Indirect proof , odd and even numbers "Show by indirect proof that if 5n + 3 is an even number then n is an odd number"
How could this be solved? I guess I'm in the right track but I don't know how to conclude.
 A: The proposition we want to prove is: "$5n + 3$ even $\implies$ $n$ odd".
This is equivalent to the contrapositive statement, "$n$ not odd $\implies$ $5n + 3$ not even". Since an integer is either odd or even, it is equivalent to saying "$n$ even $\implies$ $5n + 3$ odd.
If a number is even, then any integer multiple of it must be even. That is to say, $n$ even $\implies$ $5n$ even. 
Furthermore, if a number is even, then adding an odd number to it necessarily makes the result odd. That is to say, $5n$ even $\implies$ $5n + 3$ odd.
Hence, we have $n$ even $\implies$ $5n$ even $\implies$ $5n + 3$ odd. By proving the contrapositive statement, we have indirectly proved that $5n + 3$ even $\implies$ $n$ odd.
A: Every integer is either even or odd. An even integer is an integer divisible by 2.
If $n$ is even then $5n$ is also even, for it is divisible by 2. But since 3 is not divisible by 2, so that $5n+3$ is not divisible by 2, and hence $5n+3$ is an odd integer, qed. 
A: Without resorting to the use of modulo arithmetic...
As $5n+3$ is divisible by $2$, so too is $5n+3-4n-4$
That is $n-1$ is divisible by $2$
So $n-1=2k$ for some integer $k$
Yielding $n=2k+1$, which shows $n$ is an odd number.
Done !
A: An even number is divisible by $2$, so we can describe an even number as $2k$. 
Then
\[ 5n+3=2k \]
\[ 5n=2k-3 \]
\[ n=\frac{2k-3}{5} \]
$2k$ is even so $2k-3$ is odd and an odd number divided by $5$ is still an odd number. 
Hence, when $5n+3$ is an even number, $n$ is an odd number. 
