Q: Prove that \$\sqrt {3}\$ is an irrational number.

Let us assume on the contrary that \$\sqrt {3}\$ is a rational number. Then, there exist positive integers \$a\$ and \$b\$ such that\$\sqrt { 3 } =\dfrac { a }{ b } \$ where, \$a\$ and \$b\$, are co-prime i.e. their \$HCF\$ is \$1\$Now,\$\sqrt { 3 } =\dfrac { a }{ b } \$\$\Rightarrow \quad 3=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } } \$ \$\Rightarrow \quad 3{ b }^{ 2 }={ a }^{ 2 }\$ \$\Rightarrow \quad 3\$ divides \${ a }^{ 2 }\quad \quad \left[ \because 3 \ \text{divides}\ 3{ b }^{ 2 } \right] \$ \$\Rightarrow \quad 3\$ divides \$a\quad \quad ...\left( i \right) \$ \$\Rightarrow \quad a=3c\$ for some integer \$c\$\$\Rightarrow \quad { a }^{ 2 }=9{ c }^{ 2 }\$ \$\Rightarrow \quad 3{ b }^{ 2 }={ 9c }^{ 2 }\quad \quad \left[ \because { a }^{ 2 }=3{ b }^{ 2 } \right] \$ \$\Rightarrow \quad { b }^{ 2 }={ 3c }^{ 2 }\$ \$\Rightarrow \quad 3\$ divides \${ b }^{ 2 }\quad \quad \left[ \because 3 \ \text{divides}\ 3{ c }^{ 2 } \right] \$ \$\Rightarrow \quad 3\$ divides \$b\quad \quad ...\left( ii \right) \$ From \$(i)\$ and \$(ii),\$ we observe that \$a\$ and \$b\$ have at least \$3\$ as a common factor. But, this contradicts the fact that \$a\$ and \$b\$ are co-prime. This means that our assumption is not correct.Hence, \$\sqrt {3}\$ is an irrational number.

Q: Prove that \$\sqrt 5\$ is irrational by the method of Contradiction.

Let \$\sqrt5\$ be a rational number.then it must be in form of \$\dfrac pq\$ where, \$q\neq0\$ ( \$p\$ and \$q\$ are co-prime)\$\sqrt5=\dfrac pq\$\$\sqrt5 \times q=p\$Suaring on both sides,\$5{q}^2{}={p}^{2}\$ --------------(1)\${p}^{2}\$ is divisible by \$5\$.So, \$p\$ is divisible by \$5\$.\$p=5c\$Suaring on both sides,\${p}^{2}=25{c}^{2}\$ --------------(2)Put \${p}^{2}\$ in eqn.(1)\$5{q}^{2}=25(c)^{2}\$\${q}^{2}=5 c^{2}\$So, \$q\$ is divisible by \$5\$..Thus \$p\$ and \$q\$ have a common factor of \$5\$.So, there is a contradiction as per our assumption.We have assumed \$p\$ and \$q\$ are co-prime but here they a common factor of \$5\$.The above statement contradicts our assumption.Therefore, \$\sqrt5\$ is an irrational number.

Q: Prove that \$\sqrt {2}\$ is an irrational number.

