Question

Which among the following is true?
A.There is no rational number between two irrational numbers.
B.If \[{x^2} = 0.4\], then \[x\] is a rational number.
C.The only real numbers are rational numbers.
D.The reciprocal of an irrational number is irrational.

Answer 1

Hint: In the given question, we had to find out that which among the following four options is true. In options, we were given four options with conditions and we have to check which among the following four options is true. We will check it one after another.

Complete step-by-step answer:
In the given question, we were given four options and we have to check each option one by one, only then we will be able to tell which statement among the four is true.
Let us take option (A), that there is no rational number between two irrational numbers where rational number is the number which is written in the form of \[\dfrac{p}{q}\] where \[q \ne 0\] and irrational number is in the square root. We have to check there is no rational number between two irrational numbers. Let us take two irrational numbers \[\sqrt 2 \] and \[\sqrt 3 \] whose value is \[1.414\] and \[1.732\] respectively. So, there are many rational numbers between \[\sqrt 2 \] and \[\sqrt 3 \].
Therefore, option (A) is false.
In option (B), we are given that
If \[{x^2} = 0.4\], then \[x\] is a rational number.
We are given the value that \[{x^2} = 0.4\]
On solving it, we get
\[{x^2} = \dfrac{{04}}{{10}}\]
\[ \Rightarrow {x^2} = \dfrac{4}{{10}}\]
Dividing numerator and denominator by \[2\], we get:
\[{x^2} = \dfrac{2}{5}\]
Taking square root on both sides,
\[x = \pm \sqrt {\dfrac{2}{5}} \]
But \[x\] is an irrational number.
Therefore, option (B) is false.
In option (C), it is given that the only real numbers are rational numbers. To check it, let us take the real number \[\sqrt 2 \] whose value is \[1.414\]; since \[\sqrt 2 \] is a rational number but this is not a rational number. Therefore, real numbers are not rational numbers.
So, option (C) is false.
In option (D), we had to check that reciprocal of an irrational number is also irrational.
Take an irrational number \[\sqrt 2 \],
Taking reciprocal of it, we get \[\dfrac{1}{{\sqrt 2 }}\], which on rationalising, we get:
\[\dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}\],
which is also an irrational number.
Therefore, reciprocal of irrational numbers is also irrational.
Therefore, option (D) is correct.

Note: All the real numbers are irrational numbers but all real numbers are not rational numbers. That means irrational numbers cannot be expressed as the ratio of two irrational. Zero is also a rational number whereas the numerator is \[0\] and the denominator is any non-zero integer, say \[1\], and the result is also zero.