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Click and drag the given steps (in the right) to their corresponding step names (in the left) to prove that if a|b and b|c, then a|c. Step1 Step 2 Step 3 Suppose alb and b|c. By definition of divisibility, a|b means that a = bt for some integer t, and b|c means that b = cs for some integer s. We substitute the equation b = at into c = bs and get c = ats. By definition of divisibility, a = c(st), with ts being an integer, implies a|c. By definition of divisibility, c = a(ts), with ts being an integer, implies a|c. Suppose a|b and b|c. By definition of divisibility, a|b means that b = at for some integer t, and b|c means that c = bs for some integers. We substitute the equation b = cs into a = bt and get a = cst.

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Final Answer:

If
\(a|b\) and
\(b|c\) , then
\(a|c\) : states that if
\(a\) divides
\(b\) and
\(b\) divides
\(c\) , then
\(a\) also divides
\(c\) .

Step-by-step explanation:

The proof begins by supposing
\(a|b\) and
\(b|c\) . By the definition of divisibility,
\(a|b\) means that
\(b = at\) for some integer
\(t\) , and
\(b|c\) means that
\(c = bs\) for some integers
\(s\) . Substituting
\(b = at\) into
\(c = bs\) results in
\(c = a(ts)\). The definition of divisibility states that if
\(a = c(st)\) , with
\(ts\)being an integer, then
\(a|c\) . Thus, the initial assumption that
\(a|b\) and
\(b|c\) leads to
\(a|c\) .

In detail, assuming
\(a|b\) implies
\(b = at\) for some integer
\(t\) . Also, assuming
\(b|c\) implies
\(c = bs\) . Substituting
\(b = at\) into
\(c = bs\) gives
\(c = a(ts)\). According to the definition of divisibility, if
\(a = c(st)\) , where
\(ts\) is an integer, then
\(a|c\) . Therefore, the conditions
\(a|b\) and
\(b|c\) lead logically to
\(a|c\) , as demonstrated by the substitution and the definition of divisibility. This confirms that if
\(a|b\) and
\(b|c\) , then
\(a|c\) based on the transitive property of divisibility.

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