# Proof

• By hypothesis, $$c > 0$$ is a common divisor of two integers $$a$$ and $$b$$ and the integers $$\frac ac$$ and $$\frac bc$$ are co-prime.
• This means that $$\gcd\left(\frac ac,\frac bc\right)=1$$.
• Therefore, at least one of the integers $$\frac ac$$ and $$\frac bc$$, and so at least one of the integers $$a$$ and $$b$$ must be different from zero.
• Set $$d:=\gcd(a,b)$$, and we have $d\neq 0.$
• From the definition of the greatest common divisor we have that $$c\mid d$$. Thus, $$\frac dc$$ is an integer.
• Now, from the multiplication of rational numbers it follows that $\frac dc \frac ad=\frac ac,\quad\frac dc \frac bd=\frac bc,$ which means that $\frac dc\mid \frac ac,\quad \frac dc\mid \frac bc.$
• Because $$\gcd\left(\frac ac,\frac bc\right)=1$$, $$d > 0$$, $$c > 0$$, any divisor of $$\gcd\left(\frac ac,\frac bc\right)$$ must divide $$1$$.
• This is only true for $\frac dc=1,$ which means that $$c=d=\gcd(a,b)$$.

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### References

#### Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927