<s>[INST] 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?
 [/INST] 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.
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