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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??

Community
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Mary Star
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3 Answers3

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

Martin Brandenburg
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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 $b=d$.

Mary Star
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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.

robjohn
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