Hey so I'm doing an exercise and I got a bit confused by something.

I've learned early on that you can multiply the inverse of a fraction and would get the same result as you would if you divided, since you do the exact opposite.

Then why doesn't this hold true for:

$$\frac{10z^{1/3}} {2z^{2}}{^{}{}} = {10z^{1/3}} * {2z^{-2}}$$

Please explain in simple words.. not too math savvy ^^

  • $\begingroup$ I love the people who just downvote for no reason at all. If you are that knowledgeable to downvote questions you deem easy then why don't you move to math overflow? Thought so. $\endgroup$ – Lani Nov 16 '17 at 13:49
  • $\begingroup$ (I didn't downvote.) Note that $$\dfrac{10z^{1/3}}{2z^{2}} = 10z^{1/3}(2z^{2})^{-1} = 10z^{1/3}2^{-1}z^{-2} = 10z^{1/3}\left(\dfrac{1}{2}\right)z^{-2}$$ $\endgroup$ – Clarinetist Nov 16 '17 at 13:51
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    $\begingroup$ Upvoting to cancel a moronic downvote. :) $\endgroup$ – MPW Nov 16 '17 at 13:56

The reason is that you forgot to use the inverse of $2$ as well. So it should be $$10z^{1/3}\cdot 2^{-1}z^{-2}$$ That's because in the original expression, you are dividing by $2$ and also by $z^2$.

In other words, $(2z^2)^{-1}$ can be written as $2^{-1}z^{-2}$, but not as $2z^{-2}$.

  • $\begingroup$ I considered 2z as one single unit to the power of 2. Thanks mpw. $\endgroup$ – Lani Nov 16 '17 at 13:53
  • $\begingroup$ @Lani : Be careful to distinguish between $ab^n$ and $(ab)^n$. The former means $a\cdot b^n$ while the latter means $a^n\cdot b^n$. In the absence of grouping symbols, exponents only apply to the closest factor. $\endgroup$ – MPW Nov 16 '17 at 13:55
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    $\begingroup$ Thanks a ton, I wrote it down and will try to find suitable exercises online :) $\endgroup$ – Lani Nov 16 '17 at 13:57

The problem is the $2$. What you should do is to take the whole denominator (the thing you divide with) to $^{-1}$. Thus you should get $$\frac{10z^{1/3}}{2z^2} = 10z^{1/3}*(2z^{2})^{-1} = 10z^{1/3}*2^{-1}*z^{-2} =10z^{1/3}*\frac{z^{-2}}{2} $$


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