Solution:

 \(\displaystyle 0.555\)

\(\displaystyle \frac{\cancelto{111}{555}}{\cancelto{200}{1000}}\)

= \(\displaystyle \frac {111}{200}\)

Solution:

Let, x = \(\displaystyle 29.\overline{568}\)

∴ 1000x = 29568.568568 ...

∴ 1000x = \(\displaystyle 29568.\overline{568}\)

∴ 1000x − x = \(\displaystyle 29568.\overline{568}\) − \(\displaystyle 29.\overline{568}\)

∴ 999x = 29539

∴ \(\displaystyle x = \frac {29539}{999}\)

∴ \(\displaystyle 29.\overline{568}\) = \(\displaystyle \frac {29539}{999}\)

Solution:

Let, x = \(\displaystyle 9.315315 ...\)

∴ 1000x = \(\displaystyle 9315.315315 ...\)

∴ 1000x − x = \(\displaystyle 9315.315315 ...\) − \(\displaystyle 9.315315 ...\)

∴ 999x = 9306

∴ \(\displaystyle x = \frac{\cancelto{1034}{9306}}{\cancelto{111}{999}}\)

∴ \(\displaystyle x = \frac {1034}{111}\)

∴ \(\displaystyle 9.315315 ...\) = \(\displaystyle \frac {1034}{111}\)

Solution:

Let, x = \(\displaystyle 357.417417 ...\)

∴ 1000x = \(\displaystyle 357417.417417 ...\)

∴ 1000x − x = \(\displaystyle 357417.417417 ...\) − \(\displaystyle 357.417417 ...\)

∴ 999x = 357060

∴ \(\displaystyle x = \frac{\cancelto{119020}{357060}}{\cancelto{333}{999}}\)

∴ \(\displaystyle x = \frac {119020}{333}\)

∴ \(\displaystyle 357.417417 ...\) = \(\displaystyle \frac {119020}{333}\)

Solution:

Let, x = \(\displaystyle 30.\overline{219}\)

∴ 1000x = \(\displaystyle 30219.\overline{219}\)

∴ 1000x − x = \(\displaystyle 30219.\overline{219} ...\) − \(\displaystyle 30.\overline{219}\)

∴ 999x = 30189

∴ \(\displaystyle x = \frac{\cancelto{10063}{30189}}{\cancelto{333}{999}}\)

∴ \(\displaystyle x = \frac {10063}{333}\)

∴ \(\displaystyle 30.\overline{219}\) = \(\displaystyle \frac {10063}{333}\)

Answer:

 \(\displaystyle \frac {-\:5}{7} = -\:0.\overline{714285}\)

Answer:

 \(\displaystyle \frac {9}{11} = 0.\overline{81}\)

Answer:

 \(\displaystyle \sqrt {5} = 2.2360679\:...\)

Answer:

 \(\displaystyle \frac {121}{13} = 9.\overline{307692}\)

Answer:

 \(\displaystyle \frac {29}{8} = 3.625\)

Proof:

Let’s assume that \(\displaystyle 5 + \sqrt {7}\) is a rational number.

A rational number can be expressed in the form \(\displaystyle \frac {p}{q}\).

∴ \(\displaystyle 5 + \sqrt {7}\) = \(\displaystyle \frac {p}{q}\) ... (say)

(Here, p, q ∈ I and q ≠ 0)

∴ \(\displaystyle \sqrt {7}\) = \(\displaystyle \frac {p}{q} - 5\) ... (i)

Now, \(\displaystyle \frac {p}{q}\) is a rational number.

∴ \(\displaystyle \frac {p}{q} - 5\) is also a rational number.

Now, LHS of equation (i) is an irrational number while the RHS is a rational number.

But, this is contradictory.

∴ Our assumption that \(\displaystyle 5 + \sqrt {7}\) is a rational number is incorrect.

∴ \(\displaystyle 5 + \sqrt {7}\) is an irrational number.

Solution:

 \(\displaystyle \frac {3}{4}\:\sqrt {8}\)

= \(\displaystyle \frac {3}{4}\times \sqrt {4 \times 2}\)

= \(\displaystyle \frac {3}{4}\times \sqrt {4} \times \sqrt {2}\)

= \(\displaystyle \frac {3}{4} \times 2 \times \sqrt {2}\)

= \(\displaystyle \frac {3}{2}\:\sqrt {2}\)

Solution:

 \(\displaystyle -\:\frac {5}{9}\:\sqrt {45}\)

= \(\displaystyle -\:\frac {5}{9}\times \sqrt {9 \times 5}\)

= \(\displaystyle -\:\frac {5}{9}\times \sqrt {9} \times \sqrt {5}\)

= \(\displaystyle -\:\frac {5}{9}\times 3 \times \sqrt {5}\)

= \(\displaystyle -\:\frac {5}{3}\:\sqrt {5}\)

Solution:

 \(\displaystyle \sqrt {32} \times \sqrt {2}\)

= \(\displaystyle \sqrt {64}\)

= \(\displaystyle 8\), which is a rational number.

