$f(x)=2x+1$ produces a prime number if and only if x is prime, how can we prove this false?
I know this isn't very math-proofy, but can't we just plug in a number we know that is prime, i.e. $x=7$, and observe $f(7)=15$, which is not prime?
I'm intersting in writing a math proof for this, but I'm not exactly sure if my "proof" is good enough/proper. How should I do this?
On
Your counterexample proves that $2x+1$ need not be prime if $x$ is prime. That is the "if" part, and your counterexample contradicts it, and hence the whole statement.
You might want also to note that $2\cdot 6 +1=13$ is a counterexample to the "only if" part - in that $f(x)$ can be prime even if $x$ is not.
On
Let $\mathbb{P}$ denote the set of prime numbers.
We need to prove logical statement $A$:
$\neg\forall{x\in\mathbb{N}}:{2x+1}\in\mathbb{P}\iff{x}\in\mathbb{P}$
Or the equivalent logical statement $B$:
$\neg\forall{x\in\mathbb{N}}:({2x+1}\in\mathbb{P}\implies{x}\in\mathbb{P})\wedge({x}\in\mathbb{P}\implies{2x+1}\in\mathbb{P})$
Or the equivalent logical statement $C$:
$\exists{x\in\mathbb{N}}:\neg({2x+1}\in\mathbb{P}\implies{x}\in\mathbb{P})\vee\neg({x}\in\mathbb{P}\implies{2x+1}\in\mathbb{P})$
Or the equivalent logical statement $D$:
$\exists{x\in\mathbb{N}}:\neg[({2x+1}\not\in\mathbb{P})\vee({x}\in\mathbb{P})]\vee\neg[({x}\not\in\mathbb{P})\vee({2x+1}\in\mathbb{P})]$
Or the equivalent logical statement $E$:
$\exists{x\in\mathbb{N}}:[({2x+1}\in\mathbb{P})\wedge({x}\not\in\mathbb{P})]\vee[({x}\in\mathbb{P})\wedge({2x+1}\not\in\mathbb{P})]$
Finally, in order to prove $\exists{x}$, we only need to find such value of $x$:
$x=6\implies(2x+1\in\mathbb{P})\wedge(x\not\in\mathbb{P})\implies{E}\iff{D}\iff{C}\iff{B}\iff{A}$
To prove a statement false, all you ever need to do is exhibit a counterexample. Your counterexample $x=7$ demonstrates that (one implication direction of) the proposition is not true for some $x$, and hence it can never be true for every $x$. So yes, your proof is perfectly "math-proofy" enough.