My proof:

Let $\varepsilon >0$. By the Archimedean property, there is a $N$ in $\mathbb{N}$ such that $\dfrac {1} {N}\varepsilon < 1$. Hence, we obtain $N < \dfrac {1} {\varepsilon }$. Thus, for all $n\geq N$ $$\left| \dfrac {5} {n}-0\right|=\dfrac {5} {n}=\dfrac {1} {n(\dfrac {1} {5})}<\dfrac {1} {n}<\dfrac {1} {N}<\varepsilon. $$ We are done.

Can you check my proof and mathematical writing (proof-writing)?

  • 5
    $\begingroup$ It is not true that $\frac{5}{n} < \frac{1}{n}$. Perhaps just proceed from $|\frac{5}{n}| = \frac{5}{n} \leq \frac{5}{N}$ since $n\geq N$ and choose $N$ such that $\frac{5}{\epsilon} < N$. $\endgroup$ – Weaam Aug 10 '17 at 23:02
  • $\begingroup$ @Weaam Thanks.. $\endgroup$ – James Ensor Aug 11 '17 at 0:10

As Weaam wrote, $\dfrac {1} {n(\dfrac {1} {5})} <\dfrac {1} {n} $ is false.

You want $n$ to be large enough so that $\dfrac {5} {n} <\varepsilon $. This is equivalent to $\dfrac {5} {\varepsilon} <n $, and this is the lower bound on $n$ that is needed.

  • 1
    $\begingroup$ Then, we should say 'by the Archimedean property there is a $N$ in $\mathbb{N}$ such that $N\varepsilon > 5$. Hence, we obtain $\dfrac {5} {N } < \varepsilon$. $\endgroup$ – James Ensor Aug 11 '17 at 0:16

Let $ε>0$ By the Archimedean property,$\quad\exists n_0\in\mathbb{N},\quad n_0\varepsilon>5\iff n_0>\dfrac{5}{\varepsilon}$

We can choose $n_0=\left\lfloor\dfrac{5}{\varepsilon}\right\rfloor+1$


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