Number Systems Test - 16

Since, real numbers are the combination of rational and irrational numbers.
Hence, every rational number is a real number.

Between two rational numbers there are infinitely many rational number for example between 4 and 5 there are 4.1,4.2 .4.22,4.223............

Decimal representation of a rational number cannot be non-terminating non-repeating because the decimal expansion of rational number is either terminating or non-terminating recurring (repeating).

We know that, the product of any two irrational numbers is sometimes rational and sometimes irrational.

e.g., √2 x √2 = 2 (rational) and √2 x √3 = √6 (irrational)

The decimal expansion of the number √2 is non-terminating non-recurring. Because √2 is an irrational number.

Also, we know that an irrational number is non-terminating non-recurring.

\({\sqrt{ 4\over9}}\) = \(2\over3\) (rational)

\(\sqrt{12}\over\sqrt3\) = 2 (rational)

\(\sqrt{81}\) = 9 (rational)

but \(\sqrt7\)  is an irrational number.

Hence, \(\sqrt7\) is an irrational number.

An irrational number is non-terminating non-recurring which is 0.4014001400014….

Here, 0.14 is terminating and \(0.14\overline{16}\), \(0.\overline{1416}\) are non-terminating recurring.

We know that

\(\sqrt2\) = 1.4142135...... and \(\sqrt3\) = 1.732050807

We see that 1.5 is a rational number which lies between 1.4142135….. and 1.732050807….

let x = 1.999...

Now, 10x = 19.999....

On subtracting Eq.(i) from Eq.(ii), we get

10x - x = (19.999....) - (1.9999....)

9x = 18

\(x = {18\over9}\)

x = 2

\(2\sqrt3\, + \sqrt3\) = \((2+1)\sqrt3\) = \(3\sqrt3\)

We have

\(\sqrt{10} \times \sqrt{15}\)

= \({\sqrt{10 \times15}}\)

= \({\sqrt {5 \times 2 \times 5 \times 3}}\)

= \(5\sqrt6\)

\(1\over {\sqrt7 - 2}\) 

= \(1\over {\sqrt7 - 2}\) x \(\sqrt7 + 2\over \sqrt7 + 2\)

= \(\sqrt7 + 2\over7 - 4\)

= \(\sqrt7 + 2\over3\)

\(1\over\sqrt9-\sqrt8\\\)

= \(1\over\sqrt{3 \times3} - \sqrt{4 \times2}\)

= \(1\over3-2\sqrt2\)

= \(1\over3-2\sqrt2\) x \(3+2\sqrt2\over3+2\sqrt2\)

= \(3+2\sqrt2\over(3)^2-(2\sqrt2)^2\)

= \(3+2\sqrt2\over9-8\)

= \(3+2\sqrt2\over1\)

= \(3+2\sqrt2\)

\(7\over3\sqrt3-2\sqrt2\)

= \(7\over3\sqrt3-2\sqrt2\) x \(3\sqrt3+2\sqrt2\over3\sqrt3+2\sqrt2\)

= \(7(3\sqrt3+2\sqrt2)\over(3\sqrt3)^2 - (2\sqrt2)^2\)

= \(7(3\sqrt3+2\sqrt2)\over27-8\)

= \(7(3\sqrt3+2\sqrt2)\over19\)

Therefore, we get the denominator as 19.

\((\sqrt32+\sqrt48)\over(\sqrt8+\sqrt12)\)

= \(\sqrt{16 \times2} + \sqrt{16 \times3}\over\sqrt{4 \times2} + \sqrt{4 \times3}\)

= \(4\sqrt2 + 4\sqrt3\over2\sqrt2 +2\sqrt3\)

= \(4(\sqrt2 + \sqrt3)\over2(\sqrt2 + \sqrt3)\)

= 2

\(\sqrt{\sqrt2-1\over\sqrt2+1}\)

= \(\sqrt{(\sqrt2-1) \times(\sqrt2-1)\over(\sqrt2+1)\times(\sqrt2-1)}\)

= \(\sqrt{(\sqrt2-1)^2\over(\sqrt2)^2 - 1^2}\)

= \(\sqrt2-1\over\sqrt2-1\)

= \(\sqrt2-1\over1\)

= \(\sqrt2-1\)

= 1.4142-1

= 0.4142

\(\sqrt[3]{2}. \sqrt[4]{2}. \sqrt[12]{32}\)

=\(2^{1\over3} \times 2^{1\over4} \times (2^5)^{1\over12}\)

= \(2^{1\over3} \times 2^{1\over4} \times 2^{5\over12}\)

= \(2^{{1\over3} +{1\over4} + {5\over12}} \)

= \(2^{4+3+5\over12}\)

= \(2^{12\over12}\)

= \(2^1\)

= 2

\(\sqrt[4]{{({81}})^{-2}}\)

= \(\sqrt[4]{{({1\over81}})^2}\)

= \(\sqrt[4]{\{({1\over9})^2\}^2}\)

= \(\sqrt[4]{({1\over9})^4}\)

= \(({1\over9})^{4 \times{1\over4}}\)

= \(1\over9\)

\((256)^{0.16} \times (256)^{0.09}\)

= \((256)^{0.16+0.09}\)

= \((256)^{0.25}\)

= \((256)^{1\over4}\)

= \((4^4)^{1\over4}\)

= \(4^{4 \times{1\over4}}\)

= 4