(7\sqrt{ 5}\)

(iii) \(6 +\sqrt{ 2}\)

Solution 3:

(i)\(\dfrac{1}{\sqrt{2}}\)

Let \(\dfrac{1}{\sqrt{2}}\) is rational

Therefore, we can find two integers \(a, b (b ≠ 0)\) such that

\(\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}\)

\( \sqrt{2}=\dfrac{b}{a}\)

\(\dfrac{b}{a}\) is rational as \(a\) and \(b\) are integers. 

Therefore, \(\sqrt{2}\) is rational which contradicts to the fact that \(\sqrt{2}\) is irrational.

Hence, our assumption is false and \(\dfrac{1}{\sqrt{2}} is irrational.

(ii) \(7\sqrt{ 5}\)

Let \(7\sqrt{ 5}\) is rational.

Therefore, we can find two integers \(a, b (b ≠ 0)\) such that

\(7\sqrt{ 5}=\dfrac{a}{b}\)

\(\sqrt{ 5}=\dfrac{a}{7b}\)

\(\dfrac{a}{7b}\) is rational as \(a\) and \(b\) are integers.

Therefore, \(\sqrt{ 5}\) should be rational.

This contradicts the fact that \(\sqrt{ 5}\) is irrational. Therefore, our assumption that \(7\sqrt{ 5}\) is rational is false. Hence, \(7\sqrt{ 5}\) is irrational.

(iii) \(6 +\sqrt{ 2}\)

Let \(6 +\sqrt{ 2}\) be rational.

Therefore, we can find two integers \(a, b (b ≠ 0)\) such that

\(6 +\sqrt{ 2}=\dfrac{a}{b}\)

\( \sqrt{ 2}=\dfrac{a}{b}-6\)

Since \(a\) and \(b\) are integers,

\( \dfrac{a}{b}-6\) is also rational and hence, \( \sqrt{ 2}\) should be rational. 

This contradicts the fact that \( \sqrt{ 2}\) is irrational. 

Therefore,our assumption is false and hence, \(6 +\sqrt{ 2}\) is irrational.