Let us assume on the contrary that \$\sqrt {2}\$ is a rational number. Then, there exist positive integers \$a\$ and \$b\$ such that\$\sqrt { 2 } =\dfrac { a }{ b } \$ where, \$a\$ and \$b\$, are co-prime i.e. their \$HCF\$ is \$1\$\$\Rightarrow \quad { \left( \sqrt { 2 } \right) }^{ 2 }={ \left( \dfrac { a }{ b } \right) }^{ 2 }\$ \$\Rightarrow \quad 2=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } } \$ \$\Rightarrow \quad 2{ b }^{ 2 }={ a }^{ 2 }\$ \$\Rightarrow \quad 2|{ a }^{ 2 }\quad \quad \left[ \because 2|2{ b }^{ 2 }\ and\ 2{ b }^{ 2 }={ a }^{ 2 } \right] \$ \$\Rightarrow \quad 2|a\quad \quad ...\left( i \right) \$ \$\Rightarrow \quad a=2c\$ for some integer \$c\$\$\Rightarrow \quad { a }^{ 2 }=4{ c }^{ 2 }\$ \$\Rightarrow \quad 2{ b }^{ 2 }={ 4c }^{ 2 }\quad \quad \left[ \because 2{ b }^{ 2 }={ a }^{ 2 } \right] \$ \$\Rightarrow \quad { b }^{ 2 }={ 2c }^{ 2 }\$ \$\Rightarrow \quad 2|{ b }^{ 2 }\quad \quad \left[ \because 2|2{ c }^{ 2 } \right] \$ \$\Rightarrow \quad 2|b\quad \quad ...\left( ii \right) \$From \$(i)\$ and \$(ii),\$ we obtain that \$2\$ is a common factor of \$a\$ and \$b\$. But, this contradicts the fact that \$a\$ and \$b\$ have no common factor other than \$1\$. This means that our supposition is wrong.Hence, \$\sqrt {2}\$ is an irrational number.

Q: Prove that \$\sqrt 3 + \sqrt 5 \$ is irrational.

To prove : \$\sqrt{3}+\sqrt{5} \$ is irrational.Let us assume it to be a rational number. Rational numbers are the ones that can be expressed in \$\dfrac p q\$ form where \$p,q\$ are integers and \$q\$ isn't equal to zero.\$\sqrt{3}+\sqrt{5}=\dfrac p q\\\$\$\sqrt{3}=\dfrac p q-\sqrt{5}\\\$squaring on both sides, \$3= \dfrac {p^{2}}{q^{2}}-2.\sqrt{5} \left(\dfrac {p} {q}\right)+5\$\$\Rightarrow \dfrac{(2\sqrt{5}p)}{q}=5-3+\left(\dfrac {p^2} {q^2}\right)\$ \$\Rightarrow \dfrac{(2\sqrt{5}p)}{q}= \dfrac {2q^{2}- {p^2}} {q^2}\$\$\Rightarrow \sqrt{5}= \dfrac {2q^{2}- {p^2}} {q^2}. \dfrac {q}{2p}\\\$\$\Rightarrow \sqrt{5}=\dfrac {(2q^{2}-p^{2})} {2pq}\\\$As \$p\$ and \$q\$ are integers RHS is also rational.As RHS is rational LHS is also rational i.e \$\sqrt{5}\$ is rational.But this contradicts the fact that\$ \sqrt{5}\$ is irrational.This contradiction arose because of our false assumption.so,\$\sqrt{3}+\sqrt{5}\$ irrational.

Q: Prove that \$\sqrt {7}\$ is an irrational number.

Let us assume that \$\sqrt{7}\$ is rational. Then, there exist co-prime positive integers \$a\$ and \$b\$ such that\$\sqrt{7} = \dfrac{a}{b}\$\$\implies a = b\sqrt{7}\$Squaring on both sides, we get\$a^2 = 7b^2\$Therefore, \$a^2\$ is divisible by \$7\$ and hence, \$a\$ is also divisible by\$7\$so, we can write \$a = 7p,\$ for some integer \$p.\$Substituting for \$a\$, we get \$49p^2 = 7b^2 \implies b^2 = 7p^2.\$This means, \$b^2 \$ is also divisible by \$7\$ and so, \$b\$ is also divisible by \$7\$.Therefore, \$a\$ and \$b\$ have at least one common factor, i.e., \$7\$.But, this contradicts the fact that \$a\$ and \$b\$ are co-prime.Thus, our supposition is wrong.Hence, \$\sqrt{7} \$ is irrational.

