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I have to show the following:"If the sum of two irreducible fractions with positive denominators is an integer, then the denominators are equal." $$\frac{a}{b}+\frac{c}{d}=k, \text{ where k an integer }$$ Since the fractions are irreducible, $(a,b)=1$ and $(c,d)=1$. Right? But how can I continue??

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$ad+bc=kbd$, what does it say about divisibility of $d$ by $b$ and $b$ by $d$? –  Marcin Łoś Feb 24 at 0:03
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"Fractions" can be tricky in number-theoretic reasoning. So an almost automatic first step is to get rid of them, in this case by multiplying through by $bd$. –  André Nicolas Feb 24 at 0:05
    
@MarcinŁoś André Nicolas Could you check if my answer is correct? –  Mary Star Feb 24 at 0:19
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3 Answers 3

up vote 3 down vote accepted

The equation means $ad+bc=bd k$. It follows that $b$ divides $ad$, hence also $d$. The rest is for you ...

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you must mean $ad+bc=bdk$ :-) –  robjohn Feb 24 at 0:08
    
@MartinBrandenburg Could you check if my answer is correct? –  Mary Star Feb 24 at 0:18
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Here is a proof using Bezout's Identity.

Bezout's Identity says that since $(a,b)=1$ $$ ax+by=1\tag{1} $$ and that since $(c,d)=1$ $$ cu+dv=1\tag{2} $$ Multiplying your equation by $bd$ gives $$ ad+bc=bdk\tag{3} $$ Multiply $(1)$ by $d$ to get $$ \color{#C00000}{adx}+bdy=d\tag{4} $$ Multiply $(3)$ by $x$ to get $$ \color{#C00000}{adx}+bcx=bdkx\tag{5} $$ Solving $(5)$ for $adx$ and plugging that into $(4)$ yields $$ d=b(dy+dkx-cx)\tag{6} $$ Multiply $(2)$ by $b$ to get $$ \color{#C00000}{bcu}+bdv=b\tag{7} $$ Multiply $(3)$ by $u$ to get $$ adu+\color{#C00000}{bcu}=bdku\tag{8} $$ Solving $(8)$ for $bcu$ and plugging that into $(7)$ yields $$ b=d(bku-au+bv)\tag{9} $$ Equations $(6)$ and $(9)$ should finish things off.

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Ok!Thank you for your answer!!! –  Mary Star Feb 24 at 0:42
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It is along the same lines as the others, but I think of Bezout as a low level idea. –  robjohn Feb 24 at 0:45
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$$\frac{a}{b}+\frac{c}{d}=k \text{ with } (a,b)=1, (c,d)=1$$ $$\Rightarrow ad+bc=kbd$$ $$>ad=kbd-cb \Rightarrow ad=b(kd-c) \Rightarrow b|ad \xrightarrow{(a,b)=1} b|d (1)$$ $$>cb=kbd-ad \Rightarrow cb=d(kb-a) \Rightarrow d|cb \xrightarrow{(c,d)=1} d|b (2)$$ $$(1) \Rightarrow b \leq d$$ $$(2) \Rightarrow d \leq b$$ So $b=d$.

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Yes, that seems correct. –  Marcin Łoś Feb 24 at 0:20
    
Ok!Thanks a lot!!! :-) –  Mary Star Feb 24 at 0:20
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