Answer
a. 5 | n
b. 5 |$ n^2$
c. 5k
d. $(5k)^2$
e. 5 |$ n^2$
Work Step by Step
[The contrapositive is: For all
integers n, if 5 |n then 5 | n2.] Suppose n is any integer
such that $5|n$ .
[We must show that 5 |$ n^2$ .]
By definition of divisibility, n = $5k$ for some integer k.
By substitution, $n^2$ =$(5k)^2$ = $5(5k^2)$.
But $5k^2$ is an integer because it is a product of integers.
Hence $n^2$ = 5· (an integer), and so 5 |$ n^2$ [as was to be shown].
