Number System and Simplification

Now,

(a) Let us consider the first set of fraction $\dfrac { 7 }{ 10 } ,\dfrac { 3 }{ 4 }$ and another given fraction $\dfrac { 3 }{ 5 }$ 

(a) Let us consider the first set of fraction $-\dfrac { 1 }{ 2 } ,\dfrac { 5 }{ 6 } ,-\dfrac { 4 }{ 9 }$ 

The rational number $\dfrac {4}{-5}$ is same as $\dfrac {-4}{5}$.

This question is very easy if we solve it by verification
process.

Take (A) i.e. $\displaystyle\frac{-2}{3}\,<\,\frac{4}{-9}\, \frac{-5}{12}\, <\,\frac{7}{-18}$

First take \displaystyle\frac{-2}{3},\,\frac{-4}{9}$

$-2\,\times\,9,\, 4\,\times\,3$

-18, -12

$\because\, -12\,>\,-18$

So, $\displaystyle\frac{-4}{9}, \frac{-5}{12}$

$- 4\,\times\, 12, \, - 5\,\times\, 9$

- 48, - 45

$\because\, -45\, >\, -48$

So, $\displaystyle\frac{-5}{12}\,>\, \frac{-4}{9},\,i.e.,\frac{-4}{9}\, <\, \frac{-5}{12}$

Finally, $\displaystyle\frac{-5}{12}, \frac{-7}{18}$

$5\,\times\, 18, \, -7\,\times\, 12$

- 90, - 84

$\because\, -84\, >\, -90$

So, $\displaystyle\frac{-7}{18}\, >\, \frac{-5}{12}, i.e., \frac{-5}{12}\, <\, \frac{-7}{18}$

$\therefore\, \displaystyle\frac{-2}{3}\, <\, \frac{-4}{9}\,<\,  \frac{-5}{12}\,<\, \frac{-7}{18}$

You can identify the answer by observing the question by practicing this method.

By verification process, $A\,\rightarrow\, \displaystyle\frac{3}{4},\, \frac{-2}{1}, \, \frac{-11}{20},\, \frac{-4}{5}$

$3\,\times\, 1,\, -2\,\times\, 4$

3, - 8

Correct, $-2\,\times\, 20, \, -11\,\times\, 1$

- 40, - 11

Wrong.

$B\, \rightarrow\, \displaystyle \frac {3}{4},\, \frac{-11}{20}, \frac{-4}{5},\, \frac{-2}{1}$

$3\,\times\, 20, \, -11\,\times\, 4$

60, - 44

$\displaystyle\frac{3}{4}\, >\,\frac{-11}{20}$ $-11\,\times\, 5,\, -4\,\times\, 20$

- 55, - 80

$\because\, \displaystyle\frac{-11}{20}\, >\, \frac{-4}{5}$

$-4\,\times\, 1,\, -2\,\times\, 5$

- 4, - 10

$\because\, \displaystyle\frac{-4}{5}\, >\, -2$

When dividing an inequality by a negative number, the inequality sign must be reversed. Since 27 > 18, dividing both sides by -9 results in 27/-9 < 18/-9, which is -3 < -2.

An equivalent fraction is found by multiplying the numerator and denominator by the same number. 9/11 multiplied by 26/26 equals 234/286.

To compare fractions, make the denominators equal

$\displaystyle 49^{A}= 49^{1-\log _{7}2}= \frac{49}{49^{\log _{7}2}}= \frac{49}{7^{\log _{7}4}}= \frac{49}{4}\quad [\because a^{\log _ab}=b]$
$\displaystyle 5^{B}= 5^{-\log _{5}4} \quad [\because m\log a=\log a^m]=5^{\log _5\dfrac{1}{4}}= \frac{1}{4}$
$\therefore 49^A+5^B = \cfrac{25}{2}$

We know that,

${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$

Now, let,

$ -2ab=-i $

$ ab=i $

Now, consider $\dfrac{-17}{144}$. We can write it as,

$\dfrac{-17}{144}=\dfrac{64-81}{9\times 16}=\dfrac{4}{9}-\dfrac{9}{16}$

Thus,

$ \sqrt{\dfrac{-17}{144}-i}=\sqrt{\dfrac{4}{9}-\dfrac{9}{16}-i} $

$ =\sqrt{{{\left( \dfrac{2}{3} \right)}^{2}}+{{\left( i \right)}^{2}}{{\left( \dfrac{3}{4} \right)}^{2}}-2\times \left( \dfrac{2}{3} \right)\times \left( \dfrac{3i}{4} \right)} $

$ =\sqrt{{{\left( \dfrac{2}{3}-\dfrac{3i}{4} \right)}^{2}}} $

$ =\pm \left( \dfrac{2}{3}-\dfrac{3i}{4} \right) $

Hence, this is the required result.

$\displaystyle \frac{1\, +\, \sqrt{0.01}}{1\, -\, \sqrt{0.1}}\, =\, \displaystyle \frac{1\, +\, 0.1}{1\, -\, 0.32}\, =\, \displaystyle \frac{1.1}{0.68}\, =\, 1.6$