/sci/ - Science & Math

[Anonymous No.9672988]

> hurr lets just assume it works for the next step so it must work for every step
anons, why the fuck is this Deus ex machina allowed?

[Anonymous No.9672991]

>>9672988
it's not an assumption, fucking retard

[Anonymous No.9672996]

>>9672991
fine
>hurr we've proven its true for the next step so it must be true for all steps

[Anonymous No.9673001]

>>9672996
B-but anon, that’s how steps work...

[Anonymous No.9673003]

>>9673001
>steps are always self similar and only go in one directionat all cases
imagine being this brainlet

[Anonymous No.9673005]

>>9672988
you prove that it works for an arbitrary step. It's like a series of dominos, if you prove that if one piece falls, then the next piece will also fall for two dominos, and you prove that the first domino falls, you know every domino will fall.

[Anonymous No.9673007]

>>9673005
>muh dominos
this is exactly whats retarded about it, what if there's a domino that is of a different material/larger than the others that wont fall?

[Anonymous No.9673009]

>>9672988
Im not sure whats so fishy about it. If, for some arbitrary [math]i[/math], we have that [math]q_{i+1}[/math] holds true given that [math]i[/math] if [math]q[/math] holds true then obviously if [math]q_0[/math] holds true then does [math]q_1[/math] so n so forth..

[Anonymous No.9673011]

>>9673007
then you're unable to prove that it works for all dominos.

[Anonymous No.9673012]

>>9673009
Should be "holds true given that q holds true" without the "that i"..

[Anonymous No.9673016]

>>9673012
>>9673011
>>9673009
>one horse has a color, therefor all horses of that group are of one color
>now lets assume that n horses have the same color
>lets make that a group of n+1 horses
now we can only look at the n horses and say there are of the same color, or look at the horses without the first one and they also will have the same color
>therefore, if n horses will have the same color, n+1 horses will also the same color
>thus, all horses must be of the same color
WHERE IS YOUR INDUCTION GOD NOW

[Anonymous No.9673022]

>>9673016

[Anonymous No.9673023]

>>9673016
holy shit

[Anonymous No.9673025]

>>9673023
>>9673022
tell me pls, whats wrong about my argument

[Anonymous No.9673027]

>>9673025
your "proof" states that if you only expand your group of horses with the same color as every horse the group contains, then given a initial group of one horse every horse in that group will be of the same color.

[Anonymous No.9673029]

>>9673027
yes, but thats how induction works

[Anonymous No.9673031]

>>9673025
You assume the result. You use that every collection consisting of a single horse has the same color. That is, you first assume all horses have the same color.

[Anonymous No.9673032]

>>9673029
but you draw the wrong conclusion because you do not consider that there is a constraint on how you increase group size.

[Anonymous No.9673034]

>>9673031
thats literally how induction works
>>9673032
what constraint is there

[Anonymous No.9673035]

>>9673031
Oh wait you obviously meant the collection to have the same color as each other.

>>9673032
This is the right answer. You can go from the base case to a pair of horses.

[Anonymous No.9673036]

>>9673034
That you can only add a horse that matches the color of every horse in the group

[Anonymous No.9673037]

>>9673035
*can't

Take one red horse and one blue horse. The red horse and blue horse each have the same color when considered as collections of a single horse, but that does not imply they have the same color as each other. They don't.

[Anonymous No.9673041]

>>9673037
>when considered as collections of a single horse
exactl, but thats the premise of proof by induction

[Anonymous No.9673047]

In strictly logical terms, what you want to prove is

∀n: p(n)

And you can do this by proving

∀(i): p(i) → p(i+1)
and p(1)

[Anonymous No.9673051]

>>9672988
Then what's the smallest natural number at which the proof stops working?

[Anonymous No.9673052]

>>9673051
54 nigga

[Anonymous No.9673054]

>>9673051
Wrong. It works at 53 so it also works at 54.

