Question

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Answer 1

Answer:

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Answer 2

Answer:

a) √7

Let √7 be a rational number

√7 = $\frac{p}{q}$, where q are co-prime integer and q≠0.

√7 = p

$( { \sqrt{7}q) }^{2} = ({p})^{2} \\ 7 {q}^{2} = {p}^{2} \\ {p}^{2} = 7 {q}^{2}$

Since, 7 divides ${p}^{2}$ it will also divides p.

Let 'r' be some integer

p = 7r

$({p})^{2}=({7r})^{2}$

${p}^{2} = {49r}^{2} \\ {7 {q}^{2} = {49r}^{2} }( {p}^{2} = {7q}^{2} ) \\ {q}^{2} = {7r}^{2}$

Since, 7 divides ${q}^{2}$ it will also divides q.

So, 7 is the common factor of 'p' and 'q'.

This, contradicts the fact that√7 is an irrational number.

So, our assumption was wrong that √7 is a rational number.

Therefore, √7 is irrational number.

b) a) √3

Let √3 be a rational number

√3 = $\frac{p}{q}$, where q are co-prime integer and q≠0.

√3 = p

$( { \sqrt{3}q) }^{2} = ({p})^{2} \\ 3 {q}^{2} = {p}^{2} \\ {p}^{2} = 3 {q}^{2}$

Since, 3 divides ${p}^{2}$ it will also divides p.

Let 'r' be some integer

p = 3r

$({p})^{2}=({3r})^{2}$

${p}^{2} = {9r}^{2} \\ {3 {q}^{2} = {9r}^{2} }( {p}^{2} = {3q}^{2} ) \\ {q}^{2} = {3r}^{2}$

Since, 3 divides ${q}^{2}$ it will also divides q.

So, 3 is the common factor of 'p' and 'q'.

This, contradicts the fact that√3 is an irrational number.

So, our assumption was wrong that √3 is a rational number.

Therefore, √3 is irrational number.

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