# How do you simplify and write (-5.3)^0 with positive exponents?

Jun 29, 2016

${\left(- 5.3\right)}^{0} = 1$

#### Explanation:

Remember the identity ${a}^{m} \div {a}^{n} = {a}^{m - n}$, for all $a$, where $m$ and $n$ are two natural numbers. For example

${\left(5.3\right)}^{3} \div {\left(5.3\right)}^{2} = \frac{5.3 \times 5.3 \times 5.3}{5.3 \times 5.3}$

= $\frac{\cancel{5.3 \times 5.3} \times 5.3}{\cancel{5.3 \times 5.3}}$

= $5.3$ and is nothing but ${5.3}^{\left(2 - 1\right)} = {5.3}^{1} = 5.3$

Similarly ${\left(- 7.9\right)}^{5} \div {\left(- 7.9\right)}^{3}$

= $\frac{\left(- 7.9\right) \times \left(- 7.9\right) \times \left(- 7.9\right) \times \left(- 7.9\right) \times \left(- 7.9\right)}{\left(- 7.9\right) \times \left(- 7.9\right) \times \left(- 7.9\right)}$

= $\left(\left(- 7.9\right) \times \left(- 7.9\right)\right) = {\left(- 7.9\right)}^{2} = {\left(- 7.9\right)}^{2} = {\left(- 7.9\right)}^{\left(5 - 3\right)}$

and hence ${\left(- 2.7\right)}^{3} \div {\left(- 2.7\right)}^{3} = {\left(- 2.7\right)}^{\left(3 - 3\right)} = {\left(- 2.7\right)}^{0}$

or ${\left(- 2.7\right)}^{3} \div {\left(- 2.7\right)}^{3} = \frac{\left(- 2.7\right) \times \left(- 2.7\right) \times \left(- 2.7\right)}{\left(- 2.7\right) \times \left(- 2.7\right) \times \left(- 2.7\right)} = 1$

Hence for any number $a$,

if $m = n$, we get ${a}^{m} \div {a}^{m} = {a}^{m - m}$

or ${a}^{m} / {a}^{m} = {a}^{m - m}$

or $1 = {a}^{0}$

Hence zero power of any number $a$ is $1$

Hence ${\left(- 5.3\right)}^{0} = 1$.

Jun 30, 2016

In support of Shwetank's answer

#### Explanation:

Further demonstration of what is happening by example

Suppose we had ${3}^{2} / {3}^{4}$

Write as $\frac{3 \times 3}{3 \times 3 \times 3 \times 3}$

This is the same as: $\frac{3}{3} \times \frac{3}{3} \times \frac{1}{3} \times \frac{1}{3} \text{ "=" } 1 \times 1 \times \frac{1}{3} ^ 2$

Another way of writing $\frac{1}{3} ^ 2 \text{ }$ is $\text{ } \textcolor{m a \ge n t a}{{3}^{- 2}}$
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Lets look at this example again but in another way

write$\text{ "3^2/3^4" as } {3}^{2} \times {3}^{- 4}$

This can also be written as $\text{ "3^(2-4)" which is the same as } \textcolor{m a \ge n t a}{{3}^{- 2}}$
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$\textcolor{b l u e}{\text{Answering your question}}$

Consider ${5}^{x}$ where $x$ can by any whole number (integer)

Suppose we gave the value of 3 to $x$ then we have$\text{ } {5}^{3}$

Suppose we had ${5}^{3} / {5}^{3}$ then by the method in the example I gave you we can write this as $\text{ } {5}^{3 - 3} = {5}^{0}$

But ${5}^{3} / {5}^{3} = 1$

So $1 = {5}^{3} / {5}^{3} = {5}^{3 - 3} = {5}^{0} = 1$

$\textcolor{m a \ge n t a}{\text{So a number raised to the power of 0 equals 1}}$

$\textcolor{g r e e n}{\text{In the question the minus is inside the bracket.}}$
$\textcolor{g r e e n}{\text{The index (power) of 0 is applied to everything inside}}$
$\textcolor{g r e e n}{\text{the bracket. So } \textcolor{m a \ge n t a}{{\left(- 5\right)}^{0} = 1}}$
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Note that $- {5}^{0} \to - \left({5}^{0}\right) = - 1$
Test this out on a calculator.