I'm trying to understand what is probably a fairly simple math concept, but this is escaping me for some reason. Why are the results of these two expressions equal? Thanks for any responses.



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    $\begingroup$ What is $\frac{x-2}{2}$ ? $\endgroup$ – Kasper Mar 15 '14 at 17:49

It comes down to the distributive law: $$ \frac{x-y}{y} = \frac{1}{y}(x-y) = \frac 1y x - \frac 1y y = \frac xy - 1 $$

  • $\begingroup$ I'll read the linked article and see if I understand. $\endgroup$ – Samizdat Mar 12 '14 at 23:30
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    $\begingroup$ @Samizdat this and this might be a little clearer. $\endgroup$ – Omnomnomnom Mar 12 '14 at 23:42
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    $\begingroup$ Caveat: $y \neq 0$. $\endgroup$ – Rhymoid Mar 14 '14 at 3:00
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    $\begingroup$ @Rhymoid it applies to both expressions, so not a real problem. $\endgroup$ – Ruslan Mar 14 '14 at 8:31
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    $\begingroup$ I don't know why I missed that :$ $\endgroup$ – Rhymoid Mar 14 '14 at 10:55

If you want an informal answer rather than an algebraic proof, see if this helps.

Suppose you have $x$ lollies (or sweets, or candies, depending which country you are in) to be shared equally among $y$ children. Each will receive $x/y$ lollies.

Now suppose that before the lollies are shared, $y$ of them magically disappear. Then each child will receive one lolly fewer, that is, each will receive $$\frac{x}{y}-1$$ lollies. But looking at it another way, there are now $x-y$ lollies, so each child will receive $$\frac{x-y}{y}$$ of them. Therefore these two numbers must be equal.

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    $\begingroup$ sweet answer!!! $\endgroup$ – TooTone Mar 12 '14 at 23:49
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    $\begingroup$ while I appreciate the non-algebraic explanation, this actually confused me a little. thanks for the response, though. $\endgroup$ – Samizdat Mar 13 '14 at 3:32
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    $\begingroup$ Oh well.. sorry about that! Different answers are helpful for different people, that's one of the interesting things about teaching. (Also one of the good things about asking for help on math.se.) $\endgroup$ – David Mar 13 '14 at 4:07
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    $\begingroup$ I absolutely agree with that. I'm positive your answer will help someone. Thanks again. $\endgroup$ – Samizdat Mar 13 '14 at 18:39
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    $\begingroup$ Shouldn't the first one be that the lollies disappear after the distribution instead? $\endgroup$ – Khaur Mar 14 '14 at 15:03

Since there is subtraction in the numerator:


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    $\begingroup$ this is really the same as the first answer, except it's not justified using the distributive law. $\endgroup$ – TooTone Mar 12 '14 at 23:47
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    $\begingroup$ I think this answer is easier to follow though. $\endgroup$ – Nit Mar 14 '14 at 11:15
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    $\begingroup$ didnt understand the 1st and 2nd 40+ score, but I did on this one $\endgroup$ – ajax333221 Mar 14 '14 at 21:27

multiply both expressions by $y$ so that $$\frac{x-y}{y}$$ becomes $x-y$ and $$\frac{x}{y}-1$$ becomes $y(\frac{x}{y}-1)=x-y$.

  • $\begingroup$ you should add $y\ne 0$ $\endgroup$ – Bhaskara-III Dec 11 '16 at 11:56

As others mentioned, it's because of the "distributive property" of the operation. But why does the distributive law apply? It's more intuitive for a sum than for a difference, and I always like a material image for that:

We have a couple boards and want to split them in two. Whether we stack them and saw through the stack (that would be $\frac{a+b+c}{2}$), or whether we saw through each board individually and stack then (that would be $\frac{a}{2}+\frac{b}{2}+\frac{c}{2}$), is the same operation. If we let the distance between the individual boards shrink to zero it's hard to tell where the single boards end and the stack begins anyway.

Incidentally that resembles very nicely the identity of gravitation and inertia in masses, or concretely, why the acceleration in a gravitational field does not depend on an object's mass: Glueing boards together to a big stack shouldn't change how they fall in gravity. It's still the same boards.

  • $\begingroup$ Any indication why this was downvoted? I reread my answer and still find it insightful ;-). Comments are welcome. $\endgroup$ – Peter A. Schneider Mar 17 '14 at 10:28

Hint: What would you do if the question was:


or $$\frac{x-3}{3}$$

or $$\frac{x-4}{4}$$

etc. Try to find a pattern.


