which number in number line whose square is itself?

Question

any number whose square is itself that number
example
x^2=x
it can be any integer,fraction ,imaginary accept 0 and 1

Answer 1

There's no such number!
If x² = x then
x²-x = 0
x(x-1) = 0
so, by the zero divisor law,
x = 0 or 1.

Answer 2

It will never be an interger or fraction except for 0 or 1. I'm not sure about imaginary though

Answer 3

x^2 - x = 0

The Fundamental Theorem of Algebra states that an nth degree polynomial has exactly n complex solutions (allow multiple roots).

We're dealing with a second degree polynomial. It will have 2 roots. x=0 and x=1 are roots of the polynomial. It can have only 2 roots. We've found 2 roots for it. There can be no other roots.

Answer 4

it can only be 0 or 1

imaginary numbers? could be, but how do you prove it?

Answer 5

0 and 1 are the only acceptable numbers.
Did you mean except instead of 'accept'?

Answer 6

From what you say,this number does not exist.