∴ The simplest rationalising factor of \(\displaystyle \sqrt {32}\) is \(\displaystyle \sqrt {2}\).

Solution:

 \(\displaystyle \sqrt {50} \times \sqrt {2}\)

= \(\displaystyle \sqrt {100}\)

= \(\displaystyle 10\), which is a rational number.

∴ The simplest rationalising factor of \(\displaystyle \sqrt {50}\) is \(\displaystyle \sqrt {2}\).

Solution:

 \(\displaystyle \sqrt {27} \times \sqrt {3}\)

= \(\displaystyle \sqrt {81}\)

= \(\displaystyle 9\), which is a rational number.

∴ The simplest rationalising factor of \(\displaystyle \sqrt {27}\) is \(\displaystyle \sqrt {3}\).

Solution:

 \(\displaystyle \frac {3}{5}\:\sqrt {10} \times \sqrt {10}\)

= \(\displaystyle \frac {3}{5}\:\sqrt {100}\)

= \(\displaystyle \frac {3}{5} \times 10\)

= \(\displaystyle 6\), which is a rational number.

∴ The simplest rationalising factor of \(\displaystyle \frac {3}{5}\:\sqrt {10}\) is \(\displaystyle \sqrt {10}\).

Solution:

 \(\displaystyle 3\:\sqrt {72} \times \sqrt {2}\)

= \(\displaystyle 3\:\sqrt {144}\)

= \(\displaystyle 3 \times 12\)

= \(\displaystyle 36\), which is a rational number.

∴ The simplest rationalising factor of \(\displaystyle 3\:\sqrt {72}\) is \(\displaystyle \sqrt {2}\).

Solution:

 \(\displaystyle 4\:\sqrt {11} \times \sqrt {11}\)

= \(\displaystyle 4\:\sqrt {121}\)

= \(\displaystyle 4 \times 11\)

= \(\displaystyle 44\), which is a rational number.

∴ The simplest rationalising factor of \(\displaystyle 4\:\sqrt {11}\) is \(\displaystyle \sqrt {11}\).

Solution:

 \(\displaystyle \frac {4}{7}\:\sqrt {147} + \frac {3}{8}\:\sqrt {192} - \frac {1}{5}\:\sqrt {75}\)

= \(\displaystyle \frac {4}{7}\:\sqrt {49 \times 3} + \frac {3}{8}\:\sqrt {64 \times 3} - \frac {1}{5}\:\sqrt {25 \times 3}\)

= \(\displaystyle \frac {4}{7}\:\sqrt {49} \times \sqrt {3} + \frac {3}{8}\:\sqrt {64} \times \sqrt {3} - \frac {1}{5}\:\sqrt {25} \times \sqrt {3}\)

= \(\displaystyle \frac {4}{7} \times 7\:\sqrt {3} + \frac {3}{8} \times 8\:\sqrt {3} - \frac {1}{5} \times 5\:\sqrt {3}\)

= \(\displaystyle 4\:\sqrt {3} + 3\:\sqrt {3} - \sqrt {3}\)

= \(\displaystyle (4 + 3 - 1)\:\sqrt {3}\)

= \(\displaystyle 6\:\sqrt {3}\)

Solution:

 \(\displaystyle 5\:\sqrt {3} + 2\:\sqrt {27} + \frac {1}{\sqrt {3}}\)

= \(\displaystyle 5\:\sqrt {3} + 2\:\sqrt {9 \times 3} + \frac {1}{\sqrt {3}}\)

= \(\displaystyle 5\:\sqrt {3} + 2\:\sqrt {9} \times \sqrt {3} + \frac {1}{\sqrt {3}}\)

= \(\displaystyle 5\:\sqrt {3} + 2 \times 3\:\sqrt {3} + \frac {1}{\sqrt {3}}\)

= \(\displaystyle 5\:\sqrt {3} + 6\:\sqrt {3} + \frac {1}{\sqrt {3}}\)

= \(\displaystyle 11\:\sqrt {3} + \frac {1}{\sqrt {3}}\)

= \(\displaystyle \frac {11\:\sqrt {3} \times \sqrt {3} + 1}{\sqrt {3}}\)... (Taking L.C.M.)

