/sci/ - Science & Math

>>14699120
>assume N=0
>No horses have the same color

>>14699120
>What was your first answer to the polya fake induction?
I trusted the science, and supposed I had some sort of reverse color blindness.
#fuckTrump #fuckWhitePeople #fuckAntiVaxxers

>>14699120
My first answer to this is that it's fucking incoherent and whoever came up with it is an obvious brainlet.

>>14699120
>>Therefore all N+1 horses have the same color.
Why?

>>14699155
The proposition is trivially true for N=0, brainlet.

I don't know if my answer is correct because I'm not a mathfag, but i would say that induction can only be used for cases where it has been proven to work. Because nobody has proven that induction works on the color of horses, then you can't use induction.
Why induction can be used on real numbers, and how would you go around proving that is way above my level

>>14699120
The point of proof by induction is to show that if a statement is true for one element of a sequence, it is necessarily true for the next one by merit of the relationship between them. This just plainly doesn't apply here. Excluding the last horse will give you the previous element of the sequence, but excluding the first one (or any horse other than the last, for that matter) gives you something arbitrary that isn't part of the sequence at all.

>>14699185
I’m sorry to hear that you’re retarded.

>>14699219
His post doesn’t contradict yours, [math]\sqrt{\mathrm{brainlet}}[/math].

>>14699225
Your answer is incorrect. There is an error in the proof, and your statement about induction is nonsense.

>>14699246
You've already admitted to having an IQ of 90 and now you double down on it.

>>14699120
This argument relies on the fact that the intersection of the first n horses with the last n horses is nonempty. The reason the inductive argument doesn't work is because the inductive step relies on n being >=2.

>>14699242
>excluding the first one (or any horse other than the last, for that matter) gives you something arbitrary that isn't part of the sequence at all.
You’re wrong. The sequence consists of the numbers of horses, namely 1, 2, 3, …. When proving that the statement is true for [math]N+1[/math] horses, it’s fine to assume that it is true for any choice of [math]N[/math] horses.

>>14699255
Nigger, what are you talking about?

>>14699270
Are you OP? Every post of yours is pure mental illness. It sounds like you just found out your IQ is 90 and you're having an episode.

>>14699120
You are considering two different sets of N horses when you say they have the same color. Proof by induction requires 1 set of N elements, not two different sets. You are simply not following the induction method at a logical level

>>14699120
>Theorem. All horses are the same color.
>Proof. We’ll induct on the number of horses. Base case: 1 horse. Clearly with just 1 horse, all horses have the same color
You haven't counted all of them so you can't say that

>Well, given any set of N+1 horses, if you exclude the last horse, you get a set of N horses. By the inductive step these N horses all have the same color. But by excluding the first horse in the pack of N+1 horses, you can conclude that the last N horses also have the same color. Therefore all N+1 horses have the same color. QED
Non sequitur, the excluding horse could not be the same color

>>14699270
>The sequence consists of the numbers of horses, namely 1, 2, 3, ….
That obviously can't be the case, you dumb mongoloid, since you keep referencing the colors of the horses.

>>14699120
When you make an induction argument, you are assuming a set of N blue horses exists, not that every set of N horses are blue like is assumed in the problem.
Any is not Every, don't confuse the two.

>>14699323
/thread

>https://en.wikipedia.org/wiki/All_horses_are_the_same_color
>The argument above makes the implicit assumption that the set of n+1 horses has the size at least 3, so that the two proper subsets of horses to which the induction assumption is applied would necessarily share a common element. This is not true at the first step of induction, i.e., when n+1=2
What is this utter brainletism? You could circumvent this entirely by claiming that all groups of more than 2 horses are of the same color and the induction would still be invalid for the same reasons.

>>14699155
https://en.wikipedia.org/wiki/Vacuous_truth

>>14699246
real brainlet hours

>>14699328
that actually makes perfect sense. it is an IMMENSELY MATHEMATICALLY OBTUSE way of saying:
>there is no reason to assume that 2 horses will have the same color
since the flaw in logic so basic, the explanation becomes obtuse to the degree of unintelligibility

this is typical in math. think back to Russel proving 1+1=2 in 200 pages.

