# What is wrong with this problem

We know that:

$(a^n)^m=a^{nm}$

From this we have: $-3^3=[(-3)^2]^\frac{3}{2}=(3^2)^\frac{3}{2}=27$

Find what's wrong

-
We do not know that when $m,n$ are not integers, unless $a$ is positive. –  Thomas Andrews Jan 23 '13 at 15:29
I'm surprised no one's mentioned $-3^3=-(3^3)$, not $(-3)^3$. –  Mike Jan 23 '13 at 16:06
en.wikipedia.org/wiki/… –  lab bhattacharjee Jan 23 '13 at 16:43

$\exists m,n\in\Bbb Q$ such that $(a^n)^m \not= a^{nm}$ if $a<0$.

-
In fact the symbol $\neq$ is not really the right one. It implies 'is not the same as', but one needs a symbol that implies something like 'is not necessarily the same as' but I guess that one doesn't exist. –  barto Jan 23 '13 at 17:04
I meant that the statement is not true for all values of m and n. –  Ishan Banerjee Jan 23 '13 at 17:06

The "law of exponents" that you cite:

$$\large (a^n)^m=a^{nm}$$

applies PROVIDED $\bf{a \gt 0}$.

Here, though, we have $\,\bf{a = -3 \lt 0}$, and hence:

$$-3^3=[(-3)^2]^{\large\frac{3}{2}}=(3^2)^{\large\frac{3}{2}}=27\quad \Longleftarrow \;\text{ False}.$$ $$-3^3 = -(3^3) = -[(3^2)^{\large \frac{3}{2}}] = -(9^{\large \frac{3}{2}}) = -27\quad \Longleftarrow \; \text{ True}$$ But why the trouble? It is a very straightforward computation: $$-3^3 = -(3^3) = -(3\cdot 3\cdot 3) = -27$$

-

You can not do that. I mean $(-3)^1\neq\sqrt{(3)^2}$, since $-3<0$.

-
nice example...+1 –  amWhy Jan 23 '13 at 15:36
@amWhy: Thanks so much! –  B. S. Jan 23 '13 at 15:40