Reciprocal

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February 7th 2011, 05:19 PM
alyosha2

Negative Exponents / Algebraic Fractions / Reciprocal

$3^4*2^{-8} = \frac{3^4}{2^8}$

why not

$\frac{3^4}{\frac{1}{2^8}}$

as $\frac{1}{2^8}$ is the reciprocal of $\frac{2^{-8}}{1}$
February 7th 2011, 05:23 PM
Prove It

$\displaystyle 3^4 \times 2^{-8} = 3^4 \times \frac{1}{2^8} = \frac{3^4}{2^8}$ .
February 7th 2011, 05:23 PM
dwsmith

Quote:

Originally Posted by alyosha2

$3^4*2^{-8} = \frac{3^4}{2^8}$

why not

$\frac{3^4}{\frac{1}{2^8}}$

as $\frac{1}{2^8}$ is the reciprocal of $\frac{2^{-8}}{1}$

$\displaystyle 3^4\cdot\frac{1}{2^8}=\frac{3^4}{1}\cdot\frac{1}{2 ^8}=\frac{3^4\cdot 1}{1\cdot 2^8}=\frac{3^4}{2^8}$
February 7th 2011, 11:16 PM
alyosha2

Quote:

Originally Posted by Prove It

$\displaystyle 3^4 \times 2^{-8} = 3^4 \times \frac{1}{2^8} = \frac{3^4}{2^8}$ .

So multiplying by a given number gives the same result as multiplying it by it's reciprocal!??

I thought the idea was to invert the operation? Meaning dividing by a given number is the same as multiplying by it's reciprocal and dividing by a given number is the same as multiplying by it's reciprocal?

If that's the case, how come the operation on the number here isn't inverted when we subsitutue it for it's reciprocal?
February 7th 2011, 11:33 PM
alyosha2

Wait... $\frac{1}{2^8} =$ isn't the reciprocal of $2^{-8}$ , it's equivalent to it.

I see now.

Thanks.