Meaning of $\frac{x-y}{y}$ versus $\frac{x}{y}-1$

Question

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.

$$\frac{x-y}{y}$$

$$\frac{x}{y}-1$$

Answer 1

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.

Answer 2

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

Answer 3

Since there is subtraction in the numerator:

$$\frac{x-y}{y}=\frac{x}{y}-\frac{y}{y}=\frac{x}{y}-1$$

Answer 4

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

Answer 5

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.

Answer 6

Hint: What would you do if the question was:

$$\frac{x-2}{2}$$

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

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

etc. Try to find a pattern.

Answer 7

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.

Answer 8

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."

http://en.wikipedia.org/wiki/Fraction_(mathematics)

Examples:

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

1:
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.

2:
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.

3:
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.

4:
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.