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 {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, if x and y are odd positive integers, then \$x^2 + y^2\$ is even but not divisible by 4.

We know that any odd positive integer is of the form \$2q + 1\$, where \$q\$ is an integer. So, let \$x = 2m + 1\$ and \$y = 2n + 1\$, for some integers m and n. we have \$x^2 + y^2\$\$x^2 + y^2 = (2m + 1)^2 + (2n + 1)^2\$\$x^2 + y^2 = 4m^2 + 1 + 4m + 4n^2 + 1 + 4n = 4m^2 + 4n^2 + 4m + 4n + 2\$\$x^2+ y^2 = 4(m^2 + n^2) + 4(m + n) + 2 = 4\{(m^2 + n^2) + (m + n)\}+ 2\$\$x^2 + y2 = 4q + 2\$, when \$q = (m^2 + n^2)+(m + n)\$\$x^2 + y^2\$ is even and leaves remainder 2 when divided by 4. \$x^2 + y^2\$ is even but not divisible by 4.

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 {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 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.

Q: The decimal expansion of the rational number \$\displaystyle\frac{23}{2^{3}5^{2}}\$, will terminate after how many places of decimal?

\$\displaystyle\frac{23}{2^{3}5^{2}}\$ = \$\displaystyle\frac{23}{2(2)^{2}(5)^{2}}\$ = \$\displaystyle\frac{23}{2(2\times5)^{2}}\$ = \$\displaystyle\frac{23}{2(10)^{2}}\$ = \$\displaystyle\frac{11.5}{100} = 0.115\$Therefore, the decimal expansion of the rational number \$\displaystyle\frac{23}{2^{3}5^{2}}\$will terminate after three places of decimal.

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 $\sqrt {2}$ 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 3

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 4

In a seminar, the number of participants in Hindi, English and Mathematics are $60,\ 84$ and $108$ respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject.

Accepted Answer

The Number of room will be minimum if each room accomodates maximum number of participants. Since in each room the same number of participants are to be seated and all of them must be of the same subject. Therefore, the number of participants in each room must be the $HCF$ of $60,84$ and $108$. The prime factorisations of $60,84$ and $108$ are as under.$60={2}^{2}\times 3\times 5,84={2}^{2}\times 3\times 7$ and $108={2}^{2}\times {3}^{3}$$\therefore\ HCF$ of $60,84$ and $108$ is ${2}^{2}\times 3=12$Therefore, in each room $12$ participants can be seated.$\therefore\ Number\ of\ rooms\ required =\dfrac { Total\ number\ of\ participants  }{ 12 }$                                                          $=\dfrac { 60+84+108 }{ 12 } =\dfrac { 252 }{ 12 } =21$

Question 5

Prove that, if x and y are odd positive integers, then $x^2 + y^2$ is even but not divisible by 4.

Accepted Answer

We know that any odd positive integer is of the form $2q + 1$, where $q$ is an integer. So, let $x = 2m + 1$ and $y = 2n + 1$, for some integers m and n. we have $x^2 + y^2$$x^2 + y^2 = (2m + 1)^2 + (2n + 1)^2$$x^2 + y^2 = 4m^2 + 1 + 4m + 4n^2 + 1 + 4n = 4m^2 + 4n^2 + 4m + 4n + 2$$x^2+ y^2 = 4(m^2 + n^2) + 4(m + n) + 2 = 4\{(m^2 + n^2) + (m + n)\}+ 2$$x^2 + y2 = 4q + 2$, when $q = (m^2 + n^2)+(m + n)$$x^2 + y^2$ is even and leaves remainder 2 when divided by 4. $x^2 + y^2$ is even but not divisible by 4.

Question 6

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 7

Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:(i) $\dfrac{13}{3125}$ (ii) $\dfrac{17}{8}$ (iii) $\dfrac{64}{455}$ (iv) $\dfrac{15}{1600}$ (v) $\dfrac{29}{343}$ (vi) $\dfrac{23}{2^{3}5^{2}}$ (vii) $\dfrac{129}{2^{2}5^{7}7^{5}}$ (viii) $\dfrac{6}{15}$ (ix) $\dfrac{35}{50}$ (x) $\dfrac{77}{210}$

Accepted Answer

Theorem: Let $x = \dfrac{p}{q}$ be a rational number, such that the prime factorisation of q is of the form $2^{n}5^{m}$, where n, m are non-negative integers. Then, $x$ has a decimal expansion which terminates.(i) $\dfrac{13}{3125}$Factorise the denominator, we get$3125 = 5 \times 5  \times 5  \times 5  \times 5 = 5^5$So, denominator is in form of $5^m$ so, $\dfrac{13}{3125}$ is terminating.(ii) $\dfrac{17}{8}$Factorise the denominator, we get$8 = 2  \times 2  \times 2 = 2^3$So, denominator is in form of $2^n$ so, $\dfrac{17}{8}$ is terminating.(iii) $\dfrac{64}{455}$Factorise the denominator, we get$455 = 5  \times 7  \times 13 $So, denominator is not in form of $2^{n} 5^{m}$ so, $\dfrac{64}{455}$ is not terminating.(iv) $\dfrac{15}{1600}$Factorise the denominator, we get$1600 = 2  \times 2  \times 2  \times 2  \times 2  \times 2  \times 5  \times 5 = 2^{6} 5^{2}$So, denominator is in form of $2^{n}5^{m}$ so, $\dfrac{15}{1600}$ is terminating.(v) $\dfrac{29}{343}$Factorise the denominator, we get$343 = 7  \times 7  \times 7 = 7^{3}$So, denominator is not in form of $2^{n}5^{m}$ so, $\dfrac{29}{343}$ is not terminating.(vi) $\dfrac{23}{2^{3}5^{2}}$Here, the denominator is in form of $2^{n}5^{m}$ so, $\dfrac{23}{2^{3}5^{2}}$ is terminating.(vii) $\dfrac{129}{2^{2}5^{7}7^{5}}$Here, the denominator is not in form of $2^{n}5^{m}$ so, $\dfrac{129}{2^{2}5^{7}7^{5}}$ is not terminating.(viii) $\dfrac{6}{15}$Divide nominator and denominator both by 3 we get $\dfrac{3}{15}$So, denominator is in form of $ 5^{m}$ so, $\dfrac{6}{15}$ is terminating.(ix) $\dfrac{35}{50}$ Divide nominator and denominator both by 5 we get $\dfrac{7}{10}$Factorise the denominator, we get$10 = 2 \times 5$So, denominator is in form of $2^{n}5^{m}$ so, $\dfrac{35}{50}$ is terminating.(x) $\dfrac{77}{210}$Divide nominator and denominator both by 7 we get $\dfrac{11}{30}$Factorise the denominator, we get$30 = 2 \times 3 \times 5$So, denominator is not in the form of $2^{n}5^{m}$ so $\dfrac{6}{15}$ is not terminating.

Question 8

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 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

The decimal expansion of the rational number $\displaystyle\frac{23}{2^{3}5^{2}}$, will terminate after how many places of decimal?

Accepted Answer

$\displaystyle\frac{23}{2^{3}5^{2}}$  = $\displaystyle\frac{23}{2(2)^{2}(5)^{2}}$ = $\displaystyle\frac{23}{2(2\times5)^{2}}$ =  $\displaystyle\frac{23}{2(10)^{2}}$ =  $\displaystyle\frac{11.5}{100} = 0.115$Therefore, the decimal  expansion of the rational number $\displaystyle\frac{23}{2^{3}5^{2}}$will terminate after three places of decimal.