Direct Proof Part II

“P if and only if Q” works by breaking the statement into two different statements “if P then Q” and “if Q then P”. We prove both of these statements as if they were the statement before.

Proposition: Let $n$ be an integer, $n$ is not evenly divisible by 3 if and only if $n^2 -1$ evenly divisible by 3.

We start by dividing this into two statements. This gives us our first statement: “if $n^2-1$ is evenly divisible by 3 then $n$ is not evenly divisible by 3.”

Proof: If $n$ is not evenly divisible by 3 then $n$ is it will have a remainder of 1 or 2 when divided by 3. This means that if the remainder is 1 then $(n-1)(n+1)$ is divisible by 3 since $(n-1)$ is now a multiple of 3. If the remainder is 2 then $(n+1)$ is now a multiple of 3 so thus $(n-1)(n+1($ is divisible by 3. Thus we have that if $n$ is not evenly divisible by 3 then $n$ is not evenly divisible by 3.

Now for the second statement: “if $n^2 - 1$ is evenly divisible by 3 then $n$ is not evenly divisible by 3.”

Proof: if $n^2 - 1$ is divisible by 3 then this means either $(n-1)$ is divisible by 3 or $(n+1)$, not both since $(n+1)-(n-1)=2$. Now if $(n+1)$ is divisible by 3 then $n$ cannot be divided evenly by 3 since $n$ will give us a remainder of 1. This means that $n^2 -1$ is evenly divisible by 3 then $n$ is not evenly divisible by 3.

Now that we have proved both parts of the original statement. This means that our original statement is true.

Finally, if a theorem is a fact about a specific definition or theorem, we simply have to derive that fact from relevant definitions and theorems. Here is a quick example.

Proposition: 2 is the only even prime number.

Proof: even numbers are defined as $2n$ which means that for all $n > 1$ then $2n$ is a composite number which means 2 is the only even prime number.

Filed under : Mathematics