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 $a,b(b\ne 0)$

such that  $\sqrt{5}={\displaystyle \frac{a}{b}}$

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 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 ${\displaystyle \frac{p}{q}}$ 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 $a,b(b\ne 0)$ such that

$3+2\sqrt{5}={\displaystyle \frac{a}{b}}$

$\sqrt{5}={\displaystyle \frac{1}{2}}({\displaystyle \frac{a}{b}}-3)$

Since $a$ and $b$ are integers,

${\displaystyle \frac{1}{2}}({\displaystyle \frac{a}{b}}-3)$

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) ${\displaystyle \frac{1}{\sqrt{2}}}$

(ii) $7\sqrt{5}$

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

Solution 3:

(i)${\displaystyle \frac{1}{\sqrt{2}}}$

Let ${\displaystyle \frac{1}{\sqrt{2}}}$ is rational

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

${\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{a}{b}}$

$\sqrt{2}={\displaystyle \frac{b}{a}}$

${\displaystyle \frac{b}{a}}$ 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 $a,b(b\ne 0)$ such that

$7\sqrt{5}={\displaystyle \frac{a}{b}}$

$\sqrt{5}={\displaystyle \frac{a}{7b}}$

${\displaystyle \frac{a}{7b}}$ 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 $a,b(b\ne 0)$ such that

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

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

Since $a$ and $b$ are integers,

${\displaystyle \frac{a}{b}}-6$ 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.