Q: Prove that \$\sqrt 2+\sqrt 3\$ is irrational

Let us assume that \$\sqrt{2}+\sqrt{3}\$ is a rational numberThen. there exist coprime integers \$p\$, \$q\$,\$q\neq 0\$ such that \$\sqrt{2}+\sqrt{3}=\dfrac{p}{q}\$\$=>\dfrac{p}{q}-\sqrt{3}=\sqrt{2}\$Squaring on both sides, we get\$=>(\dfrac{p}{q}-\sqrt{3})^2=(\sqrt{2})^2\$\$=>\dfrac{p^2}{q^2}-2\dfrac{p}{q}\sqrt{3}+(\sqrt{3})^2=2\$\$=>\dfrac{p^2}{q^2}-2\dfrac{p}{q}\sqrt{3}+3=2\$\$=>\dfrac{p^2}{q^2}+1=2\dfrac{p}{q}\sqrt{3}\$\$=>\dfrac{p^2+q^2}{q^2}\times \dfrac{q}{2p}=\sqrt{3}\$\$=>\dfrac{p^2+q^2}{2pq}=\sqrt{3}\$Since, \$p\$,\$q\$ are integers, \$\dfrac{p^2+q^2}{2pq}\$ is a rational number.\$=>\sqrt{3}\$ is a rational number.This contradicts the fact that \$\sqrt{3}\$ is irrational.Thus, our assumption is incorrect.Therefore, \$\sqrt{2}+\sqrt{3}\$ is a irrational.

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

Prove \$ 3+2\sqrt{5}\$ is irrational.\$ \rightarrow \$ let take that \$3+2\sqrt{5}\$is rational number\$ \rightarrow \$ so, we can write this answer as\$\Rightarrow\$\$ 3+2\sqrt{5} = \dfrac ab\$ Here \$a\$ & \$b\$ use two coprime number and \$ b \neq 0.\$\$\Rightarrow\$\$ 2\sqrt{5} = \dfrac{a}{b}-3 \$\$\Rightarrow\$\$ 2\sqrt{5} = \dfrac{a-3b}{b}\$\$ \therefore \sqrt{5} = \dfrac{a-3b}{2b}\$Here \$a\$ and \$b\$ are integer so \$ \dfrac{a-3b}{2b}\$ is a rational number so \$ \sqrt{5}\$ should be rational number but\$ \sqrt{5}\$ is a irrational number so it is contradict- Hence \$ 3+2\sqrt{5}\$ is irrational.

Q: Prove that \$3\sqrt { 2 } \$ is irrational.

Let us assume, to the contrary, that \$3 \sqrt {2}\$ is rational. Then, there exist co-prime positive integers \$a\$ and \$b\$ such that\$3\sqrt { 2 } =\cfrac { a }{ b } \$\$\Rightarrow\$ \$\sqrt { 2 } =\cfrac { a }{ 3b } \$\$\Rightarrow \$ \$\sqrt {2}\$ is rational ...\$[\$\$\because 3,a\$ and \$b\$ are integers\$\therefore \cfrac { a }{ 3b }\$is a rational number\$]\$This contradicts the fact that \$\sqrt {2}\$ is irrational. So, our assumption is not correct.Hence, \$3 \sqrt {2}\$ is an irrational number.

