/sci/ - Science & Math

>>1987795
The only differnce is in the inductive step


regular induction:
assume works for (n=m )=> (works for n=m+1)

strong induction:
assume works for (n<=m) => (works for n=m+1)

\thread

>>1987815
ok, ok, so instead of proving it at the (k+1) level [strong induction]

I'd prove it for some integer between k and k+1?
wouldn't that just be direct proof? Wouldn't I already have proven that in my basis step, since if it's true for k it would be true for any k up to k+1?

>I'd prove it for some integer between k and k+1?
> integer between k and k+1
stopped reading there

>>1987827
why?

by k+1 I mean, proving it for any integer (which is the induction step in normal mathematical induction)

I guess I should have said prove it for an m between n and n+1 (this is what the first poster said)

Basically this

Induction: Because domino k falls, domino k+1 must fall.

Strong induction: All the dominos from 0 to k have fallen, therefore the k+1 domino must fall.

>>1987824
Your statement is fucking ridiculous !!!
U Troll ???

It doenst sound like you even know what regular induction is.

Please prove you understand regular induction, and then I will help you understand strong induction.

>>1987846
regular induction: if p(k) is true, then p(k+1) is also true

no troll

>>1987855
No, that is not regular induction!


Regular induction

Step 1)
Show p(n) is true

Step 2)
Assume p(k) is true
Show that if p(k) is true, it imples p(k+1) is true

Understand?

Wait, what I'm taught in school is along the lines of

IF:
-P(1) is true
-Assuming P(K) is true, we can show that P(K+1) is true

Then P(K) is true for all positive integers

So what did I leard

>>1987855
What? No. If p(k) is true and p(k+1) is true, then p is true for all k. That's induction.

>>1987855
Regular induction:
I am allowed to assume only p(k) to be true, I must then show this imples p(k+1) to be true

Strong Induction:
I am allowed to assume that p(k) is true for all k<=n, I must then show this imples p(n+1) is true

TITS NAO!!!

>>1987871
regular induction

>>1987870
yes, I understand. I just didn't word it well.

basically, since you're assuming p(k) is true, you show that p(k+1) is true, which means it is true for every k

but how would the induction step go in strong induction?

let's say P(n) is the statement
P(1) is true

so, for each k>=1, if P(m) is true (m < k) then P(k) is true [how would I prove this part]

and then, if P(k) is true P(n) is true for n=>1?

>>1987879
>>1987890
keep posting dude ! A+

>>1987895
WTF?

just stop! YOU ARE DOING IT WRONG!
Listen!

let's say P(n) is the statement

Step 1):
Show P(1) is true

Step 2):
For some m, assume that P(k) is true for k<=m
Hence we assume
P(2), P(3), P(4), P(5),....,P(m)
are true

Now, with these assumtpions I must show that
P(m+1) is true


Get it?

>>1987921
FYI: Jennique Adams is her name

No hardcore of her though :(
Cant find anything other then her giving a handjob though :(

Fun fact: Adventure Time is based in a post-apocalyptic time where humans have been wiped out

>>1987959
I thought it was a child in a coma?

>>1987975
Nope. Even little skeletons in sittin down watching some tv too.

>>1987928
ok, I think I'm catching on here

so, instead of in normal induction where, we assume p(k) is true, and use that assumption to prove p(k+1) is true (true for every k)

we assume p(k) holds true up to some m [k<=m]

then we can use this assumption to prove p(m+1) is true

>>1987959
yes, and Dungeons and Dragons

>>1988012
YES! YOU GOT IT!

>>1987824
You think theres an integer between two consecutive integers? You're hopeless.

>>1988027
I think OP already figured out his mistake
OP is a pretty cool guy

>>1988030
I am loving these and I don't care for hardcore, double win!

>>1988030
>>1988030
this is going in my schoolgirls folder..

>>1988041
This is Jana Defi

OP goes to GSU..... maybe.