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I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?

the_candyman
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wa7d
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6 Answers6

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$\frac{x}{\frac{1}{2}}=\frac{x}{\frac{1}{2}}\cdot1=\frac{x}{\frac{1}{2}}\cdot\frac{2}{2}=\frac{x\cdot 2}{\frac{1}{2}\cdot2}=\frac{2x}{1}$

Generally,

$\frac{a}{\frac{b}{c}}=\frac{a}{\frac{b}{c}}\cdot\frac{c}{c}=\frac{ac}{b}=a\cdot\frac{c}{b}$

Hasan Saad
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Several answers focused on the fact that division is really just multiplication with the inverse, and that is perfectly fine.

Here's another aproach:

When you calculate what $\frac{a}{b}$ is, what are you really doing? You are asking yourself

"How many pieces do I get if I divide $a$ into $b$ equal pieces?"

so you are asking yourself

"by how much must I multiply $b$ to get $a$?"

For example, if you are calculating $\frac{10}2$, you want to know how many groups of $2$ you can make out of $10$ people, or, equivalently, what you need to multiply by $2$ to get $10$, and the answer is $5$ because $2\cdot 5=10$.

So, the bottom line here is that $x=\frac{a}{b}$ is defined as the unique number that satisfies the equation $bx = a$.

Using this definition, you can now see what happens if $b\frac12$. Well, in that case, by how much must I multiply $\frac 12$ to get $a$? In other words, what is the solution to the equation $$\frac12 \cdot x = a?$$

You can easily see that the answer is obvious when you multiply the equation by $2$ to get $x=2a$.

5xum
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Unfortunately, nobody tells you the definition of $x/y$ nowadays. (Did they ever?) The simplest definition is surely that if $x$ is a real number and $y$ is a non-zero real number, then $x/y$ is--by definition-- a mere shorthand for the expression $x y^{-1}.$ So whenever $y$ is non-zero, we have:

$$\frac{x}{y} = x/y = xy^{-1}$$

This viewpoint is very powerful; for example, your question now becomes very easy to answer. Observe that if $a$ is a real number and $c$ and $d$ are non-zero real numbers, then we may argue as follows:

$$\frac{a}{c/d} = \frac{a}{c d^{-1}} = a(cd^{-1})^{-1} = a(c^{-1}d) = adc^{-1} = \frac{ad}{c}$$

The above chain of reasoning makes use of the following rules about raising to the power of $-1$, which hold for all real numbers distinct from $0$:

  1. $(xy)^{-1} = x^{-1} y^{-1}$
  2. $(x^{-1})^{-1} = x$

Other important laws (that I didn't use) include:

  1. $x x^{-1} = 1$
  2. $1x = x$ (this last one holds even if $x$ equals $0$.)

In summary:

Theorem. For all real numbers $a$ and all non-zero real numbers $c$ and $d$, we have: $$\frac{a}{c/d} = \frac{ad}{c}$$

Applying this to your problem:

$$\frac{x}{1/2} = \frac{x \cdot 2}{1} = 2x$$

The take home message is that before trying to understand division, you should try to understand raising to the power of $-1$. The expressions $x/y$ and $$\frac{x}{y}$$ are best viewed as nifty notation for $x y^{-1}$, nothing more.

goblin GONE
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  • Did somebody really just go through and downvote every answer? – goblin GONE May 05 '15 at 14:06
  • Remark that $,x/y,$ can exist without $,y^{-1},$ existing, e.g. in $,\Bbb Z[x/y]\subset \Bbb Z[x,y]\ $ (This is meant as a general remark, not a reply to your comment about downvotes, about which I have no knowledge). – Bill Dubuque May 05 '15 at 16:37
  • @BillDubuque, good point. Yet another case where my lack of knowledge about commutative monoids shows :) Edit. However I don't think this is the right question to post a detailed technical answer in, since the OP may not understand it. – goblin GONE May 05 '15 at 16:43
  • @BillDubuque, a related comment: I tend to think that $x/y=xy^{-1}$ is a good definition whenever $y$ is a regular element of our commutative ring, since we can always simply pass to the localization and then pull our result back into the original ring. The only time (in commutative algebra) when $x/y$ needs to be defined more carefully is when $y$ is a zero divisor. Do you agree? – goblin GONE May 05 '15 at 16:54
  • Sometimes one desires to work in subrings of fraction fields that are not localizations, e.g. see here. – Bill Dubuque May 05 '15 at 17:08
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Suppose you have $2$ pies and $x$ people at a party each want a piece, so each pie is cut into pieces, each a fraction $2/x$ of a pie. Half of the party leaves and takes their pieces with them, thus taking away a whole pie. The remaining people are now sad and lonely, but they all still get the same size piece of pie, $1$ pie $/$ $(x/2)$ people.

Jair Taylor
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Recall that if $ABC$ and $A'B'C'$ are similar triangles, then the ratio of $BC$ to $AB$ is the same as $B'C'$ to $A'B'$. If the length of $AB$ is $\frac{1}{2}$, and the length of $A'B'$ is $1$, then the length of $B'C'$ is twice the length of $BC$.

Kyle Miller
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  • Do people really not like the geometric interpretation of division that much? :-) In case it wasn't clear, the length of $BC$ represents $x$ and so the length of $B'C'$ represents $2x$. (I was trying to give a non-arithmetic explanation.) – Kyle Miller May 05 '15 at 18:02
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Division can be defined as multiplication by the inverse. Explicitly, for any associative division algebra (including real numbers, rational numbers, complex numbers, quaternions, etc.) we can define division by

$$a \div b = \frac{a}{b} := ab^{-1}$$

for any $b \neq 0$. If the algebra is commutative (that is $ab = ba$ for any $a, b$), which the real numbers, rationals, and complex numbers are, then we must have

$$\frac{a}{b} \div \frac{c}{d}= ab^{-1}\div (cd^{-1})= ab^{-1}(cd^{-1})^{-1}=ab^{-1}(dc^{-1})=adb^{-1}c^{-1}=(ad)(bc)^{-1}= (ad) \div (bc) =\frac{ad}{bc}= \frac{a}{b} \cdot \frac{d}{c}.$$

Thus, when you divide fractions you "flip and multiply".

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