Answer
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Hint: We will first find the reciprocal of the first number and then rationalize it to compare to the given second number. If they are the same, then we know that the statement is true.
Complete step-by-step answer:
The first number given to us is \[\dfrac{{\sqrt 3 }}{2}\].
We know that the reciprocal of a number $x$ is $e$, then $x \times e = 1 = e \times x$.
So, we get:- $e = \dfrac{1}{x}$.
Using this concept by taking $x = \dfrac{{\sqrt 3 }}{2}$, then we have:-
Reciprocal of $x$ is $\dfrac{1}{x} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$
We know that $\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{1}{a} \times b = \dfrac{b}{a}$
So, we get:- $\dfrac{1}{x} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{2}{{\sqrt 3 }}$
We know that multiplying a number by 1 doesn’t change its value because $a \times 1$ = $1 \times a$ = $a$.
So, $\dfrac{2}{{\sqrt 3 }} = \dfrac{2}{{\sqrt 3 }} \times 1$
We can rewrite it as:- $\dfrac{2}{{\sqrt 3 }} \times 1 = \dfrac{2}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$.
This is done because we know that $\dfrac{a}{a}$ = 1 for all a.
This implies that:- $\dfrac{1}{x} = \dfrac{{2 \times \sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2}}}$
We know that square root and square cancel each other. So, we get:-
$\dfrac{1}{x} = \dfrac{{2\sqrt 3 }}{3}$.
We clearly see that this is equal to the second number given to us.
Hence, the statement is true.
Note: We can think of a whole number as being $\dfrac{{number}}{1}$, so the reciprocal is just like "flipping it over" and by that we receive $\dfrac{1}{{number}}$.
We need to remember the fact that sometimes there might be a possibility that the second number is not rationalized and the denominator already contains a number with a square root. So, then we would not have to rationalize the reciprocal we got from the first number.
We must read the question properly as there might be cases that we are already given no in the question. For example:- 1 is not reciprocal of 1. Then its answer will be false because reciprocal of 1 is always 1.
Fun Fact:- We cannot find a reciprocal of 0 because denominator a number cannot be 0. So, its reciprocal is not defined but for a bit ease, we say it to be $\infty $.
Complete step-by-step answer:
The first number given to us is \[\dfrac{{\sqrt 3 }}{2}\].
We know that the reciprocal of a number $x$ is $e$, then $x \times e = 1 = e \times x$.
So, we get:- $e = \dfrac{1}{x}$.
Using this concept by taking $x = \dfrac{{\sqrt 3 }}{2}$, then we have:-
Reciprocal of $x$ is $\dfrac{1}{x} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}}$
We know that $\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{1}{a} \times b = \dfrac{b}{a}$
So, we get:- $\dfrac{1}{x} = \dfrac{1}{{\dfrac{{\sqrt 3 }}{2}}} = \dfrac{2}{{\sqrt 3 }}$
We know that multiplying a number by 1 doesn’t change its value because $a \times 1$ = $1 \times a$ = $a$.
So, $\dfrac{2}{{\sqrt 3 }} = \dfrac{2}{{\sqrt 3 }} \times 1$
We can rewrite it as:- $\dfrac{2}{{\sqrt 3 }} \times 1 = \dfrac{2}{{\sqrt 3 }} \times \dfrac{{\sqrt 3 }}{{\sqrt 3 }}$.
This is done because we know that $\dfrac{a}{a}$ = 1 for all a.
This implies that:- $\dfrac{1}{x} = \dfrac{{2 \times \sqrt 3 }}{{{{\left( {\sqrt 3 } \right)}^2}}}$
We know that square root and square cancel each other. So, we get:-
$\dfrac{1}{x} = \dfrac{{2\sqrt 3 }}{3}$.
We clearly see that this is equal to the second number given to us.
Hence, the statement is true.
Note: We can think of a whole number as being $\dfrac{{number}}{1}$, so the reciprocal is just like "flipping it over" and by that we receive $\dfrac{1}{{number}}$.
We need to remember the fact that sometimes there might be a possibility that the second number is not rationalized and the denominator already contains a number with a square root. So, then we would not have to rationalize the reciprocal we got from the first number.
We must read the question properly as there might be cases that we are already given no in the question. For example:- 1 is not reciprocal of 1. Then its answer will be false because reciprocal of 1 is always 1.
Fun Fact:- We cannot find a reciprocal of 0 because denominator a number cannot be 0. So, its reciprocal is not defined but for a bit ease, we say it to be $\infty $.