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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?

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  • $\begingroup$ Actually, $\dfrac{ab}{\gcd(a,b)^2} \mid c \mid \dfrac{ab}{\gcd(a,b)}$. $\endgroup$ – lhf Jan 13 '12 at 1:22
  • $\begingroup$ @lhf $\:$ Indeed $\rm\ c\:(a,b)\ |\ c\ (b\ x + a\ y) = zab\ \Rightarrow\ c\ | zab/(a,b)\ \Rightarrow\ c\ | ab/(a,b)\ $ by $\rm\ (c,z) = 1\:.$ $\quad\qquad$ $\endgroup$ – Bill Dubuque Jan 13 '12 at 1:44
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$\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.

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Yes, that is true. To see this, let $g = \gcd(a,b)$ so we can rewrite the sum as $$ \dfrac x {ga} + \dfrac y { gb} = \dfrac z c $$ where $\gcd(a, b) = \gcd(a,x) = \gcd(b,y) = 1$. Then $$ \dfrac{ bx + ay } {gab} = \dfrac z c. $$ But $\gcd(a,bx) = 1$ so the factor of $a$ cannot cancel, and similarly neither can the factor of $b$. Thus $$ ab = \dfrac{ ga \cdot gb }{g^2} | c $$ as desired.

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    $\begingroup$ You re-defined $a$ and $b$! So I have to keep track in my mind of which $a$'s and $b$'s are the original ones, and which are your re-defined ones. That's terrible! Call them $e$ and $f$. Please. $\endgroup$ – TonyK Jan 13 '12 at 0:52
  • $\begingroup$ @TonyK, $a'$ and $b'$ are pretty usual for that role. $\endgroup$ – lhf Jan 13 '12 at 1:23
  • $\begingroup$ @lhf: As long as it's not $a$ and $b$. $\endgroup$ – TonyK Jan 13 '12 at 1:52

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