Proving/disproving $∀n ∈ \text{positive integers}$, $\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5$? Is my proof getting anywhere for proving/disproving $∀n ∈ \text{postive integers}$, $$\left\lceil{\frac{4n^2+1}{n^2}}\right \rceil = 5\text{?}$$
$$\left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil = 5$$
$$\left\lceil{\frac{1}{n^2}}\right \rceil = 5 - 4$$
Therefore, for $n \in \mathbb R$, where $n \neq 0$
$$4 < \left\lceil{\frac{4n^2}{n^2}}+\frac{1}{n^2}\right \rceil \le 5$$
$$4 < \left\lceil{4}+\frac{1}{n^2}\right \rceil \le 5$$
$$0 < \left\lceil\frac{1}{n^2}\right \rceil \le 1$$
I feel kind of without purpose once I hit point. Should I be scrapping the inequality and solving for $n$ instead?
 A: For all $ n \in \mathbb{N}$, we know that $0 < \frac{1}{n^2} \leq 1$, hence $\left\lceil \frac{1}{n^2}\right\rceil=1.$
$$\left\lceil  \frac{4n^2+1}{n^2} \right\rceil=\left\lceil 4 +  \frac{1}{n^2} \right\rceil= 4 + \left\lceil  \frac{1}{n^2} \right\rceil=4+1=5$$
A: $$\left\lceil  \frac{4n^2+1}{n^2} \right\rceil=5\tag{1}$$
means
$$4 < \frac{4n^2+1}{n^2} \le 5\tag{2}$$
so try to prove (2).  This should be less error prone that transforming expressions with ceilings in them.
A: For any $n\in\mathbb{N}$ we have that 
$$4=\dfrac{4n^2}{n^2}<\color{red}{\dfrac{4n^2+1}{n^2}}\le\dfrac{4n^2+n^2}{n^2}=\dfrac{5n^2}{n^2}=5.$$ So, we have that
$$\left\lceil  \frac{4n^2+1}{n^2} \right\rceil=5$$ and 
$$\left\lfloor  \frac{4n^2+1}{n^2} \right\rfloor=4.$$
A: Your proof is probably okay if you specify all statements are $\iff$ results. That way one can reverse direction.  Otherwise you are assuming what you need to prove.
$[\frac {4n^2 + 1}{n^2}] = 5 \iff$
$[4 + \frac 1{n^2}] = 5 \iff$
$[\frac 1{n^2}] = 5 - 4$  (actually do you know if that is true?  I don't feel comfortable with that step.  I'd suggest you first prove that if $k\in \mathbb Z$ then $[k + y] = k + [y]$.  [$*$].)
$\iff 0< \frac 1{n^2} \le 1$. 
And then continue.
$\iff  1 \le n^2$ which as $n$ is a non-zero integer is true.
And we would be done.
But this is a backwards proof and only works because every step was $\iff$. If any one step was not reversible, the proof would fail[$**$]. To do a direct proof, we take a new sheet of paper and work forward.
$n\ne 0; n\in \mathbb Z \implies$
$n^2 > 0; n^2\ne 0; n^2 \in \mathbb Z \implies$
$n^2 \ge 1 \implies$
$0 < \frac 1{n^2} \le 1 \implies$
$[\frac 1{n^2}] = 1 \implies$
$[4 + \frac 1{n^2}] = 4 + 1 \implies$ [$*$]
$[\frac {4n^2 + 1}{n^2}] = 5$.
======
[$*$]
Let $[k + y] = m$ were $k \in \mathbb Z$.
then $m - 1 < k +y \le m$
so $(m-k) -1 < y \le (m-k)$
so $[y] = m-k = [k+y] - k$.
====
P.S.
As GEdgar points out a more elegant proof would be 
$[\frac {4n^2 +1}{n^2}] = 5 \iff$
$4 < \frac {4n^2 + 1}{n^2} \le 5 \iff$
$4n^2 < 4n^2 + 1 \le 5n^2 \iff$
$0 <  1 \le n^2$ which is true.
To make this a forward proof:
$n \ne 0 \implies n^2 \ge 1$ so
$0< 1  \le n^2$
$4n^2 < 4n^2 + 1 \le 5n^2$
$4 < \frac{4n^2 + 1}{n^2} \le 5$
$[\frac {4n^2 + 1}{n^2} ] = 5$.
[$**$]
A proof that $1 = 5$:
$1 = 5 \implies$
$1 -3 = 5-3 \implies$
$-2 = 2\implies$
$(-2)^2 = 2^2\implies$
$4 =4 $ which is true.
So $1 = 5$.  Notice every step of the proof is valid except the conclusion.
$-2 = 2 \implies (-2)^2 = 2^2$ but
$-2 = 2 \not \Leftarrow (-2)^2 = 2^2$.
A Conclusion $\to $ true statement is only valid if we do: conclusion $\Leftarrow ... \Leftarrow$ true statement.
But conclusion $\implies ... \implies$ true statement, is NEVER valid.
A: Hint: The ceiling of a real number is always an integer. So $0<\lceil x \rceil\leq 1$ is equivalent to $\lceil x\rceil = 1$, or $0<x\leq 1$.
