Question

Find the number whose reciprocal is itself.

Answer 1

Hint: We first take the general forms and numbers to express the concept of reciprocal numbers. We then use mathematical forms to find the equation. We solve the equation and find its solution.

Complete step-by-step answer:
Let us take an arbitrary number $ x $ .
The reciprocal of the number $ x $ is $ z $ then we have $ xz=1 $ which gives $ z=\dfrac{1}{x} $ .
The condition for the reciprocal to exist is that $ x\ne 0 $ .
We have to find the number whose reciprocal is itself. Let the number be $ a $ .
Then using the theorem, we get $ a=\dfrac{1}{a} $ . We get a simplified form as $ {{a}^{2}}=1 $ .
We now solve the equation $ {{a}^{2}}=1 $ to find the value of $ a $ .
So, $ a=\pm \sqrt{{{a}^{2}}}=\pm 1 $ . We got two values.
Therefore, the numbers whose reciprocals are themselves are \[-1,1\] respectively.
So, the correct answer is “ \[-1,1\] ”.

Note: The opposite number is free of conditions. It is defined for every real number. But the reciprocal number has only one condition where the number is non-zero. 0 has no reciprocal number.