Mathematical Induction on a Subset of the Natural Numbers I am given a strict inequality of the form
$$
2n - 8 < n^2-8n+14,
$$
where $n$ belongs to the set of natural numbers $\mathbb{N}$ (in this case $n$ does not equal 0).
I am asked, for what values of $n$ is the above inequality true?
Upon tabulating the various inputs and outputs, I find that it appears that the inequality is true for $\mathbb{N} \setminus \{4,5,6\}$.
Am I supposed to then proceed with induction with the new set of natural numbers, and if so, how many base cases do I need to establish, that is, do I go up to $n=3$ and then to $n = 7$, for instance? 
I hope that makes sense.
 A: No induction necessary. The equation $x^2 -10x +22 = 0$ has the roots $5-\sqrt 3$ and $5+\sqrt 3$. We know that this function is positive outside the roots, which means on $(-\infty, 5-\sqrt 3) \cup (5+\sqrt 3, \infty)$. The first interval contains the natural numbers $0,1,2,3$, while the second one all the natural numbers starting from $7$ (use that $\sqrt 3 \approx 1.7328$).
A: HINT: Solve this inequality for real numbers and restrict the solution to naturals.
A: Completing the square, we have the quivalences:
$$2n−8<n^2−8n+14 \Leftrightarrow 3<n^2−8n+14-2n+11  \Leftrightarrow 3<(n-5)^2 $$
and here you can conclude about the values for which the inequality is true
A: First, $2n-8<n^2-8n+14$ is equivalent to $0<n^2-10n+22$.  Now, the parabola $y=x^2-10x+22$ opens up and has x-intercepts at $5\pm \sqrt{3} \approx 3.3,6.7$. Hence, for any $n\geq 7$ your inequality holds.
A: Without solving on $\mathbf R$: The inequality is equivalent to $n^2-10n+22>0$.
Now the function$f(x)=x^2-10x+22$ attains its minimum at $x=5$, and $f(x)=-3$
Furthermore, $f(3)=1,\enspace f(4)=-2$, hence by the symmetry properties of the function, $f(6)=-2,\enspace f(7)=1$. We conclude at once that:
$$f(n)>0\iff 0\le n\le 3\enspace\text{or}\enspace n\ge 7.$$
