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Estou com uma grande lista de exercícios de PROPRIEDADES DO MDC (MÁXIMO DIVISOR COMUM), e não estou conseguindo entender quais os passos que tenho que seguir nas demonstrações, e gostaria muito de aprender este conteúdo, alguém me ajuda em uma questão para ver se eu consigo entender as outras?

Questão: Mostre que, se (a, b) = 1, a|c e b|c, então a · b|c

Added Translation from Portuguese

I have a large list of exercises PROPERTIES MDC (greatest common divisor), and I am not able to understand what steps you have to follow in the statements, and would love to learn this content, someone help me on a question to see if I I can understand the other?

question: Show that if $(a, b) = 1, a | c$ and $b | c$, then $a · b | c$

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marked as duplicate by Zander, Start wearing purple, Cameron Buie, Dan Rust, m_t_ Jul 9 '13 at 12:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

up vote 0 down vote accepted

Another way is as follows. If $(a,b)=1$ then there are integers $\alpha, \beta$ such that $\alpha a + \beta b = 1$. Now $c = c\cdot 1 = c\alpha a + c \beta b$ But $c\alpha a$ is divisible by $ab$ (why?), as is $c \beta b$, so $c = c\alpha a + c\beta b$ is divisible by $ab$ too.

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Este Jeito E o Que eu Mais me interesso los entendre, na Verdade Curso Matemática, EO professora explica UO Mais Menos ASSIM, Que Só trabalhamos com UMA definição parecida, mas that TEM UM Sinal Negativo, Tipo αa - βb = 1 – marcelolpjunior Jul 8 '13 at 17:25
Consegui...:) Obrigado pela ajuda – marcelolpjunior Jul 8 '13 at 18:18
de nada!........ – m_t_ Jul 8 '13 at 18:23

Proof Sketch:

Take the prime decomposition $c = p_1 \cdots p_n$.

Since $(a,b) = 1$, we know that $a$ and $b$ have no common factors. $(*)$

Since $a | c$, we know that $a$ is a product of $p_i$'s; similarly, $b|c$, so $b$ is a product of $p_i$'s.

By $(*)$ we know that $a$ and $b$ don't share any prime factors; thus, multiplying them together will give a product that divides $c$. "QED"

$$c = \prod p_i$$

$$a = \prod_{i \in A} p_{i}$$

$$b = \prod_{i \in B} p_{i}$$

$$(a,b) = 1 \implies A \cap B = \emptyset$$

$$a \cdot b = \prod_{i \in A \cup B} p_i \big| \prod p_i = c$$

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Realmente interessante... Só que não fui capaz de compreender bem... Pode ser pela tradução!!! – marcelolpjunior Jul 8 '13 at 17:26
Okay: I added a version with no English! – Benjamin Dickman Jul 8 '13 at 17:32

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