>>14699278
I’m not OP.
>Every post of yours is pure mental illness.
If you e.g. mean the OP, this is obviously not true. Regardless of its formulation (which isn’t even bad, anyway), it’s a nice ‘fake induction’ because the error (explained by >>14699267) is easy to miss if you’re not used to thinking through these things carefully. If you are at all familiar with proofs by induction, you should easily understand what OP is trying to say. Even if it had been very poorly written, though, it obviously wouldn’t be pure mental illness (even if it had been written by someone mentally ill), because there would still be something of value in there (namely the fake induction). If you say that it’s indistinguishable from mental illness, you are either greatly exaggerating (to the extent that it would apply to many things, such as your own post, probably) or stupid enough that you don’t understand the idea of this ‘fake induction’ trick regardless of how it’s explained (although you delude yourself into thinking that any explanation of it is simply incoherent).

>>14699313
Nigger, the horses are quantified locally for every element of the sequence. How the hell do you propose that logical reasoning should work? With your way, you couldn’t prove e.g. ‘every horse is a horse’, because you apparently have to choose a particular horse one at a time. And before you say that ‘every horse is a horse’ doesn’t need to be proved, or whatever, the point is that there is clearly a problem if such a simple true statement cannot be proved in your way.

>>14699323
[math]\exists[/math] vs. [math]\forall[/math] being used in the statement is orthogonal to induction itself. In this case, the statement is using [math]\forall[/math].

>>14699328
How would you prove the base case of 2 horses, though? The thing is that it’s true for 1 horse but not for 2 horses, so you won’t be able to do so.

>>14699353
It makes no sense. Using this logic, I could prove that if all groups of 3 horses have the same color, then all groups of more than 3 horses do as well, which is obviously a total nonsequitur.

>>14699353
>think back to Russel proving 1+1=2 in 200 pages.
Was he successful at least?

>>14699361
>It makes no sense. Using this logic, I could prove that if all groups of 3 horses have the same color, then all groups of more than 3 horses do as well, which is obviously a total nonsequitur.

no you wouldnt, because then you would lose the connection to the base case, N=0. The fact that you fail to have a connection to the base case causes the further steps to become incoherent. This is not a bug, it is a feature. The steps in an inductive argument cannot be expected to be sensible if it has no connection to the base case.

>>14699356
No, >>14699267 is a common false (non)-explanation, of course a set of 2 horses of the same color exist. >>14699323 is the correct explanation.

>>14699365
he generates a logical frame work, and stipulates a certain logical structure corresponds to "addition operation" which is consistent with the underlying/simpler logical axioms. so he doesnt so much prove that 1+1=2, he rather defines the addition operation in a more basic logical structure.

>>14699370
>no you wouldnt, because then you would lose the connection to the base case, N=0.
Again, a total nonsequitur.

>>14699380
>Again, a total nonsequitur.
Again, a total nonsequitur.

>>14699386
Nothing to do with N=0 or any base cases. You can forget about base cases completely. They have nothing to do with what I said, brainlet.

>>14699388
>They have nothing to do with what I said, brainlet.
Again, a total nonsequitur.

>>14699388
>it doesnt matter if an inductive argument connects to initial case

>>14699394
I accept your full and direct concession. Base cases have nothing to do with this. Call me back when you finish middle school and understand the actual logic behind proof by induction, brainlet.

>>14699399
Read and re-read >>14699361 until you grasp what it says, trog.

>>14699380
prove the base case for N=3.

the whole point of the base case is that the base case is trivially true for N=0 and N=1. If you want to start the base case at N=3, lets see you prove the base case for N=3, and then you may expect a serious response from me.

>>14699361
>I could prove that if all groups of 3 horses have the same color
THE POINT IS THAT IT IS IMPOSSIBLE TO PROVE SUCH A BASE CASE. attempt it, I'm waiting.

>>14699408
>>14699412
>but muh base case
Absolutely braindamaged. Read >>14699361 over and over until you wrap your bot "minds" around what it says. It clearly doesn't involve proving any base cases.