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

(i) \$\dfrac{1}{\sqrt{2}}\$Let us assume \$\dfrac{1}{\sqrt{2}}\$ is rational.So we can write this number as\$\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}\$ ---- (1)Here, \$a\$ and \$b\$ are two co-prime numbers and \$b\$ is not equal to zero.Simplify the equation (1) multiply by \$\sqrt{2}\$ both sides, we get\$1 = \dfrac{a\sqrt{2}}{b}\$Now, divide by \$b\$, we get\$b = a\sqrt{2}\$ or \$\dfrac{b}{a}= \sqrt{2}\$Here, \$a\$ and \$b\$ are integers so, \$\dfrac{b}{a}\$ is a rational number, so \$\sqrt{2}\$ should be a rational number.But \$\sqrt{2}\$ is a irrational number, so it is contradictory.Therefore, \$\dfrac{1}{\sqrt{2}}\$ is irrational number.(ii) \$7\sqrt{5}\$Let us assume \$7\sqrt{5}\$ is rational.So, we can write this number as\$7\sqrt{5} = \dfrac{a}{b}\$ ---- (1)Here, \$a\$ and \$b\$ are two co-prime numbers and \$b\$ is not equal to zero.Simplify the equation (1) divide by 7 both sides, we get\$\sqrt{5} = \dfrac{a}{7b}\$Here, \$a\$ and \$b\$ are integers, so \$\dfrac{a}{7b}\$ is a rational number, so \$\sqrt{5}\$ should be a rational number.But \$\sqrt{5}\$ is a irrational number, so it is contradictory.Therefore, \$7\sqrt{5}\$ is irrational number.(iii) \$6+\sqrt{2}\$Let us assume \$6+\sqrt{2}\$ is rational.So we can write this number as\$6+\sqrt{2} = \dfrac{a}{b}\$ ---- (1)Here, \$a\$ and \$b\$ are two co-prime number and \$b\$ is not equal to zero.Simplify the equation (1) subtract 6 on both sides, we get\$\sqrt{2} = \dfrac{a}{b}-6\$\$\sqrt{2} = \dfrac{a-6b}{b}\$Here, \$a\$ and \$b\$ are integers so, \$\dfrac{a-6b}{b}\$ is a rational number, so \$\sqrt{2}\$ should be a rational number.But \$\sqrt{2}\$ is a irrational number, so it is contradictory.Therefore, \$6+\sqrt{2}\$ is irrational number.

Question 1

Prove that $\sqrt {3}$ is an irrational number.

Accepted Answer

Let us assume on the contrary that $\sqrt {3}$ is a rational number. Then, there exist positive integers $a$ and $b$ such that$\sqrt { 3 } =\dfrac { a }{ b } $ where, $a$ and $b$, are co-prime i.e. their $HCF$ is $1$Now,$\sqrt { 3 } =\dfrac { a }{ b } $$\Rightarrow \quad 3=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } } $ $\Rightarrow \quad 3{ b }^{ 2 }={ a }^{ 2 }$ $\Rightarrow \quad 3$ divides ${ a }^{ 2 }\quad \quad \left[ \because 3 \ \text{divides}\ 3{ b }^{ 2 } \right] $ $\Rightarrow \quad 3$ divides $a\quad \quad ...\left( i \right) $ $\Rightarrow \quad a=3c$ for some integer $c$$\Rightarrow \quad { a }^{ 2 }=9{ c }^{ 2 }$ $\Rightarrow \quad 3{ b }^{ 2 }={ 9c }^{ 2 }\quad \quad \left[ \because { a }^{ 2 }=3{ b }^{ 2 } \right] $ $\Rightarrow \quad { b }^{ 2 }={ 3c }^{ 2 }$ $\Rightarrow \quad 3$ divides ${ b }^{ 2 }\quad \quad \left[ \because 3 \ \text{divides}\ 3{ c }^{ 2 } \right] $ $\Rightarrow \quad 3$ divides $b\quad \quad ...\left( ii \right) $ From $(i)$ and $(ii),$ we observe that $a$ and $b$ have at least $3$ as a common factor. But, this contradicts the fact that $a$ and $b$ are co-prime. This means that our assumption is not correct.Hence, $\sqrt {3}$ is an irrational number.

Question 2

Prove that $\sqrt 5$ is irrational by the method of Contradiction.

Accepted Answer

Let $\sqrt5$ be a rational number.then it must be in form of  $\dfrac pq$  where,  $q\neq0$     ( $p$ and $q$ are co-prime)$\sqrt5=\dfrac pq$$\sqrt5 \times q=p$Suaring on both sides,$5{q}^2{}={p}^{2}$           --------------(1)${p}^{2}$ is divisible by $5$.So, $p$ is divisible by $5$.$p=5c$Suaring on both sides,${p}^{2}=25{c}^{2}$         --------------(2)Put ${p}^{2}$ in eqn.(1)$5{q}^{2}=25(c)^{2}$${q}^{2}=5 c^{2}$So, $q$ is divisible by $5$..Thus $p$ and $q$ have a common factor of $5$.So, there is a contradiction as per our assumption.We have assumed $p$ and $q$ are co-prime but here they a common factor of $5$.The above statement contradicts our assumption.Therefore, $\sqrt5$ is an irrational number.

