Find the lowest $$N \in \mathbb{N} $$ meeting the requirements: $$|\frac{ n-7 }{ n+7 }-1|\le \frac{ 1 }{ 1000 } \forall n \ge N$$

Question

Find the lowest $$N \in \mathbb{N} $$ meeting the requirements: $$|\frac{ n-7 }{ n+7 }-1|\le \frac{ 1 }{ 1000 }  \forall n \ge N$$

ganeshie8 · Answer

$\left|\dfrac{n-7}{n+7}-1\right|$

$=\left|\dfrac{(n+7)-14}{n+7}-1\right|$

$=\dfrac{14}{n+7}$

You want this expression to not exceed $\frac{1}{1000}$ : 
$\dfrac{14}{n+7}\le \dfrac{1}{1000}$

see if you can solve $n$

anonymous · Answer

If i solve n, i get:
$$n \ge 13993, n <-7$$ What does this tell me of N?

ganeshie8 · Answer

you're given that $N\in\mathbb{N}$
so we only consider positive values of $n$

anonymous · Answer

$$\frac{ 1 }{ \frac{ n+7 }{ 14 } }\le \frac{ 1 }{ 1000 }$$
$$\frac{ n+7 }{ 14 }\ge1000$$
$$n+7\ge14000,n \ge 13993$$

anonymous · Answer

Ohh yes, that is correct. That gives the interval:
$$[13993, infinity)$$

ganeshie8 · Answer

$n \ge 13993$

tells us that the given statement is true when ever $n$ is at least $13993$.
In other words $N = 13993$

anonymous · Answer

But I need to find the lowest N, and if $$n \ge N$$, couldnt N be 1 then? In this case it would still be lower than n and in the natural numbers.

ganeshie8 · Answer

$N$ is the lowest $n$ that satisfies the given inequality.

ganeshie8 · Answer

since $13993$ is the lowest $n$ that satisfies the given inequality, we have $N = 13993$

anonymous · Answer

Ohh, okay. That makes sense, thank you so much :)

ganeshie8 · Answer

np :)