>>14699418
https://en.wikipedia.org/wiki/Principle_of_explosion
>>14699361
>I could prove that if all groups of 3 horses have the same color

>>14699376
No, >>14699267 is correct. The fact that 2 horses of the same colour exist is irrelevant, because the ‘question’ is whether every 2 horses have the same colour. >>14699323 is fucking-stupidly wrong, because it is exactly right that you use the same kind of quantifier in the induction step as in what you’re proving. Also, I think he has (funnily enough) confused the two things he’s trying to distinguish, because ‘any’ as used in the OP is in fact synonymous to ‘every’ ([math]\forall[/math]), though it is true that it is in some contexts synonymous to ‘some’ ([math]\exists[/math]). I’ll give an example of each:

‘For every [math]x\in A[/math], it holds that …’ means the same as ‘For any [math]x\in A[/math], it holds that …’.
‘If there is any [math]x\in A[/math] for which …, then …’ means the same as ‘If there exists an [math]x\in A[/math] for which …, then …’.

(This is really just a quirk of the English word ‘any’, though.)

>>14699380
Shut up.

>>14699120
>I was studying apostol calculus book on my own. my answer was that it was based on a false assumption because not all girls are blonde (apostol version). When I saw the answer I felt so embarrassed that I left self studying forever and signed up for virtual courses at Indian universities.
It's not really related to analysis, it's maybe foundations of analysis, or set theory. You should continue your study of analysis and just keep in the back of your mind in case you start making complicated induction arguments.

>>14699419
LOL. Absolute mongoloid.

Reminder to drink water and have some citrus.

>>14699419
>>14699435
I think I see where you two are kind of talking past each other. The thing is that the induction would actually be valid if it started at [math]N\ge2[/math] (and is therefore not ‘countered’ by the quoted sentence from Wikipedia, if it doesn’t start at [math]N=1[/math]). One of you is complaining about this because the statement would still be false so you think the induction must still be invalid. However, the reason for the statement being false in this case is not that the induction step is invalid (and it isn’t), but rather that the base case is ‘invalid’ (by which I just mean that you can’t prove it). This distinction is probably taken for granted by the other one of you, which is why you bring up the base case, because although the base case is not relevant to the induction step, it is obviously relevant to the proof as a whole, and so its invalidity for [math]N\ge2[/math] explains why it’s not a problem that the induction works then.

>>14699444
Thanks.

>>14699463
>the reason for the statement being false in this case is not that the induction step is invalid (and it isn’t), but rather that the base case is ‘invalid’ (by which I just mean that you can’t prove it)
Okay. You live in a universe where it just so happens that all groups of 3 horses are always of the same color. Does that mean all groups of more than 3 horses must be of the same color in this universe? Even someone as dense as you should be able to see that it's a nonsequitur.

>>14699465
>You live in a universe where it just so happens that all groups of 3 horses are always of the same color. Does that mean all groups of more than 3 horses must be of the same color in this universe?
Yes. Otherwise, I could find horses of at least two different colours by gathering more than 3 horses, and then I could remove a few of the horses until there are only 3 left while keeping at least one horse of each of at least two colours. I would then have a group of 3 horses that are not all the same colour, which would contradict the fact that every group of 3 horses has a single colour within the group.
>Even someone as dense as you should be able to see that it's a nonsequitur.
I can see that someone as dense as you are is probably a nigger.

>>14699484
>Yes.

>>14699484
You're the dumb nigger here, clearly, since all you're really doing is alter the scenario into a universe where every subset of three horses has the same color, which is the only one where your false induction step applies.

>>14699488
>>14699465
>Okay.

>>14699501
Selfies go on /soc/. Call me back when you finish middle school. The amount of retards on this board who lack basic education is staggering.

>>14699494
If you look at the post that I replied to, you’ll see that the scenario was given by the other poster. So what the hell are you talking about?

>>14699505
You’re a funny troll, I’ll give you that.

