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Let $a,b \in \mathbb{Z^+},\ a<b,\ d=\gcd(a,b)\ $ and $\ 1<d<a,\ x=\frac ad,\ y=\frac bd,\ x,y \in \mathbb{Z^+}.$

Suppose $a=a_1+a_2,\ b=b_1+b_2,\ a_1<b_1,\ d_1=\gcd(a_1,b_1)$ and $1<d_1<a_1,$

if $\frac{a_1}{d_1}=x\ $ and $\ \frac{b_1}{d_1}=y,\ $ then $\ \gcd(a_2,b_2)=d-d_1$.

It seems this question is very simple,but I dont know if this question is right(include grammar etc.).Could there's a proof for this question?

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Hint: You have $$a=xd,\; b=yd,\; a_1=x d_1,\; b_1=y d_1$$ and therefore $$a_2=x(d-d_1), \quad b_2=y(d-d_1)$$ So $d-d_1$ is a common divisor of $a_2$ and $b_2$. What you know about $\gcd(x,y)$?

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  • $\begingroup$ @gammatester,Thanks!Maybe $gcd(x,y)=1$. $\endgroup$ – miket May 15 '14 at 8:16
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    $\begingroup$ @miket: Yes, $g=\gcd(x,y)=1,\,$ otherwise $g\cdot d\,$ would be a common divisor of $a\,$ and $b\,$ greater than $d$. $\endgroup$ – gammatester May 15 '14 at 8:22
  • $\begingroup$ @gammatester ,Your cool and atractive answer I like.How about this question:math.stackexchange.com/questions/791311/… at least one number not in the two matrices can be strictly proved,and for two numbers there's a unreadable proof,maybe there's shortcut. $\endgroup$ – miket May 15 '14 at 11:33
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$\begin{eqnarray}{\bf Hint}\ \ && \ \ (a,b) &=& \ \ (d\,x,d\,y),\ \ {\rm where}\ \ d = \gcd(a,b),\ \ {\rm so}\ \ \color{#c00}{\gcd(x,y)= 1}\\ && (a_1,b_1) &=& (d_1 x,d_1 y) \\ && (a_2,b_2) &=& (a,b) - (a_1,b_1)\\ && &=& (d_2 x,d_2 y),\ \ {\rm for}\,\ \ d_2 = d - d_1\\ \Rightarrow\ \ \ \ \ &&\!\!\!\!\!\!\!\! \gcd\!\!(a_2,b_2) &=& \gcd\!\!(d_2 x, d_2 y) = d_2\color{#c00}{\gcd(x,y)} = d_2 \end{eqnarray}$

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