Adding fractions is not at all obvious 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?
 A: 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
$$ \frac{5}{4} + \frac{2}{3} = \frac{23}{12}.$$
A: 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. 
A: 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:


*

*$(a, 1) + (b, 1) = (a + b, 1)$

*$(a, 1)(b, 1) = (ab, 1)$

*$(a, 1)(1, b) = (a, b)$

*$(ac, bc) = (a, b)$

*Multiplication to distribute over addition.

*Multiplication and addition to be distributive and commutative.


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.
A: Good question.
Write $x'$ for the reciprocal of $x$. Then from a purely algebraic standpoint, we have the following equations.


*

*$\frac{5}{4}+\frac{2}{3} = 5 \cdot 4'+2\cdot 3'$

*$\frac{15}{8}+\frac{12}{8} = 15 \cdot 8'+12\cdot 8'$


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.
A: 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.
