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To remove parentheses from $x(y-z)$ I reword it to $xy-xz$.

How do I remove parenthesis from $x/(y-z)$?

pjs36
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Al Klimov
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    You can write $\frac{x}{y-z}$, where there are no parentheses. – Vaneet Jun 10 '16 at 15:24
  • The parentheses are there for clarification, to differentiate $\frac{x}{y} - {z}$ from $\frac{x}{y-z}$. –  Jun 10 '16 at 15:24
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    The short answer to your question is that you cannot. The distributive law does not apply for division in the same way as it does for multiplication. – almagest Jun 10 '16 at 15:25
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    Note, however, that there is a right-distributive law for division over addition as there is for multiplication over addition:$$(a+b)\cdot c=(a\cdot c) + (b\cdot c)$$ and $$(a+b)\div c=(a\div c) + (b\div c)$$ which can be written $$(a+b)\cdot c=a\cdot c + b\cdot c$$ and $$(a+b)\div c=a\div c+ b\div c$$ – MPW Jun 10 '16 at 16:07

2 Answers2

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N.B. I will assume that we are only dealing with real numbers here.

I can very well imagine what you mean to ask, eventhough "removing parntheses form $x/(y-z)$" has no conventional meaning.

What you are dealing with is called the distributive property. We say that multiplication ($*$) is left-distributive over addition ($+$), because $$a*(b+c)=ab+bc\tag L\label l$$ always holds. Likewise, we have right-distributivity because $$(a+b)*c=ac+bc\tag R\label r$$ always holds. We say that addition is not left-distributive over multiplication, because $$a+(b*c)=a+b*a+c$$ does not always hold.

Notice how the word "because" is emphasised. The distibutive property is not something we deliberately construct. It either holds, or it doesn't. This also means that the way to see if a property is (left-, or right-)distributive, is to simply check if $\eqref l$ and/or $\eqref r$ always hold(s).

Now for your case, division ($/$) is not left-distributive over substraction ($-$). Look for instance at $$12/(4-6)\neq 12/ 4-12/ 6.$$ However, it is right-distributive: $$(a- b)/ c=a/ c-b/ c,$$ always.

So to sum up:

  • removing the parentheses from $x/(y-z)$ has no clear meaning
  • we cannot use the distributive property here to find an expression, which equals $x/(y-z)$ and contains no parentheses.

As far as I can see, the only thing we could say is that $x/(y-z)=\frac{x}{y-z}$, which is just another notational convention for writing division (so it wouldn't really count as algebraic manipulation).

Community
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gebruiker
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    I would just like to add that removing the parenthesis from $x/(y-z) $ does have a clear meaning, it just isn't the same meaning. $x/y-z=(x/y)-z$ – Will Fisher Jun 10 '16 at 17:56
  • After a quick search I saw that the term removing indeed appears to be used in some English literature. I thought this wasn't the case (not a native English speaker), so I stand corrected. – gebruiker Jun 11 '16 at 08:22
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How about $$x/(y-z) = \frac{x}{y} \frac{1}{(1-\frac{z}{y})} = \frac{x}{y} (1 + \frac{z}{y} + \frac{z^2}{y^2} + \cdots) = \sum\limits_{i=0}^{\infty} \frac{xz^i}{y^{i+1}}$$

D_S
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