Why is $-32^{\frac{1}{5}} = 2$

Question

When you factorize $-32$, you get:

$-32 = (-16) \cdot 2$

$-16 = (-8) \cdot 2$

$-8 = (-4) \cdot 2$

$-4 = (-2) \cdot 2$

$-32^{\frac{1}{5}} = -2$

The reason I am asking is because you get $-4 = -2 \cdot 2$ where all along we have been multiplying by positive 2 which would lead me to believe that $-32^{\frac{1}{5}}$ $ = 2$; but that is not so - then I got to thinking: Is $-2$ positive or negative? I.e. the 2 that we were multiplying to get $-16, -32$, etc.

That is my question, maybe it should be rephrased because I dont know all the terminology so any help would be appreciated.

Answer 1

Observe that $$-(32^\frac15)=-(2^{5\cdot\frac15})=-2$$ And $$(-32)^\frac15=(-2)^{5\cdot\frac15}=-2$$

I hope your doubt is resolved now.

Answer 2

$$(-2)\cdot(-2)\cdot(-2)\cdot(-2)\cdot(-2)=-32$$ Then $$(-2)^5=-32 \Rightarrow -32^{\frac15}=-\sqrt[5]{32}=-2$$

Answer 3

$$-32^{\frac{1}{5}}=(-32)^{\frac{1}{5}}=(-2)^{{5}^{\frac{1}{5}}}=(-2)^{5\cdot \frac{1}{5}}=-2$$

Answer 4

Your example shows that factorizationis not agood way to find the $n-$rooth of a negative integer number, because the factorization of negative number is not unique (as it is for positive integers).

Also excuding the factor $-1$, you have: $$ -32=(-2)(2)^4=(-2)^3(2)^2=(-2)^5 $$ and always the last factorization gives a number such that its $5-$power is $-32$.