Multiply Me

Let my number be $x$ .

Let's check the given statements one by one.

If I multiply it by itself, then it will become larger than before.

Here,

$x^2 > x \\ \implies x^2 - x > 0 \\ \implies x \in (-\infty, 0) \cup (1, \infty)$

If I multiply this new number by the original number one more time, then it will become smaller than ever before.

Here,

$x^3 < x \\ \implies x^3 - x < 0 \\ \implies x(x+1)(x-1) < 0 \\ \implies x \in (-\infty,-1) \cup (0,1)$

From the set of two obtained ranges of $x$ , we conclude that,

$x \in (-\infty,-1)$

Thus, $x$ is a negative real number less than $-1$ .

Since the even power of a negative real number is always positive and $|x|>1$ , therefore when I multiply my number by itself four times I will obtain the number $x^4$ which will be positive as well as it will be greater than all of $x, \ x^2$ and $x^3$ .

The number is neither 0 nor 1, neither of which is greater when it is squared (#1). Nor is it greater than 1, because then cubing it would result in a greater number (#2). Nor is it between negative one and zero, because cubing it would also result in a greater number (#2). Nor is it -1, because cubing it works result in a number equal to the original (#2). Therefore it is a number less than -1 in order to fulfill there three conditions. Raising such a number to the fourth power results in as negative number less than all the others