= \(\displaystyle \frac {11 \times 3 + 1}{\sqrt {3}}\)

= \(\displaystyle \frac {33 + 1}{\sqrt {3}}\)

= \(\displaystyle \frac {34}{\sqrt {3}}\)

Now, we will rationalize the denominator.
Multiplying the numerator and denominator by \(\displaystyle \sqrt {3}\), we get:

= \(\displaystyle \frac {34}{\sqrt {3}} \times \frac {\sqrt {3}}{\sqrt {3}}\)

= \(\displaystyle \frac {34}{3}\: \sqrt {3}\)

Solution:

 \(\displaystyle \sqrt {216} - \:5\:\sqrt {6} + \sqrt {294}\: - \:\frac {3}{\sqrt {6}}\)

= \(\displaystyle \sqrt {36 \times 6} - \:5\:\sqrt {6} + \sqrt {49 \times 6}\: - \:\frac {3}{\sqrt {6}}\)

= \(\displaystyle \sqrt {36} \times \sqrt {6} - \:5\:\sqrt {6} + \sqrt {49} \times \sqrt {6}\: - \:\frac {3}{\sqrt {6}}\)

= \(\displaystyle 6\:\sqrt {6}\:- \:5\:\sqrt {6} + 7\:\sqrt {6}\: - \:\frac {3}{\sqrt {6}}\)

= \(\displaystyle (6 - 5 +7)\:\sqrt {6}\: - \:\frac {3}{\sqrt {6}}\)

= \(\displaystyle 8\:\sqrt {6}\: - \:\frac {3}{\sqrt {6}}\)

Now, we will rationalize the denominator of the second term.

 \(\displaystyle 8\:\sqrt {6}\:- \:\frac {3}{\sqrt {6}} \times \frac {\sqrt {6}}{\sqrt {6}}\)

= \(\displaystyle 8\:\sqrt {6}\: - \:\frac {3 \times \sqrt {6}}{6}\)

= \(\displaystyle 8\:\sqrt {6}\:- \:\frac {1}{2}\:\sqrt {6}\)

= \(\displaystyle \left[\:8 - \:\frac {1}{2}\:\right]\:\sqrt {6}\)

= \(\displaystyle \left[\:\frac {16 - 1}{2}\:\right]\:\sqrt {6}\) ... (Taking L.C.M.)

= \(\displaystyle \frac {15}{2}\:\sqrt {6}\)

Solution:

 \(\displaystyle 4\:\sqrt {12} - \:\sqrt {75} - \: 7\sqrt {48}\)

= \(\displaystyle 4\:\sqrt {4 \times 3} - \:\sqrt {25 \times 3} - \: 7\sqrt {16 \times 3}\)

= \(\displaystyle 4\:\times \sqrt {4} \times \sqrt {3} - \:\sqrt {25} \times \sqrt {3} - \: 7\sqrt {16} \times \sqrt {3}\)

= \(\displaystyle 4 \times 2\:\sqrt {3} - 5\:\sqrt {3} - \: 7 \times 4\:\sqrt {3}\)

= \(\displaystyle 8\:\sqrt {3} - 5\:\sqrt {3} - 28\:\sqrt {3}\)

= \(\displaystyle (8 - 5 - 28)\:\sqrt {3}\)

= \(\displaystyle -\:25\:\sqrt {3}\)

Solution:

 \(\displaystyle 2\:\sqrt {48}\: - \:\sqrt {75} \: - \:\frac {1}{\sqrt {3}}\)

= \(\displaystyle 2\:\sqrt {16 \times 3} - \:\sqrt {25 \times 3} - \:\frac {1}{\sqrt {3}}\)

= \(\displaystyle 2\:\sqrt {16} \times \:\sqrt {3} - \:\sqrt {25} \times \:\sqrt {3} - \:\frac {1}{\sqrt {3}}\)

= \(\displaystyle 2 \times 4\:\sqrt {3} - 5\:\sqrt {3} - \:\frac {1}{\sqrt {3}}\)

= \(\displaystyle 8\:\sqrt {3}\:-\:5\:\sqrt {3}\: - \:\frac {1}{\sqrt {3}}\)

= \(\displaystyle 3\:\sqrt {3}\:- \:\frac {1}{\sqrt {3}}\)

Now, we will rationalize the denominator of the second term.

 \(\displaystyle 3\:\sqrt {3} - \:\frac {1}{\sqrt {3}} \times \frac {\sqrt {3}}{\sqrt {3}}\)

= \(\displaystyle 3\:\sqrt {3} - \:\frac {\sqrt {3}}{3}\)

= \(\displaystyle \left[\:3 - \:\frac {1}{3}\:\right]\:\sqrt {3}\)

= \(\displaystyle \left[\:\frac {9 - 1}{3}\:\right]\:\sqrt {3}\) ... (Taking L.C.M.)

