elementary number theory - How much can a fraction reduce?

Assume $x/a$ and $y/b$ are positive fractions in it's reduced form.

If $x/a+y/b=z/c$, where $z/c$ is also reduced. What can we say about $c$?

Does $\frac{ab}{\gcd(a,b)^2}|c$?

If it's not true. Is it true when they are all between 0 and 1?

Answer

$\rm(1)\ \ abz\ =\ bcx + acy\ \Rightarrow\ a,b\ |\ ac,bc\ \Rightarrow\ lcm(a,b)\ |\ (a,b)\ c\ \Rightarrow\ lcm(a,b)/(a,b)\ |\ c\ \ $ QED

Note $\rm\ lcm(a,b)/(a,b) =\: ab/(a,b)^2\ $ by the LCM $\cdot$ GCD law $\rm\ lcm(a,b)\ (a,b)\ =\ a\:b\:.$

Above we used $\rm\ a,b\ |\ x,y\ \iff\ lcm(a,b)\ |\ x,y\ \iff lcm(a,b)\ |\ (x,y)\:.$

Below are two more proofs, the first a bit more conceptual and more detailed.

$(2)\ $ Suppose $\rm\ c\ $ is a denominator for $\rm\ \dfrac{x}a\: +\: \dfrac{y}b\ $ where $\rm\ (a,x) = 1 = (b,y)\:.\ $ Then we infer

$\rm\phantom{\quad\Rightarrow\quad}\: c\ \left(\dfrac{x}a\: +\: \dfrac{y}b\right) =\: z \in \mathbb Z\ \ \Rightarrow\ \ abz\ =\ cbx + cay\ \Rightarrow\ \ a\ |\ cbx,\ \ b\ |\ cay$

$\rm\quad\Rightarrow\quad c\ \left\{\dfrac{b}a,\ \dfrac{a}b\right\}\subset\: \mathbb Z\ \ $

by $\rm\ (a,x) = 1,\ \ a\ |\ cbx\ \Rightarrow\ a\ |\ cb\:.\ $ Similarly $\rm\ b\ |\ ca\:.$

$\rm\quad\Rightarrow\quad c\ \left\{\dfrac{\beta}\alpha,\ \dfrac{\alpha}\beta\right\}\subset\: \mathbb Z\ \ $ by cancelling out $\rm\ (a,b),\:$ with $\rm\ \alpha = a/(a,b),\ \ \beta = b/(a,b)\:.\ $

$\rm\quad\Rightarrow\quad c\ \left\{\dfrac{1}\alpha,\ \dfrac{1}\beta\right\}\subset\: \mathbb Z\ \ $ by $\rm\ (\alpha,\beta) = 1,\ \ \alpha\ |\ c\:\beta\ \Rightarrow\ \alpha\ |\ c\:.\ $ Similarly $\rm\ \beta\ |\ c\:. \ $

$\rm\quad\Rightarrow\quad c\ \:\left(\dfrac{1}\alpha\ \:\dfrac{1}\beta\right)\ \in\ \mathbb Z\ $ by $\rm\: \alpha,\beta\ |\ c\ \Rightarrow\ lcm(\alpha,\beta)\ |\ c\:.\:$ $\rm lcm(\alpha,\beta) = \alpha\: \beta\ $ by $\rm\: (\alpha,\beta) = 1.\:$ QED

$(3)\ $ Finally, here is another proof based upon squaring a gcd and applying basic gcd laws.

$\rm\ \ abz\ =\ bcx\: +\: acy\ \ \Rightarrow\ \ a\ |\ bcx\ \ \Rightarrow\ \ a\ |\ bc\ $ by $\rm\ (a,x) = 1\:.\ $ Similarly $\rm\ b\ |\ ac\:.$

$\rm Thus\quad \rm a\ |\ bc,\ b\ |\ ac\ \ \Rightarrow\ \ ab\ |\ a^2c,\ abc,\ b^2 c$

$\rm \phantom{Thus\quad \rm a\ |\ bc,\ b\ |\ ac\ \ } \Rightarrow\ \ ab\ |\ (a^2c,\ abc,\ b^2 c)\ =\ (a,b)^2\:c\quad\ $ QED

Note that the above proof uses only basic gcd laws (associative, commutative, distributive, etc) therefore it holds true in any GCD domain. Below is some further detail using said laws

$$\rm\ (a,b)\ (a,b)\ =\ (a(a,b),b(a,b))\ =\ ((a^2,ab),(ab,b^2))\ =\ (a^2,ab,ab,b^2)\ =\ (a^2,ab,b^2)$$

For some further discussion see here.