Question 3

Prove that $\sqrt {2}$ is an irrational number.

Accepted Answer

Let us assume on the contrary that $\sqrt {2}$ is a rational number. Then, there exist positive integers $a$ and $b$ such that$\sqrt { 2 } =\dfrac { a }{ b } $ where, $a$ and $b$, are co-prime i.e. their $HCF$ is $1$$\Rightarrow \quad { \left( \sqrt { 2 }  \right)  }^{ 2 }={ \left( \dfrac { a }{ b }  \right)  }^{ 2 }$ $\Rightarrow \quad 2=\dfrac { { a }^{ 2 } }{ { b }^{ 2 } } $ $\Rightarrow \quad 2{ b }^{ 2 }={ a }^{ 2 }$ $\Rightarrow \quad 2|{ a }^{ 2 }\quad \quad \left[ \because 2|2{ b }^{ 2 }\ and\ 2{ b }^{ 2 }={ a }^{ 2 } \right] $ $\Rightarrow \quad 2|a\quad \quad ...\left( i \right) $ $\Rightarrow \quad a=2c$ for some integer $c$$\Rightarrow \quad { a }^{ 2 }=4{ c }^{ 2 }$ $\Rightarrow \quad 2{ b }^{ 2 }={ 4c }^{ 2 }\quad \quad \left[ \because 2{ b }^{ 2 }={ a }^{ 2 } \right] $ $\Rightarrow \quad { b }^{ 2 }={ 2c }^{ 2 }$ $\Rightarrow \quad 2|{ b }^{ 2 }\quad \quad \left[ \because 2|2{ c }^{ 2 } \right] $ $\Rightarrow \quad 2|b\quad \quad ...\left( ii \right) $From $(i)$ and $(ii),$ we obtain that $2$ is a common factor of $a$ and $b$. But, this contradicts the fact that $a$ and $b$ have no common factor other than $1$. This means that our supposition is wrong.Hence, $\sqrt {2}$ is an irrational number.

Question 4

Prove that $\sqrt 3  + \sqrt 5 $ is irrational.

Accepted Answer

To prove : $\sqrt{3}+\sqrt{5} $ is irrational.Let us assume it to be a rational number. Rational numbers are the ones that can be expressed in $\dfrac p q$ form where $p,q$ are integers and $q$ isn't equal to zero.$\sqrt{3}+\sqrt{5}=\dfrac p q\$$\sqrt{3}=\dfrac p q-\sqrt{5}\$squaring on both sides, $3= \dfrac {p^{2}}{q^{2}}-2.\sqrt{5} \left(\dfrac {p} {q}\right)+5$$\Rightarrow \dfrac{(2\sqrt{5}p)}{q}=5-3+\left(\dfrac {p^2} {q^2}\right)$ $\Rightarrow \dfrac{(2\sqrt{5}p)}{q}= \dfrac {2q^{2}- {p^2}} {q^2}$$\Rightarrow \sqrt{5}= \dfrac {2q^{2}- {p^2}} {q^2}. \dfrac {q}{2p}\$$\Rightarrow \sqrt{5}=\dfrac {(2q^{2}-p^{2})} {2pq}\$As $p$ and $q$ are integers RHS is also rational.As RHS is rational LHS is also rational i.e $\sqrt{5}$ is  rational.But this contradicts the fact that$ \sqrt{5}$ is irrational.This contradiction arose because of our false assumption.so,$\sqrt{3}+\sqrt{5}$ irrational.

Question 5

Prove that $\sqrt {7}$ is an irrational number.

