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A rational number a/b with (a, b) = 1 is called a reduced fraction. If the sum of two reduced fractions is an integer, say (a/b) + (c/d) = n, prove that |b| = |d|.

I am having trouble to understand the solution i found of this exercise:

Note that (a/b) + (c/d) = (ad + bc)/(bd). This is an integer implies bd|(ad + bc), which by linearity implies d|bc and b|ad. However, since (a, b) = (c, d) = 1, this implies that d|b and b|d. Hence |b| = |d|

I don't understand how does the linearity apply here, i mean, it does not make sense to me to use it here. The only things i could conclude is that:

$ ad = r$ mod $(bd)$

$bc = bd - r$ mod $(bd)$

Let me know if i am wrong, but linearity is not this?: if $a|m, a|n => a|(k_{1}n + k_{1}m)$

I mean, the solution used in the contrary way, the right side implying the left side.

But, even if what the autor of the solutions did is right, why does " bd|(ad + bc), which by linearity implies d|bc and b|ad"?

J. W. Tanner
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    $bd\mid ad+bc\implies d\mid ad+bc\implies d\mid bc$ – J. W. Tanner Jan 13 '21 at 21:44
  • in other words, $bc$ is a linear combination of $ad+bc$ and $ad$, which are both multiples of $d$, so $bc$ is divisible by $d$ – J. W. Tanner Jan 13 '21 at 21:49
  • With $,g=(b,d),$ the dupe $\Rightarrow (b/g)(d/g)\mid 1,$ so $,b = g = d\ \ $ (wlog $,b,d> 0)\ \ $ – Bill Dubuque Jan 13 '21 at 22:53
  • Here "linearity" refers to the linear structure of the set of all integer multiples of a given integer $,d,,$ i.e. $,d\mid j,k\Rightarrow d\mid mj!+!nk,$ for all integers $,m,n.,$ So $,d\mid bc!+!ad, ad\Rightarrow d\mid (bc!+!ad)!-!ad = bc$. Similarly sets of common multiples are closed under arbitrary integer-linear combinations. This linear structure will be brought to the fore if you study abstract algebra and learn about ideals and $R$-modules. – Bill Dubuque Jan 13 '21 at 23:04

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$bd|ad+bc\implies d|\color{blue}{ad+bc}.$

Since $d|\color{green}{ad}$, it follows that the linear combination $(\color{blue}{ad+bc})-\color{green}{ad}=bc$ is also divisible by $d$.

A similar argument shows that $b|ad$.

J. W. Tanner
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