Mathematical Induction

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[late tinsel]

You have $3^{k+1}>2k\cdot 3=6^k$ in your part a

[tardy dove]

Omegabet_

[vague cliff]

oh good catch thank you

[scarlet tide]

@vague cliff has given 1 rep to @late tinsel

[vague cliff]

otherwise is my work correct?

[late tinsel]

but you also just conclude 6k>2k+2 without justification, depending on your prof might want to add the small justification there

$k\geq1\implies 4k\geq4>2$ to which it follows

[tardy dove]

Omegabet_

[late tinsel]

your b step is poor as well

you assumed the thing you're trying to prove, then went from there

[vague cliff]

so could I state 2k + 2k + 2k > 2k + 2 because n >= 1

[late tinsel]

if k>=1, then clearly 4k>=4

and 4>2

so 4k>2

hence 6k-2k>2, etc

[vague cliff]

Im just kinda confused is stating for part b k^2 + 2k + 1 > 4k + 2 not enough?

because we know k^2 > 2k + 1

so 2k + 1 + 2k + 1 = 4k + 2

[late tinsel]

for b, you started with assuming $P(k+1)$ was true

[tardy dove]

Omegabet_

[late tinsel]

since you have $(k+1)^2>2(k+1)+1$ then did work starting from that

[tardy dove]

Omegabet_

[vague cliff]

would I instead do $2^{k+1}>4k+2$

idk how this works lol

[tardy dove]

Sef

[vague cliff]

sorry wrong one

$(k+1)^2>4k+2$

[tardy dove]

Sef

[late tinsel]

why would you do that...?

nor has that been proved true

you start with one side of the inequality, and work until you get the other

just like trig identities in hs

[vague cliff]

well $(k+1)^2$ is $k^2+2k+1$ and we showed that $k^2>2k+1$

[tardy dove]

Sef

[late tinsel]

you assumed that as your IH

so yes, $(k+1)^2=k^2+2k+1>(2k+1)+2k+1=4k+2$, that's a valid start of the inductive step

[tardy dove]

Omegabet_

[vague cliff]

then from there am I looking to make $4k+2$ like $2k+3$?

[tardy dove]

Sef

[late tinsel]

you're trying to show 4k+2>2k+3 at least

[vague cliff]

so can i say we know that $k>=3$ so because of that can we say it is at least $2k+9$

[tardy dove]

Sef

[late tinsel]

$4k+2>2k+3\iff 2k>1$

[tardy dove]

Omegabet_

[vague cliff]

so i just need to state when it is larger?

[late tinsel]

you need to prove 2k>1 whenever k>=1

[vague cliff]

Would something like this be valid

[late tinsel]

you havent said why 4k+2>2k+3

you again just asserted it to be true

[vague cliff]

how can show that though other than stating $n>=3$

[tardy dove]

Sef

[late tinsel]

the obvious way?

$k\geq 3\implies 2k\geq 6>1$

[tardy dove]

Omegabet_

[late tinsel]

hence $2k>1\iff 4k+2>2k+3$

[tardy dove]

Omegabet_

[late tinsel]

when you first start on proof writing, you need to understand that even small 'obvious' details play a key role

[vague cliff]

im still confused on where the 2k is coming from

[late tinsel]

.

from the hs algebra I omitted

cause I assume you can do that algebra

[vague cliff]

lol yes I can however the way you approached it I guess is just going right over my head

[late tinsel]

you can argue I just 'simplified' the inequality

we tend to work better with simpler objects

2k>1 is much simpler than 4k+2>2k+3

[vague cliff]

oh wait did you just substract the 2k from both sides and 2 from both sides

[late tinsel]

yes

again, the high school algebra lol

[vague cliff]

Yeah that went right over my head lol

so from there how does that prove that $k^2+2k+1>2k+3$

[tardy dove]

Sef

[vague cliff]

or is that all

that just proves it

oh wait $2k>1$ is always true because $k>=3$

[tardy dove]

Sef

[vague cliff]

or is it because since this is true then $(k+1)^2>2k+3$

[tardy dove]

Sef

[late tinsel]

you showed by the IH that $(k+1)^2>4k+2$

[tardy dove]

Omegabet_

[late tinsel]

and I showed $2k>1$ is true, hence $4k+2>2k+3$

[tardy dove]

Omegabet_

[late tinsel]

so $(k+1)^2>2k+3$, to which it follows

[tardy dove]

Omegabet_

[vague cliff]

ahhh okay okay so I was just missing showing why $4k+2>2k+3$

[tardy dove]

Sef

[late tinsel]

yes

general rule of thumb, if something's not brutally obvious, you need some amount of justification. The problem with this rule of thumb is that at the beginning of a proofs course, you generally should belabor points

[vague cliff]

Okay I really appreciate the help

in #helper-nominations can I vote for you or not possible?

also just to make sure I have the concept down this is correct right?

[late tinsel]

looks fine, just could be formatted better to make it nicer to read. For example, proofs should have very little logic symbols generally, logic symbols are for 'drafts'

$(k+1)!=(k+1)k!>(k+1)2^k$ by the inductive hypothesis. Since $k\geq 4$, $k+1>2$, so $(k+1)2^k>2^{k+1}$, hence $(k+1)!>2^{k+1}$

as an example

[tardy dove]

Omegabet_

[vague cliff]

+close