Mathematical Induction: Involving Inequalities

[ku1005]

Hey guys, with these Mathematical Induction probs, i seem to be able to do those involving "=" signs, but find it really hard to follow the methods I have seen where inequalitites are involved?...for example:

RQP: 6n+12<3^n for every integer n>= 4

I did:

Let S(n) be the statement "6n+12<3^n for every integer n>= 4"

1) Prove S(4) is true:

6(4)+12< 3^4
36<81 Therefore S(4) is true


2) Let k be in the set of positive integers where k>=4 and assume S(k) to be true, therefore

S(k) = 6k+12<3^k

Now proove k+1 is true

6(k+1) + 12 < 3^(k+1)

~ (6k+12) + 6 < (3^k) . 3

In this last step, I recognise that on one side we have added 6 to the original LHS while on the other side we have multiplied RHS of original by 3

BUT....im unsure as to how I use this efficeintly to prove?

any pointers as to where to go from here would be appreciated greatly !!!

cheers

rhys

 

[daon]

I think this problem may belong in the "Advanced" section for future knowledge.

Assume $\displaystyle 6k+12 \,\, < \,\, 3^k$ then,

$\displaystyle 6(k+1) + 12 = (6k+12) + 6 \,\, \le \,\, 3^k+6$

Since $\displaystyle k \,\, \ge \,\, 4$, $\displaystyle 3^k \,\, \ge \,\, 3^4 \,\, = 81 > 6$

Thus, $\displaystyle 3^k+6 \,\, < \,\, 3^k + 3^k + 3^k = 3^k(1+1+1) = 3^{k+1}$

 

[soroban]

Hello, ku1005!

Prove: $\displaystyle \,6n\,+\,12\:<\:3^n$ for every integer $\displaystyle n\,\geq\,4$

I did:
. . Let $\displaystyle S(n)$ be the statementL $\displaystyle \:6n\,+\,12\:<\:3^n$ for every integer $\displaystyle n\,\geq\,4$

1) Prove S(4) is true:
. . $\displaystyle 6(4)\,+\,12\:<\: 3^4$
. . $\displaystyle 36\,<\,81\;$ . . . Therefore $\displaystyle S(4)$ is true.


2) Let $\displaystyle k$ be in the set of positive integers where $\displaystyle k\,\geq\,4$ and assume $\displaystyle S(k)$ to be true.

Hence, $\displaystyle \:S(k):\;6k\,+\,12\:<\:3^k$

Add $\displaystyle 6$ to both sides: $\displaystyle \:6k\,+\,12\,+\,6\:<\:3^k\,+\,6$

. . And we have: $\displaystyle \:6(k\,+\,1)\,+\,12\;< \;3^k\,+\,6\;$ [1]


Consider: $\displaystyle \:3 \:< \:3^k$ for $\displaystyle k\,\geq\,4$

Multiply both sides by $\displaystyle 2:\;6\:<\:2\cdot3^k$

Add $\displaystyle 3^k$ to both sides: $\displaystyle \:3^k\,+\,6\:< \:3^k\,+\,2\cdot3^k \:=\:3\cdot3^k\:=\:3^{k+1}$

Then [1] becomes: $\displaystyle \:6(k\,+\,1)\,+\,12\:<\:3^k\,+\,6\:<\:3^{k+1}$

. . Therefore: $\displaystyle \;6(k\,+\,1)\,+\,12\:<\:3^{k+1}$


We have proved $\displaystyle S(k+1)$ . . . the induction proof is complete.

 

[ku1005]

sorry daon...i didnt realise was advanced.....also...how did you know to do what you did???....like does that come with practice???...irrespective...thanks for you help!!...i understand your workings fine!