Let's assume our fraction $\dfrac{x-y}{y}$ equals $a$. $$\frac{x-y}{y}=a$$ Multiplying by $y$ on both sides... $$x-y=ay$$ Isolating $x$... $$x=ay+y$$ Factoring the right hand side:... $$x=y(a+1)$$ Dividing by $y$ on both sides... $$\frac{x}{y}=a+1$$ Moving $1$ back to the left hand side... $$\frac{x}{y}-1=a$$ Because $\dfrac{x-y}{y}=a$ and $\dfrac{x}{y}-1=a$: $$\displaystyle \therefore \frac{x-y}{y}=\frac{x}{y}-1$$ Thank you for reading.

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    $\begingroup$ A bit of a complicated way to reach a simple conclusion, but it does work, so I cant argue. $\endgroup$ – Asimov May 26 '14 at 22:16

I would like to take a crack at explaining this...

Using the distributive property, we can rearrange the equation... $$\frac{x-y}{y} \rightarrow \frac{x-y}{y..y} \rightarrow\frac{x}{y}-\frac{y}{y}$$ We can DISTRIBUTE clones of the 'y' on the bottom to become denominators for the values on the top.

From Wikipedia:

"The numerator represents a number of equal parts, and the denominator, which cannot be zero, indicates how many of those parts make up a unit or a whole. For example, in the fraction 3/4, the numerator, 3, tells us that the fraction represents 3 equal parts, and the denominator, 4, tells us that 4 parts make up a whole."



Number of parts in a kit = x
Number of parts missing = y
Number of parts required to make a widget = z

One widget requires 4(z) parts to build. 6(y) of the 8(x) parts in the kit are missing. That gives you this equation:

$\frac{x-y}{z}$ or $\frac{8-6}{4}$ ... $\text{Which solves to}\rightarrow \frac{2}{4}\text{ or } \frac{1}{2}$

You have only enough parts to make half a widget.

One widget requires 4(z) parts to build. the kit has all 8(x) parts. You put the parts together. Someone takes 1 widget and 2 parts from the other[total of 6(y) parts]. This equation:

$\frac{x}{z}-\frac{y}{z}$ which gives us $\frac{8}{4}$ - $\frac{6}{4}= \frac{2}{4}$ which reduces to $\frac{1}{2}$

You have half a widget left.

The 4 represents the same thing in each equation. It is a value that signifies how many parts make up the whole.

There is a change. One widget now requires 6(z) parts to build. A kit will build 1 widget and provide 2 spare parts. Again 6(y) of the 8(x) parts in the kit are missing. We still have the same equation:

$\frac{x-y}{z}$ which gives us$\frac{8-6}{6}$ . z = y, so we can change all z's to y's and get our equation $\frac{x-y}{y}$ . $\text{It solves to}\rightarrow \frac{2}{6}\text{ or } \frac{1}{3}$

You have less than half the parts to make a widget.

One widget requires 6(z) parts to build. the kit has all 8(x) parts. You put the parts together. Someone takes the widget[total of 6(y) parts]. The equation is:

$\frac{x}{z}-\frac{y}{z}$ ... $\frac{8}{6}$ - $\frac{6}{6}= \frac{2}{6}$ which reduces to $\frac{1}{3}$.

Mathematically, if we wanted to, we could do this:

$\frac{8}{6}-\frac{6}{6} \rightarrow \frac{8}{6}-1$ which looks like our other equation $ \frac{x}{y}-1$.

The manipulation of the 4 and the 6 (as the denominator only) in the scenarios above should help demonstrate the basic mechanics of the distributive property in this situation.

We used the distributive property to rearrange the equation... $$\frac{x}{y}-\frac{y}{y}$$ We can simplify by canceling $\frac{y}{y}$ to give us 1. We cannot simplify the other side, so that gives us:$$\frac{x}{y}-1$$

This form of the equation is also a simplification of the first equation and preferred because you are getting rid of the second instance of 'y', therefore only calling each variable once.

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    $\begingroup$ That was just too much writing for such a short and simple answer. Technically correct, and a respectable effort, but tl;dr for most people while a short answer suffices. $\endgroup$ – Asimov May 26 '14 at 22:17

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