Question

Positive integer domain.

When is: n+6 < (n^2-8n)/16

and prove by mathematical induction.

Answer 1

What do you mean by positive integer domain? You mean n=1,2,3,.. or that the expressions n+6 and (n^2-8n)/16 are both positive integers?

Answer 2

n>0 ; but seeing how this is not true for all n>0 we have to find out when it is to begin with in order to establish a basis step.

Answer 3

Oh I see. So you want first to find the domain at which the inequality is valid, and then prove that by induction.

Answer 4

correct

Answer 5

finding the domain is simple enough with a quadratic; and turns out to be n>27

Answer 6

its the induction par that the teacher had a problem with :)

Answer 7

I did it, but not entirely sur7e about it. As you said this is true only for n>27. We can see that it's true for n=28, since 34<35. Now, let's assume it's true for n=k and write
\[k+6<{k^2-8k \over 16} \implies k^2-24k>96.\]
Now we should prove based on our assumption about n=k that the relation is true for n=k+1.
\[k+1<{(k+1)^2-8(k+1) \over 16} \implies k^2-6k>16k+112 \implies k^2-24k+2k>96+16.\]
We already have k^2-24k>96, so we should have 2k>=16 which gives k>=8. This is true since we've said from the beginning that n>27.

Answer 8

There's a missing piece there that says
\[\implies k^2-24k+2k>96+16\]

Answer 9

(k+1) +6 < ..... right?

Answer 10

your on the right track at least ;)

Answer 11

Yeah of course, just a typo :D

Answer 12

the rhs is what needs to be proven; so the teach said that its best not to mess with that end. but to work out the lhs

Answer 13

the inequality you have given can be simplified to:\[n^2-24n-96>0\]using induction, we can show this is true for n=28 and then assume its true for n=k to get:\[k^2-24k-96>0\]we then try and prove its also true for n=k+1. so for n=k+1, we get:\[(k+1)^2-24(k+1)-96=(k^2-24k-96)+(2k-23)\]but we already know \(k^2-24k-96>0\), so we are left to show \(2k-23>0\) which is true for \(k>11\)

Answer 14

assume its true for k
\[k+6<\frac{k^2-8k}{16}\]
work it for (k+1)
\[(k+1)+6<\frac{k^2-8k}{16}+C\]
\[(k+6)+1<\frac{k^2-8k}{16}+1\]

Answer 15

the proofing then amounts to working out the "add on" parts to make sure they are always in cahoots

Answer 16

I think you could extend the proof to show that it is not true for n=27, assume its not true for n=k and then prove that its also not true for n=k-1. a kind of "reverse induction".

Answer 17

thereby proving its not true for n<28

Answer 18

proof by reduction? lol

Answer 19

You can write the RHS as \(\frac{k^2-6k}{16}=\frac{k^2-8k}{16}+\frac{2k}{16}\), which implies the condition \(\frac{2k}{16}\ge 1 \implies k\ge8\). SAME THING!

Answer 20

\[(k+6)+1<\frac{k^2-8k}{16}+1\le \frac{(k+1)^2-8(k+1)}{16}\]
\[(k+6)+1<\frac{k^2-8k}{16}+1\le \frac{k^2-6k-7}{16}\]
\[\frac{k^2-6k-7}{16}+\frac{23-2k}{16}\le \frac{k^2-6k-7}{16}+0\]
\[\frac{23-2k}{16}\le 0\]
\[23-2k\le 0\]
\[23\le2k;\ k>27\]
\[23 \le 2(28)\text{ is true}\]

Answer 21

i wonder if wed have to prove that 23<= 2k by MI :)

Answer 22

I don't think so, since we want to prove it for n>27.

Answer 23

23/2 <= k

11.5 <= k ; and k is always greater than 11.5 would suffice for me