Accepted Answer

Let us assume that $\sqrt{7}$ is rational. Then, there exist co-prime positive integers $a$ and $b$ such that$\sqrt{7} = \dfrac{a}{b}$$\implies a = b\sqrt{7}$Squaring on both sides, we get$a^2 = 7b^2$Therefore, $a^2$ is divisible by $7$ and hence, $a$ is also divisible by$7$so, we can write $a = 7p,$ for some integer $p.$Substituting for $a$, we get $49p^2 = 7b^2 \implies b^2 = 7p^2.$This means, $b^2 $ is also divisible by $7$ and so, $b$ is also divisible by $7$.Therefore, $a$ and $b$ have at least one common factor, i.e., $7$.But, this contradicts the fact that $a$ and $b$ are co-prime.Thus, our supposition is wrong.Hence, $\sqrt{7} $ is irrational.

Question 6

Prove that $\sqrt 2+\sqrt 3$ is irrational

Accepted Answer

Let us assume that $\sqrt{2}+\sqrt{3}$ is a rational numberThen. there exist coprime integers $p$, $q$,$q\neq 0$ such that      $\sqrt{2}+\sqrt{3}=\dfrac{p}{q}$$=>\dfrac{p}{q}-\sqrt{3}=\sqrt{2}$Squaring on both sides, we get$=>(\dfrac{p}{q}-\sqrt{3})^2=(\sqrt{2})^2$$=>\dfrac{p^2}{q^2}-2\dfrac{p}{q}\sqrt{3}+(\sqrt{3})^2=2$$=>\dfrac{p^2}{q^2}-2\dfrac{p}{q}\sqrt{3}+3=2$$=>\dfrac{p^2}{q^2}+1=2\dfrac{p}{q}\sqrt{3}$$=>\dfrac{p^2+q^2}{q^2}\times \dfrac{q}{2p}=\sqrt{3}$$=>\dfrac{p^2+q^2}{2pq}=\sqrt{3}$Since, $p$,$q$ are integers, $\dfrac{p^2+q^2}{2pq}$ is a rational number.$=>\sqrt{3}$ is a rational number.This contradicts the fact that $\sqrt{3}$ is irrational.Thus, our assumption is incorrect.Therefore, $\sqrt{2}+\sqrt{3}$ is a irrational.

Question 7

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

Accepted Answer

Prove $ 3+2\sqrt{5}$ is irrational.$ \rightarrow $ let take that $3+2\sqrt{5}$is rational number$ \rightarrow $ so, we can write this answer as$\Rightarrow$$ 3+2\sqrt{5} = \dfrac ab$ Here $a$ & $b$ use two coprime number and $ b \neq 0.$$\Rightarrow$$ 2\sqrt{5} = \dfrac{a}{b}-3 $$\Rightarrow$$ 2\sqrt{5} = \dfrac{a-3b}{b}$$ \therefore \sqrt{5} = \dfrac{a-3b}{2b}$Here $a$ and $b$ are integer so $ \dfrac{a-3b}{2b}$ is a rational number so $ \sqrt{5}$ should be rational number but$ \sqrt{5}$ is a irrational number so it is contradict- Hence $ 3+2\sqrt{5}$ is irrational.

Question 8

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

Accepted Answer

Let us assume, to the contrary, that $3 \sqrt {2}$ is rational. Then, there exist co-prime positive integers $a$ and $b$ such that$3\sqrt { 2 } =\cfrac { a }{ b } $$\Rightarrow$ $\sqrt { 2 } =\cfrac { a }{ 3b } $$\Rightarrow $ $\sqrt {2}$ is rational       ...$[$$\because 3,a$ and $b$ are integers$\therefore \cfrac { a }{ 3b }$is a rational number$]$This contradicts the fact that $\sqrt {2}$ is irrational. So, our assumption is not correct.Hence, $3 \sqrt {2}$ is an irrational number.

