Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

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

2 Answers

up vote 5 down vote accepted

$\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.

share|improve this answer
add comment

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.

share|improve this answer
3  
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. –  TonyK Jan 13 '12 at 0:52
    
@TonyK, $a'$ and $b'$ are pretty usual for that role. –  lhf Jan 13 '12 at 1:23
    
@lhf: As long as it's not $a$ and $b$. –  TonyK Jan 13 '12 at 1:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.