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Using induction, Prove for every natural number $n$, with $n ≥ 7$, that $\frac{(2n−18)}{(n^2−8n+ 8)} < 1$. I cannot get the original equation to match the k+1 equation, and im not sure what I am messing up. I would post my work but I would rather see it done out completely from scratch. Any help appreciated!

  • Why not write out your work? Second, you can further simplify the denominator to just $10 - 8n$, as what's written. The numerator can be re-written as $-16n$, if that's what you mean. – John W. Smith Apr 16 '18 at 03:58
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    There are very probably many thing wrong with the way you typed the question. If you really meant $2n-18n$ for the numerator... then why wouldn't you have just simplified it as $-16n$ after having combined the terms. Maybe you mean to have $2^n-18n$? Maybe something else entirely? Similarly you should check the denominator. Formatting tips. – JMoravitz Apr 16 '18 at 04:01
  • Oops yes my bad, the denominator should be n^2 -8n+8. Just fixed it. My bad –  Apr 16 '18 at 04:03
  • How about the numerator? – JMoravitz Apr 16 '18 at 04:04
  • Jeez, yes the numerator should be 2n-18. My bad... again! –  Apr 16 '18 at 04:04
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    Okay then. So... with large enough $n$ you have both the numerator and the denominator are positive. In such a case, you have $\frac{(2n-18)}{(n^2-8n+8)}<1\iff (2n-18)<(n^2-8n+8)$, and then by moving everything to one side... this is true iff $0<$_________. Continue. Remember that you can use the hypothesis that $n\geq 7$ (or if you desire to make the induction argument easier, that $n\geq 8$ after having shown that it is true for both $7$ and $8$ individually) – JMoravitz Apr 16 '18 at 04:07
  • Your question was put on hold, the message above (and possibly comments) should give an explanation why. (In particular, this link might be useful.) You might try to edit your question to address these issues. Note that the next edit puts your post in the review queue, where users can vote whether to reopen it or leave it closed. (Therefore it would be good to avoid minor edits and improve your question as much as possible with the next edit.) – Martin Sleziak Apr 17 '18 at 01:43

4 Answers4

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Alt. hint (without induction): $\;\dfrac{2n−18}{n^2−8n+ 8} = \dfrac{2(n-7)-4}{(n-1)(n-7)+1} \le \dfrac{2(n-7)-4}{2(n-7)+1}\,$ for $\,n \ge 7\,$.

dxiv
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Denominator $n^2-8n+8=n^2-8n+16-8=(n-4)^2-8$. If $n\ge7$ then $(n-4)^2-8\ge 3^2-8=1>0$, so then (as suggested in a comment) $\frac{(2n-18)}{(n^2-8n+8)}<1\iff (2n-18)<(n^2-8n+8)$.

That is, prove that $n^2-10n+26>0$ for $n\ge7$, by induction. True for $n=7$, we get $49-70+26=5>0$. Assume it is true for some $n$, consider $n+1$. $(n+1)^2-10(n+1)+26=n^2+2n+1-10n-10+26=(n^2-10n+26)+2n-9$. Since $2n-9\ge14-9=5>0$, and using the induction hypothesis that $n^2-10n+26>0$ we obtain that $(n^2-10n+26)+2n-9>0$.

Mirko
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Partial Answer as Method of Induction is Not Used.

For curiosity, I have provided an answer that does not use induction.

Multiply the fraction by $1=\dfrac{2n+18}{2n+18}$.

$$\begin{align} \frac{2n-18}{n^2-8n+8}\times\frac{2n+18}{2n+18}&=\frac{4n^2-324}{2n^3 - 16n^2 + 16n + 18n^2 -144n + 144} \\ \\ &=\frac{2n^2-162}{n^3-8n^2+8n+9n^2-72n+72} \\ \\ &= \frac{2n^2-162}{n^3+n^2-64n+72}.\end{align}$$ So now, given that $n\in\mathbb{N}_{\geqslant 7}$, we want to show that $$\begin{align} 2n^2 - 162 &< n^3 +n^2-64n+72 \\ \\ \Leftrightarrow n^2-234&<n^3-64n \\ \\ \Leftrightarrow n^2+64n-234&<n^3.\end{align}$$ So, since $n\geqslant 7$ then there will always exist a constant $c_n$ such that $$n^2 + 64n - (234+n^3-c_n)=0$$ iff (if and only if) $$\begin{align} n&=\frac{-64\pm\sqrt{4096+4(234+n^3-c_n)}}{2} \\ \\ &= 2\left(-16\pm\sqrt{1258+n^3-c_n}\right).\end{align}$$ But since $n>0$ then $$n=2\left(\sqrt{1258+n^3-c_n}-16\right).$$ Of course, to prove the inequality, we need to show that $c_n>0$. Of course $c_n\neq0$ because the constant must always be changing, depending on the value of $n$. Also, given the strict inequality, $c_n\not<0$, thus $c_n>0$.

This proves the inequality.

Mr Pie
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As the denominator $n^2-8n+8$ is positive for $n\ge7,$ the inequality $$\frac{2n-18}{n^2-8n+8}\lt1$$ is equivalent to the inequality $$2n-18\lt n^2-8n+8$$ which is equivalent to $$0\lt n^2-10n+26$$ or $$0\lt(n-5)^2+1$$ which is true.


(As presented, this is not a proof by induction. However, any proof can easily be converted into a proof by induction, in which the proof of the induction step makes no use of the inductive hypothesis.)

bof
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