[Anonymous No.9673057]

>>9673054
Meant to quote this guy >>9673052

[Anonymous No.9673058]

>>9673016
while proving n horse has the same color implies n+1 horses have the same color, you are assuming n>2. note your method of proving doesn't work for a group of two horses

[Anonymous No.9673059]

>>9673058
why shouldnt it? if it works for one, then it should for two? isnt that how induction works?

[Anonymous No.9673065]

>>9673059
it doesn't work for 2

[Anonymous No.9673066]

>>9673059
your proof of n=>n+1 is wrong. proof doesn't work if n=2

[Anonymous No.9673068]

>>9672988
>another millenial thinks he thought of something no one else has
part of induction is the assumption that the problem is of the type for which induction is valid

>>9673016
proof breaks immediately at n=2 because when you "look at the horses without the first one" you will be looking only at the single horse which was added.
the n+1=2nd horse can be purple with the n=1st horse being grey without violating the assumption that n horses have the same color.

[Anonymous No.9673084]

>>9673068
>>9673066
>>9673065
if you fags are gonna argue against faulty induction, at least know who induction works

[Anonymous No.9673088]

>>9673084
how*

[Anonymous No.9673100]

>>9673084
So we are assuming k horses is a sub-group of k+n horses (the last k horses) and since k+1 falls between k and k+n all k+1 horses are the same color?

I don't really understand. It sort of makes sense in a way but not completely. I can't decide if that's a reasonable assumption or faulty logic so I'm leaning towards faulty.

[Anonymous No.9673105]

>>9673084
You're creating two different groups with at least one horse in common. That's obviously impossible at n = 2

[Anonymous No.9673107]

>>9673100
Consider two horses and try to divide the group in the same way as the picture.

[Anonymous No.9673108]

>>9673105
>>9673107
you got me anons

[Anonymous No.9673111]

>>9673108
but the induction in itself isnt faulty

[Anonymous No.9673259]

>>9672996
>we've proven it's true for the next step from an arbirtary step
>therefore it's true for the step beyond that
>therefore it's true for all steps, because all steps are the "next step on" from another arbitrary step

[Anonymous No.9673276]

>>9673084
>he's this far into the semester and still struggling with chapter 0 of sipser
you're fucked faggot

>>9673111
my turn to ask you a question since you "understand" induction: is the proof valid if we set the base case as n=2 ?

[Anonymous No.9673314]

>>9672996
If you actually don't understand this you're a fucking brainlet and need to get out of STEM

[Anonymous No.9673417]

>>9673276
>>9673314
>retards who are taiking the bait
ITT op demonstrated that he understands the concept of induction, how it works, and even jokingly brought up paradoxes associated with it. do you unironically believe he's struggling with it?

[Nigger Business No.9673963]

>>9673025
Every class that teaches induction as a part of its curriculum uses THIS. EXACT. EXAMPLE as a common error that retards commit when using induction for the first time. It doesn't work because if there are n horses on Earth total, then there aren't n+1 horses.
You think this is a """""""""""""""flaw""""""""""""""" because you are a retard.

[Anonymous No.9673968]

>>9672988
Nothing succeeds like success

[Anonymous No.9674021]

It's an axiom

[Anonymous No.9674034]

>>9673417
>Damage controlling this hard
>Not answering the question

[Anonymous No.9674037]

>>9672988
Oh I see. You happen to be a brainlet.
Don't sweat it, billions of humans suffer the same fate.

[Anonymous No.9674236]

>>9672988
I love threads like these
Thanks OP

[Anonymous No.9674302]

>>9674021
Well ordering for natural numbers shouldn't cause that much controversy.

[Anonymous No.9674307]

>>9672988
is this just induction or am i as dumb as a black person

[Anonymous No.9674782]

>>9673417
Yes, because its a lot more likely he is actually that much of an idiot.

[Anonymous No.9674797]

>>9673003
What are definitions by recursion?

[Anonymous No.9674805]

>mathematical induction is the same as induction
>you can’t prove mathematical induction from axioms

[Anonymous No.9674839]

>>9672988
You forgot the base case, nigger