Question

Can someone answer this pls....
Pls as soon as possible
Pls.....

Answer 1

1. Every Rational Number is a Real Number

Because :

✿  Every Rational Number is not a Natural Number

$\mathsf{Example : \frac{5}{7}\;is\;a\;Rational\;Number.\;But, \frac{5}{7}\;is\;not\;a\;Natural\;Number}$

✿  Every Rational Number is not an Integer

$\mathsf{Example : \frac{5}{7}\;is\;a\;Rational\;Number.\;But, \frac{5}{7}\;is\;not\;an\;Integer}$

✿  Every Rational Number is not a Whole Number

$\mathsf{Example : \frac{5}{7}\;is\;a\;Rational\;Number.\;But, \frac{5}{7}\;is\;not\;a\;Whole\;Number}$

As Rational Numbers are Subset of Real Numbers, Every Rational Number is Real Number

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2. Between two rational numbers there are infinitely many rational numbers

$\mathsf{Consider\;Two\;Rational\;Numbers\;\frac{1}{5}\;and\;\frac{2}{5}}$

$\mathsf{The\;Rational\;Number\;between\;\frac{1}{5}\;and\;\frac{2}{5}\;is\;(\frac{\frac{1}{5} + \frac{2}{5}}{2}) = (\frac{\frac{3}{5}}{2}) = \frac{3}{10}}$

$\mathsf{Now,\;The\;Rational\;Number\;between\;\frac{1}{5}\;and\;\frac{3}{10}\;is\;(\frac{\frac{1}{5} + \frac{3}{10}}{2}) = (\frac{\frac{5}{10}}{2}) = \frac{1}{4}}$

In this way, We can find Infinite Rational Numbers between Two Rational Numbers.

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3. Decimal Representation of a Rational Number cannot be non terminating non repeating

Because, non terminating non repeating is the Decimal Representation of a Irrational Number.

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4. The product of any two rational numbers is sometimes rational, sometimes irrational

$\mathsf{Consider\;Two\;Irrational\;Numbers\;\sqrt{8}\;and\;\sqrt{2}}\\\\\mathsf{\implies Product\;of\;these\;Irrational\;Numbers : \sqrt{8} \times \sqrt{2} = \sqrt{16} = 4}\\\\\mathsf{\implies 4\;is\;a\;Rational\;Number}$

$\mathsf{Consider\;Two\;Irrational\;Numbers\;\sqrt{3}\;and\;\sqrt{2}}\\\\\mathsf{\implies Product\;of\;these\;Irrational\;Numbers : \sqrt{3} \times \sqrt{2} = \sqrt{6}}\\\\\mathsf{\implies \sqrt{6}\;is\;a\;Irrational\;Number}$

So, The Product of any two rational numbers can be Rational or Irrational

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5. The decimal expansion of the number $\mathsf{\sqrt{2}}$ is non terminating non-recurring

$\mathsf{Because,\;\sqrt{2}\;is\;an\;Irrational\;Number}$

The Decimal Expansion of an Irrational Number is non terminating non-recurring

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$\mathsf{19.\;Value\;of\;\sqrt[4]{(81)^-^2}\;is :}$

$\mathsf{\implies \sqrt[4]{[(9)^2]^-^2}}$

$\mathsf{\implies \sqrt[4]{[9]^-^4}}$

$\mathsf{\implies \sqrt[4]{\dfrac{1}{9^4}}}$

$\mathsf{\implies (\dfrac{1}{9^4})^\frac{1}{4}}$

$\mathsf{\implies (\dfrac{1}{9^\frac{4}{4}})}$

$\mathsf{\implies \dfrac{1}{9}}$

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$\mathsf{20.\; Value\;of\;(256)^0^.^1^6 \times (256)^0^.^0^9}$

$\mathsf{\implies (256)^0^.^1^6^+^0^.^0^9}$

$\mathsf{\implies (256)^0^.^2^5}$

$\mathsf{\implies (4^4)^0^.^2^5}$

$\mathsf{\implies (4)^(^4^\times^0^.^2^5^)}$

$\mathsf{\implies (4)^1 = 4}$

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21.

✿  Consider Option (A) :

$\mathsf{(x^\frac{12}{7} - x^\frac{5}{7}) \neq x}$

✿  Consider Option (B) :

$\mathsf{\sqrt[12]{(x^4)^\frac{1}{3}} = \sqrt[12]{x^\frac{4}{3}} = (x^\frac{4}{3})^\frac{1}{12} = x^(^\frac{4}{3}^\times^\frac{1}{12}^) = x^\frac{1}{9} \neq x }$

✿  Consider Option (D) :

$\mathsf{(x^\frac{12}{7}).(x^\frac{7}{12}) = x^(^\frac{12}{7}^+^\frac{7}{12}^) = x^(^\frac{144 + 49}{84}^) = x^\frac{193}{84} \neq x}$

✿  Consider Option (C) :

$\mathsf{(\sqrt{x^3})^\frac{2}{3} = (x^\frac{3}{2})^\frac{2}{3} = x^(^\frac{3}{2}^\times^\frac{2}{3}^) = x}$

Hence, Option (C) is the Answer

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18.

$\mathsf{Given : \sqrt[3]{2}.\sqrt[4]{2}.\sqrt[12]{32}}\\\\\mathsf{\implies 2^\frac{1}{3}.2^\frac{1}{4}.(2^5)^\frac{1}{12}}\\\\\mathsf{\implies 2^(^\frac{1}{3}^+^\frac{1}{4}^).2^\frac{5}{12}}\\\\\mathsf{\implies 2^(^\frac{4 + 3}{12}^).2^\frac{5}{12}}\\ \\\mathsf{\implies 2^\frac{7}{12}.2^\frac{5}{12}}\\\\\mathsf{\implies 2^(^\frac{7}{12}^+^\frac{5}{12}^)}\\\\\mathsf{\implies 2^(^\frac{7 + 5}{12}^)}\\\\\mathsf{\implies 2^\frac{12}{12}}\\\\\mathsf{\implies 2}$