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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.

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  • $\begingroup$ One way to do it would be to show explicitly that it holds for $n=1,2,3,7$, and then show that if it holds for some $n>7$ then it must hold for $n+1$. $\endgroup$
    – TorsionSquid
    Apr 19, 2015 at 17:18
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    $\begingroup$ Why exactly are you trying to use induction here? $\endgroup$
    – Daniel W. Farlow
    Apr 19, 2015 at 17:19
  • $\begingroup$ True...you could just graph the parabola $x^2-8x+14-2x+8$ . $\endgroup$
    – TorsionSquid
    Apr 19, 2015 at 17:20
  • $\begingroup$ If you really want to do induction, the correct approach is the following: first check the "isolated" cases $0,1,2,3$; next, consider the "base case" $7$, and then perform the induction step. $\endgroup$
    – Alex M.
    Apr 19, 2015 at 17:35

5 Answers 5

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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 $\sqrt 3 \approx 1.7328$).

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HINT: Solve this inequality for real numbers and restrict the solution to naturals.

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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

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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.

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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: $$f(n)>0\iff 0\le n\le 3\enspace\text{or}\enspace n\ge 7.$$

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