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Question:

Prove for $n$ $ \in \mathbb N $ , $ 2 n\ -\ 18\ <\ n^2-8n\ +8 $

My attempt:

$ Base\ Case:\ n\ =\ 1,\ it\ holds. $

$I.H:\ Suppose\ 2k-18\ <\ k^2-8k+8,\ where\ k\ is\ a\ natural\ number.$

$ Then,\ \left(k+1\right)^2-8\left(k+1\right)+8\ =\ k^2+2k+1-8k-8+8\ >2k-18+2k+1-8$

I am stuck here. Any help would be appreciated. Thanks.

Daniel R
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  • You can't do induction because induction works for discrete, increasing unitary, values of $n$. Here we have $n \in \mathbb{R}$. – Ahmed S. Attaalla Jun 17 '17 at 01:11
  • In other words, induction works for integers only. – Toby Mak Jun 17 '17 at 01:11
  • Sorry it was suppose to be natural –  Jun 17 '17 at 01:11
  • There is a much more efficient way to prove this, by showing that the maximum value of $y$ in $(n^2-8n+8) - (2n-18)$ is less than zero. You can do this by using the vertex $x$-coordinate formula $-b/2a$, and then substituting the numbers and plugging them in the quadratic to find the $y$-value. This works for real numbers, so it must work for natural numbers as well. – Toby Mak Jun 17 '17 at 01:25

4 Answers4

3

Hint:  rewrite the given inequality as:

$$2 n - 18 \lt n^2- 8n + 8 \quad\iff\quad 0 \lt n^2 - 10 n + 26 \quad\iff\quad 0 \lt (n-5)^2+1$$

You can prove the latter by induction if it really needs be, but of course, it's painfully obvious what the direct proof is.

dxiv
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2

Notice, $2n-18<n^2-8n+8 \implies n^2-10n+26=(n-5)^2+1>0$.

Let $(n-5)=u$, so that we have $u^2+1>0$.

Moving over the $1$ gives us $u^2>-1$. We know this is true because all squares of real (and natural) numbers are greater than $-1$.

Therefore, we have $(n-5)^2>-1$, and $(n-5)^2+1>0$, and $n^2-10n+26>0$.

Adding $2n-18$ to both sides gives us $n^2-8n+8>2n-18 \implies 2n-18<n^2-8n+8$.

Thus, our proof is complete.

Saketh Malyala
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2

Assuming $\quad2n-18 < n^2-8n+8\quad $ then

$\begin{array}{ll} 2(n+1)-18 = (2n-18)+2 < & (n^2-8n+8)+2\\ &=[n^2-6n+1]-2n+9\\ &=[(n+1)^2-8(n+1)+8]-(2n-9)\\ \end{array}$

So the induction works when $2n-9\ge 0$ that is when $n\ge 5$

You'll get to verify the initial step for $n=5\ :\ \begin{cases}2n-18=-8\\ n^2-8n+8=25-40+8=-7\end{cases}$

For the cases $n\le 4$ you have to verify them manually.

Or go for the algebraic solution $n^2-10n+26=(n-5)^2+1>0$.

zwim
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1

When we group the terms on both sides, the expression will become clearer. Therefore:

$$2(k+1) - 18 < (k+1)^2 - 8(k+1) + 8$$ $$(2k + 2) - 18 < (k^2 + 2k + 1) + (-8k-8) + 8$$ $$(2k) + (2-18) < (k^2) + (2k-8k) + (1-8+8)$$ $$2k - 16 < k^2 - 6k + 15$$ $$0 < k^2 - 8k + 31.$$

Now convert $k^2 - 8k + 31$ into vertex form and show that the roots are not integers, which means that no integers satisfy the condition. Additionally, what is the discriminant the of the quadratic? What does this tell you?

Toby Mak
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