Z, if n2 is odd, then n is odd. 2. We are asked to prove, using the method of contraposition, the
statement: "nZ, if n2 is odd, then n is odd.
The contrapositive of the above statement is: "n
Z, if n is not odd, then n2 is not
odd, or equivalently "nZ,
if n is even, then n2 is even.
Now prove "nZ,
if n is even, then n2 is even by a direct proof.
Proof Suppose that n is an even integer. Thus, n = 2·k for some integer k.
n2 |
= | (2·k)2 |
| = | 4·k2 | |
| = | 2(2·k2) |
Hence n2 = 2·s for some integer s (namely 2k2). Therefore n2 is even.
Since the contrapositive statement is equivalent to the original
statement, we can now conclude that "nZ, if n2 is
odd, then n is odd.