# How do you simplify (-6)^0?

Mar 20, 2016

color(blue)(=1

#### Explanation:

• As per property:
color(blue)(a^0= 1

So, applying the above:

${\left(- 6\right)}^{0} = 1$

Mar 20, 2016

1

#### Explanation:

${\left(- 6\right)}^{0} = {\left(- 6\right)}^{n - n} = {\left(- 6\right)}^{n} / {\left(- 6\right)}^{n} = 1$

Mar 20, 2016

${\left(- 6\right)}^{0} = 1$

#### Explanation:

If $n$ is a positive integer, then for any number $a$ we can define:

${a}^{n} = {\overbrace{a \times a \times . . \times a}}^{\text{n times}}$

This has the pleasing property that if $m$ and $n$ are positive integers then:

${a}^{m} \cdot {a}^{n} = {\overbrace{a \times a \times . . \times a}}^{\text{m times" xx overbrace(a xx a xx .. xx a)^"n times}}$

$= {\overbrace{a \times a \times . . \times a}}^{\text{m + n times}} = {a}^{m + n}$

If $a \ne 0$ then we can also define ${a}^{- n} = \frac{1}{\underbrace{a \times a \times . . \times a}} _ \text{n times}$

With this definition we find that $a$'s in numerators and denominators combine or cancel in such a way that:

${a}^{m} \cdot {a}^{n} = {a}^{m + n}$

for any integers $m$ and $n$ (positive or negative).

In particular. If $a \ne 0$ then ${a}^{0} = {a}^{1 - 1} = {a}^{1} \cdot {a}^{- 1} = \frac{a}{a} = 1$

In fact, in the case $n = 0$ we can think of ${a}^{0}$ as being an empty product of $0$ copies of $a$. The empty product has the value $1$ - the identity under multiplication.

$\textcolor{w h i t e}{}$
In particular, if $a = - 6$ then ${a}^{0} = 1$