Proving $\frac{1}{4n}$ converges to $0$ in epsilon-delta form.

Question

$$(\forall\varepsilon>0)(\exists K)(\forall x)(x>K\Rightarrow\frac{1}{4x}<\varepsilon)$$

Here is my attempted proof: Let $\varepsilon>0$ be a real number. Let $K$ be $\frac{1}{2\varepsilon}$. Assume $x$ is a real number and $x>K$. Then $x>\frac{1}{2\varepsilon}$, so $x>\frac{1}{4\varepsilon}$. Therefore, $4x\varepsilon>1$, so $\frac{1}{4x}<\varepsilon$.

Is it correct? According to my pen friend who is a math major there's a fault in it but he didn't tell me which, leaving me to figure it out, but I couldn't

EDIT: $K$ is a natural number, I think that's the error. So do you have any ideas for a better proof?

Answer 1

The following rewrites your proof so that $K$ is a natural number.

Let $\varepsilon>0$ be a real number. Let $K$ be $\lceil \frac{1}{2\varepsilon} \rceil$; note that $K \geq \frac{1}{2\varepsilon}$, therefore. Assume $x$ is a real number and $x>K$. Then $x>\frac{1}{2\varepsilon}$, so $x>\frac{1}{4\varepsilon}$. Therefore, $4x\varepsilon>1$, so $\frac{1}{4x}<\varepsilon$.

Answer 2

In the epsilon arguments, it does not matter if $K$ is integer or not.

"There exists an integer $K$ such that for all $x > K$, fact $F_x$ holds"

is logically equivalent to the same sentence with "integer" replaced by real number.

That is true whether or not $x$ is restricted to integers.

I changed the epsilon formula in the title to something more comprehensible. If your goal is not to prove that $1/4n$ goes to $0$ the edit is easy to roll back.