Question

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.

Answer 1

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.