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jeffreydk
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Hello, I'm working out of Hungerford's Abstract Algebra text and this proof has been bothering me because I think I know why it works and it's so simple but I can't figure out how you would show a rigorous proof of it...
If a=p_1^{r_1}p_2^{r_2}p_3^{r_3} \cdots p_k^{r_k} and b=p_1^{s_1}p_2^{s_2}p_3^{s_3} \cdots p_k^{s_k}
where p_1,p_2, \ldots ,p_k are distinct positive primes and each r_i,s_i \geq 0 ,
then prove that GCD(a,b)=p_1^{n_1}p_2^{n_2}p_3^{n_3} \cdots p_k^{n_k}, where for each i \text{, } n_i=\min(r_i,s_i).
I had thought I might be able to show it through using the definition of GCD as a linear combination--where the GCD, d, is the smallest positive element in the set
S=\big\{ d=am+bn \text{ } \vert \text{ } m,n \in \mathbb{Z} \big\}
and therefore I could use that to show that the GCD(a,b) must be the minimum of each r_i,s_i. But that just isn't working out and seems like it's making the proof too complicated anyway. I apologize that I don't have more of a solution worked out--any hint or help would be greatly appreciated. Thank you.
If a=p_1^{r_1}p_2^{r_2}p_3^{r_3} \cdots p_k^{r_k} and b=p_1^{s_1}p_2^{s_2}p_3^{s_3} \cdots p_k^{s_k}
where p_1,p_2, \ldots ,p_k are distinct positive primes and each r_i,s_i \geq 0 ,
then prove that GCD(a,b)=p_1^{n_1}p_2^{n_2}p_3^{n_3} \cdots p_k^{n_k}, where for each i \text{, } n_i=\min(r_i,s_i).
I had thought I might be able to show it through using the definition of GCD as a linear combination--where the GCD, d, is the smallest positive element in the set
S=\big\{ d=am+bn \text{ } \vert \text{ } m,n \in \mathbb{Z} \big\}
and therefore I could use that to show that the GCD(a,b) must be the minimum of each r_i,s_i. But that just isn't working out and seems like it's making the proof too complicated anyway. I apologize that I don't have more of a solution worked out--any hint or help would be greatly appreciated. Thank you.
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