Why does $\frac{5}{4} + \frac{2}{3}$ need to be rewritten as $\frac{15}{12} + \frac{8}{12}$ to be added? It's not obvious.
I'm looking towards the fact that any integer can be rewritten as $x=qy$ but these work for rational numbers as well.
Can anyone clarify why exactly fraction addition works only when you find a common denominator?
Think of the denominator as the units you're using to measure the fraction. $\frac{2}{3}$ is, literally, "two thirds", in the same way that $2$ cm is "two centimeters." The unit here is a third.
If you want to add two measures of distance, like "two meters plus fifteen decimeters", you have to find a common unit of measurement. In this case, you know that a meter is 100 centimeters, so two meters is 200 centimeters. You also know that a decimeter is 10 centimeters, so 15 decimeters is 150 centimeters. All in all you have 350 centimeters.
Same with fractions. If you want to add $\frac{5}{4} + \frac{2}{3}$, you have to find a common unit of measurement -- a bigger denominator that can be used to measure both fourths and thirds. There are many, and the best one is twelfths. Three twelfths is a fourth. Four twelfths is a third.
So, converting to this common unit of measurement, five fourths is fifteen twelfths and two thirds is eight twelfths. Now you have in total twenty-three twelfths, so $$ \frac{5}{4} + \frac{2}{3} = \frac{23}{12}.$$
If you are looking for why adding rational numbers must work this way, rather than looking for a natural motivation for this structure, here you go.
Assume the usual addition and multiplication methods and rules for integers, and the multiplication rules and distributive property for fractions.
Theorem: For $a, c$ any integers and $b, d$ nonzero integers,$\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$
Proof: Suppose not. Then for some such $a, b, c, d$ defined as above, $$\frac{a}{b}+\frac{c}{d} \not = \frac{ad+bc}{bd}$$
This means $$(bd)(\frac{a}{b}+\frac{c}{d}) \not = \frac{ad+bc}{bd}(bd)$$
Applying distribution, we have that $$da+cb \not = ad+bc$$
a contradiction.
First, we have to define: what is a fraction? For now, we will just say it is an ordered pair $(a, b)$, where $b \ne 0$. We identify $(a, 1)$ with the integer $a$. (The notation doesn't matter, I just chose this so that we don't accidentally use identities we already know)
Here is what we want:
This is sufficient to define everything! First, we show that $(a, 1)(1, a) = (1, 1)$: $$(a, 1)(1, a) = (a, a) = (a1, a1) = (1, 1)$$ Next, we show $(1, a)(1, b) = (1, ab)$: $$(ab, 1) \cdot (1, a) \cdot (1, b) = (ab, a) \cdot (1, b) = (b, 1) \cdot (1, b) = (1, 1)$$ Because $(ab, 1) \cdot \left( (1, a) \cdot (1, b) \right) = (1, 1)$, we know that $(1, a) \cdot (1, b) = (1, ab)$.
So now we want to find $(a, b) \cdot (c, d)$. We factor: $$(a, 1) \cdot (1, b) \cdot (c, 1) \cdot (1, d) = (ac, 1) \cdot (1, bd) = (ac, bd)$$
Now we move on to addition: $$ \begin{align*} (a, b) + (c, d) &= (a, 1)(1, b) + (c, 1)(1, d) \\ &= (a, 1)(d, bd) + (c, 1)(b, bd) \\ &= (a, 1)(d, 1)(1, bd) + (c, 1)(b, 1)(1, bd) \\ &= (ad, 1)(1, bd) + (bc, 1)(1, bd) \\ &= [ (ad, 1) + (bc, 1) ] (1, bd) \\ &= (ad + bc, 1)(1, bd) \\ &= (ad + bc, bd) \end{align*} $$
So, just starting from that bullet list, we have determined that $(a, b) + (c, d)$ must equal $(ad + bc, bd)$. If we change our notation, this gives us what you want.
This is a great question which I feel was not celebrated enough at the time of posting. Anyways, here is how I'd do it. To begin with, you hopefully agree with me what the expression $\frac{a}{b}$ means conceptually: $a$ parts of $b$. Now, suppose we want to add $a$ parts of $b$ to $c$ parts of $d$, then one way to it is to size up everything by $bd$ then immediately size down by it. Why does this work?
$$ bd \cdot \frac{1}{bd}=1$$
So, we have:
$$ \frac{bd}{bd} ( \frac{a}{b} + \frac{c}{d}) = \frac{1}{bd} ( bd\frac{a}{b}+ bd \frac{c}{d})= \frac{ad+cb}{bd}$$
Second equality follows from distributivity of multiplication.
Good question.
Write $x'$ for the reciprocal of $x$. Then from a purely algebraic standpoint, we have the following equations.
Observe that (2) is just the "common denominator" form of (1). Observe also that its easy to apply the distributive law to the RHS of (2); we see that the RHS equals
$(15+12) 8' = 27 \cdot 8' = 27/8$
On the other hand, there's no obvious way of applying the distributive law to the RHS of (1). So you might say: "We translate from form (1) into (2) in order to apply the distributive law."
Hope that helps.
Fractions are similar: You can think of the fraction $p/q$ as measuring $p$ pieces of $1$ cut into $q$ pieces. In order to measure the total, we need comparable units first; the denominator can be though of as specifying what units one is using.
– Feb 16 '14 at 00:25