
Which of these four numbers are rational:
$\sqrt{{{\pi }^{2}}},\sqrt[3]{0.8},\sqrt[4]{0.00016},\sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}$
A. none
B. all
C. the first and fourth
D. only the fourth
E. only the first
Answer
500.1k+ views
Hint: At first, we simplify the numbers by expressing the decimals as fractions and carrying out the roots, keeping the fractions intact. If in the end product we get something which cannot be expressed in the $\dfrac{p}{q}$ form, then the number is irrational else not.
Complete step by step solution:
A rational number is one which can be expressed in the form $\dfrac{p}{q}$ where p and q are integers and $q\ne 0$ .
If we check the first option, we can see that,
$\Rightarrow \sqrt{{{\pi }^{2}}}=\pi $
We know that $\pi $ is a transcendental number and thus, not a rational number.
If we check the second option, we can see that,
$\Rightarrow \sqrt[3]{0.8}=\sqrt[3]{\dfrac{8}{10}}=\dfrac{\sqrt[3]{8}}{\sqrt[3]{10}}=\dfrac{2}{\sqrt[3]{10}}$
We can see that the cube root of $8$ is $2$ , but the cube root of $10$ is not a rational number. So, $\sqrt[3]{0.8}$ is irrational.
If we check the third option, we can see that,
$\Rightarrow \sqrt[4]{0.00016}=\sqrt[4]{\dfrac{16}{100000}}=\dfrac{\sqrt[4]{16}}{\sqrt[4]{100000}}=\dfrac{2}{\sqrt[4]{100000}}$
We can see that the fourth root of $16$ is $2$ , but the fourth root of $100000$ is not a rational number. So, $\sqrt[4]{0.00016}$ is irrational.
If we check the fourth option, we can see that,
$\Rightarrow \sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}=\left( -1 \right).\sqrt{\dfrac{100}{9}}=-\dfrac{\sqrt{100}}{\sqrt{9}}=-\dfrac{10}{3}$
We can see that the cube root of $-1$ is $-1$ , and the square root of ${{\left( 0.09 \right)}^{-1}}$ is $\dfrac{10}{3}$ . Both of them are rational numbers. So, $\sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}$ is rational.
Thus, we can conclude that out of the four numbers, only $\sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}$ is rational, which is option D.
Note: In order to solve this problem, we must have a clear knowledge about numbers, especially rational, irrational and transcendental numbers. Only then can we solve this problem. One thing to remember in the fourth option is that we should not involve the imaginary cube roots of infinity.
Complete step by step solution:
A rational number is one which can be expressed in the form $\dfrac{p}{q}$ where p and q are integers and $q\ne 0$ .
If we check the first option, we can see that,
$\Rightarrow \sqrt{{{\pi }^{2}}}=\pi $
We know that $\pi $ is a transcendental number and thus, not a rational number.
If we check the second option, we can see that,
$\Rightarrow \sqrt[3]{0.8}=\sqrt[3]{\dfrac{8}{10}}=\dfrac{\sqrt[3]{8}}{\sqrt[3]{10}}=\dfrac{2}{\sqrt[3]{10}}$
We can see that the cube root of $8$ is $2$ , but the cube root of $10$ is not a rational number. So, $\sqrt[3]{0.8}$ is irrational.
If we check the third option, we can see that,
$\Rightarrow \sqrt[4]{0.00016}=\sqrt[4]{\dfrac{16}{100000}}=\dfrac{\sqrt[4]{16}}{\sqrt[4]{100000}}=\dfrac{2}{\sqrt[4]{100000}}$
We can see that the fourth root of $16$ is $2$ , but the fourth root of $100000$ is not a rational number. So, $\sqrt[4]{0.00016}$ is irrational.
If we check the fourth option, we can see that,
$\Rightarrow \sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}=\left( -1 \right).\sqrt{\dfrac{100}{9}}=-\dfrac{\sqrt{100}}{\sqrt{9}}=-\dfrac{10}{3}$
We can see that the cube root of $-1$ is $-1$ , and the square root of ${{\left( 0.09 \right)}^{-1}}$ is $\dfrac{10}{3}$ . Both of them are rational numbers. So, $\sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}$ is rational.
Thus, we can conclude that out of the four numbers, only $\sqrt[3]{-1}.\sqrt{{{\left( 0.09 \right)}^{-1}}}$ is rational, which is option D.
Note: In order to solve this problem, we must have a clear knowledge about numbers, especially rational, irrational and transcendental numbers. Only then can we solve this problem. One thing to remember in the fourth option is that we should not involve the imaginary cube roots of infinity.
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