= \(\displaystyle \frac {8}{3}\:\sqrt {3}\)

Solution:

 \(\displaystyle \frac {1}{\sqrt {5}}\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt {5}\), we get:

 \(\displaystyle \frac {1}{\sqrt {5}} \times \frac {\sqrt {5}}{\sqrt {5}}\)

= \(\displaystyle \frac {\sqrt {5}}{\sqrt {25}}\)

= \(\displaystyle \frac {\sqrt {5}}{5}\)

Solution:

 \(\displaystyle \frac {2}{3\:\sqrt {7}}\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt {7}\), we get:

 \(\displaystyle \frac {2}{3\:\sqrt {7}} \times \frac {\sqrt {7}}{\sqrt {7}}\)

= \(\displaystyle \frac {2\:\sqrt {7}}{3\:\sqrt {49}}\)

= \(\displaystyle \frac {2\:\sqrt {7}}{3 \times 7}\)

= \(\displaystyle \frac {2\:\sqrt {7}}{21}\)

Solution:

 \(\displaystyle \frac { 1 }{ \sqrt { 3 } - \sqrt { 2 }}\)

Multiplying the numerator and denominator by \(\displaystyle \sqrt { 3 } + \sqrt { 2 }\),

 \(\displaystyle \frac { 1 }{ \sqrt { 3 } - \sqrt { 2 }} \times \displaystyle \frac {\sqrt { 3 } + \sqrt { 2 }} {\sqrt { 3 } + \sqrt { 2 }}\)

= \(\displaystyle \frac { 1(\sqrt { 3 } + \sqrt { 2 })}{(\sqrt { 3 } - \sqrt { 2 }){(\sqrt { 3 } + \sqrt { 2 })} }\)

= \(\displaystyle \frac { \sqrt { 3 } + \sqrt { 2 }}{(\sqrt { 3 })^2 - {(\sqrt { 2 })^2} }\) ... [∵ (a − b)(a + b) = a² − b² ]

= \(\displaystyle \frac { \sqrt { 3 } + \sqrt { 2 }}{3 - 2}\)

= \(\displaystyle \frac { \sqrt { 3 } + \sqrt { 2 }}{1}\)

= \(\displaystyle \sqrt { 3 } + \sqrt { 2 }\)

Solution:

 \(\displaystyle \frac { 1 }{ 3 \sqrt { 5 } + 2 \sqrt { 2 }}\)

Multiplying the numerator and denominator by \(\displaystyle 3 \sqrt { 5 } - 2 \sqrt { 2 }\),

 \(\displaystyle \frac { 1 }{ 3 \sqrt { 5 } + 2 \sqrt { 2 }} \times \frac {3 \sqrt { 5 } - 2 \sqrt { 2 }}{3 \sqrt { 5 } - 2 \sqrt { 2 }}\)

= \(\displaystyle \frac { 1(3 \sqrt { 5 } - 2 \sqrt { 2 })}{(3 \sqrt { 5 } + 2 \sqrt { 2 }) {(3 \sqrt { 5 } - 2 \sqrt { 2 })} }\)

= \(\displaystyle \frac { 3 \sqrt { 5 } - 2 \sqrt { 2 }}{(3 \sqrt { 5 })^2 - {(2 \sqrt { 2 })^2} }\) ... [∵ (a − b)(a + b) = a² − b² ]

= \(\displaystyle \frac { 3 \sqrt { 5 } - 2 \sqrt { 2 }}{9 \times 5 - 4 \times 2}\)

= \(\displaystyle \frac { 3 \sqrt { 5 } - 2 \sqrt { 2 }}{45 - 8}\)

= \(\displaystyle \frac { 3 \sqrt { 5 } - 2 \sqrt { 2 }}{37}\)

Solution:

 \(\displaystyle \frac { 12 }{ 4 \sqrt { 3 } - \sqrt { 2 }}\)

Multiplying the numerator and denominator by \(\displaystyle 4 \sqrt { 3 } + \sqrt { 2 }\),

 \(\displaystyle \frac { 12 }{ 4 \sqrt { 3 } - \sqrt { 2 }} \times \frac {4\sqrt { 3 } + \sqrt { 2 }}{4\sqrt { 3 } + \sqrt { 2 }}\)

= \(\displaystyle \frac { 12(4 \sqrt { 3 } + \sqrt { 2 })}{(4 \sqrt { 3 } - \sqrt { 2 }) {(4 \sqrt { 3 } + \sqrt { 2 })}}\)

= \(\displaystyle \frac { 12(4 \sqrt { 3 } + \sqrt { 2 })}{(4 \sqrt { 3 })^2 - {(\sqrt { 2 })^2} }\) ... [∵ (a − b)(a + b) = a² − b² ]

= \(\displaystyle \frac { 12(4 \sqrt { 3 } + \sqrt { 2 })}{16 \times 3 - 2}\)

= \(\displaystyle \frac { 12(4 \sqrt { 3 } + \sqrt { 2 })}{48 - 2}\)

= \(\displaystyle \frac { 12(4 \sqrt { 3 } + \sqrt { 2 })}{46}\)

= \(\displaystyle \frac { 6(4 \sqrt { 3 } + \sqrt { 2 })}{23}\)