Question 9

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

Accepted Answer

(i) $\dfrac{1}{\sqrt{2}}$Let us assume $\dfrac{1}{\sqrt{2}}$ is rational.So we can write this number as$\dfrac{1}{\sqrt{2}} = \dfrac{a}{b}$ ---- (1)Here, $a$ and $b$ are two co-prime numbers and $b$ is not equal to zero.Simplify the equation (1) multiply by $\sqrt{2}$ both sides, we get$1 = \dfrac{a\sqrt{2}}{b}$Now, divide by $b$, we get$b = a\sqrt{2}$ or $\dfrac{b}{a}= \sqrt{2}$Here, $a$ and $b$ are integers so, $\dfrac{b}{a}$ is a rational number, so $\sqrt{2}$ should be a rational number.But $\sqrt{2}$ is a irrational number, so it is contradictory.Therefore, $\dfrac{1}{\sqrt{2}}$ is irrational number.(ii) $7\sqrt{5}$Let us assume $7\sqrt{5}$ is rational.So, we can write this number as$7\sqrt{5} = \dfrac{a}{b}$ ---- (1)Here, $a$ and $b$ are two co-prime numbers and $b$ is not equal to zero.Simplify the equation (1) divide by 7 both sides, we get$\sqrt{5} = \dfrac{a}{7b}$Here, $a$ and $b$ are integers, so $\dfrac{a}{7b}$ is a rational number, so $\sqrt{5}$ should be a rational number.But $\sqrt{5}$ is a irrational number, so it is contradictory.Therefore, $7\sqrt{5}$ is irrational number.(iii) $6+\sqrt{2}$Let us assume $6+\sqrt{2}$ is rational.So we can write this number as$6+\sqrt{2} = \dfrac{a}{b}$ ---- (1)Here, $a$ and $b$ are two co-prime number and $b$ is not equal to zero.Simplify the equation (1) subtract 6 on both sides, we get$\sqrt{2} = \dfrac{a}{b}-6$$\sqrt{2} = \dfrac{a-6b}{b}$Here, $a$ and $b$ are integers so, $\dfrac{a-6b}{b}$ is a rational number, so $\sqrt{2}$ should be a rational number.But $\sqrt{2}$ is a irrational number, so it is contradictory.Therefore, $6+\sqrt{2}$ is irrational number.

Question 10

Prove that $5 -\sqrt 3 $ is irrational

Accepted Answer

Let us assume the given number be rational and we will write the given number in p/q form$\Rightarrow 5-\sqrt {3}=\dfrac {p} {q} $$\Rightarrow \sqrt {3}=\dfrac {5q-p}{q}$We observe that LHS is irrational and RHS is rational, which is not possible. This is contradiction. Hence our assumption that given number is rational is false $\Rightarrow 5-\sqrt {3}$  is irrational

Square Root of Prime Number is Irrational

Language

Rate

REVISE WITH CONCEPTS

Proof of the Irrationality of Sqrt (2) and Other Surds

QUICK SUMMARY WITH STORIES

The Irrationality Of √2

The Irrationality of 5-√3.

Theorem on Prime Numbers

Prove that $3$ is an irrational number.

Prove that $5$ is irrational by the method of Contradiction.

Prove that $2$ is an irrational number.

Prove that $3 + 5$ is irrational.

Prove that $7$ is an irrational number.

Prove that $2 + 3$ is irrational

Prove that $3 + 25$ is irrational.

Prove that $32$ is irrational.

Prove that the following are irrational:
(i) $\frac{1}{2}$ (ii) $75$ (iii) $6 + 2$

Prove that $5 - 3$ is irrational

Square Root of Prime Number is Irrational

Language

Rate

REVISE WITH CONCEPTS

Proof of the Irrationality of Sqrt (2) and Other Surds

QUICK SUMMARY WITH STORIES

The Irrationality Of √2

The Irrationality of 5-√3.

Theorem on Prime Numbers

Prove that 3​ is an irrational number.

Prove that 5​ is irrational by the method of Contradiction.

Prove that 2​ is an irrational number.

Prove that 3​+5​ is irrational.

Prove that 7​ is an irrational number.

Prove that 2​+3​ is irrational

Prove that 3+25​ is irrational.

Prove that 32​ is irrational.

Prove that the following are irrational: (i) 2​1​ (ii) 75​ (iii) 6+2​

Prove that 5−3​ is irrational