>>14699513
Alright. Here's a scenario for you, you dumb niggers:
There is exactly two groups of horses in this universe:
Black, black, black
Black, black, black, white
All groups with three horses are of the same color. Why does your retarded induction step not work?
>b-b-b-b-but I can rearrange the horses into different groups, therefore not all groups of 3 are the same color
Kill yourself. Your retarded proof is about groups of horses, not every possible permutation of groups of horses.

>>14699520
>Your retarded proof is about groups of horses, not every possible permutation of groups of horses.
You mean subset, not permutation. And no, that is not a valid universe in this context, because you don’t get to specify the groups of horses additionally to the horses themselves. Obviously, the universe should consist of a set of horses, and they can be grouped however one pleases. This is really clear if you look at what is actually being proved. Intuitively, we think of it as grouping horses, but the proof actually concerns all horses in the universe, and regardless of how you have ‘grouped’ the horses in your universe, we can reach your universe via induction by adding one horse at a time to each group in your universe, so to speak. Note that we are not really changing ‘your groups’, we’re just considering different universes with groups going from empty to what yours look like. It is then easy to find a counter-example to ‘all groups of three are single-coloured’, for example the universe with two black horses and one white horse. I agree that you can make up groups that satisfy certain conditions and then decide that no other groups are allowed, but that is much less interesting and it is not what is being discussed here.

Induction is just horseshit. Pun slightly intended.

>>14699484
>>14699505
the person you are arguing with is probably mentally retarded. i am so so sorry you had to waste your life trying to argue with him like I did

All horses ARE the same color. I don't need a fancy math proof to tell me what my lived reality is

>>14699242
only correct answer

Jfc, this board is so terrible. Imagine not being able to answer a basic first class high school level discrete mathematics question

>>14699612
Thanks (I think), but, to clarify:

Me:
>>14699195
>>14699246
>>14699270
>>14699356
>>14699420
>>14699463
>>14699484
>>14699501
>>14699513
>>14699549

Probably him:
>>14699465
>>14699488
>>14699494
>>14699505
>>14699520
… and other posts farther up in the thread, I’m guessing.

Is this you?:
>>14699267
>>14699353
>>14699370
>>14699386
>>14699408

>>14699627
yes, the last 4 is me, but not the 5th from bottom.

>>14699650
Ah.

>>14699365
Not really. He sneaks in a couple of unspoken assumptions along the way.

>>14699465
>You live in a universe where it just so happens that all groups of 3 horses are always of the same color. Does that mean all groups of more than 3 horses must be of the same color in this universe?
it depends on the definition.
if I have 6 horses I can divide them into two groups of 3 horses, which therefore must be of the same color. this would work not for all numbers but at least for the multiples of three.
the problem with your example is that you are not using mathematical notation so it's not clear what you mean.

>>14699549
>And no, that is not a valid universe in this context, because you don’t get to specify the groups of horses additionally to the horses themselves.
Stopped reading. Notice how your moronic "proof" requires you to do all of these mental gymnastics, while valid ones work in scenarios like the one I presented? What does it tell you?

>>14699120
>>Well, given any set of N+1 horses, if you exclude the last horse, you get a set of N horses. By the inductive step these N horses all have the same color.
Doesn't follow.

>>14699549
Okay. You're right. I was the dumb nigger all along and somehow got into my head that it was trying to prove something other than what it was. Sorry for wasting your time.

>>14699120
>What was your first answer to the polya fake induction?
I saw it right away that the induction step does not work for n=2. not everyone is a brainlet here.

>>14701413
I don't know, you sound like a gigabrainlet to me.

>>14699120
I think the error is that you’re assuming that by saying these N horses all have the same color, you can say any group of N horses all have the same color, which is obviously wrong unless you’re dealing with strictly interchangeable objects (which horses obviously aren’t) and at that point the whole proof would be pointless anyway because it would be tautologically true.

>>14702093
It's actually a valid test. Think about it outside the context of this proof. The idea is to apply it recursively. It just breaks down when N=2 because then it is no longer applicable.

>>14699242
Filtered.

>>14699120
The two sets of N do not necessarily have a non-empty intersection. Therefore, the set of N+1 is not necessarily the same color.

>>14700316
That’s okay, we’re both white. :) I can see how it would seem wrong based on how you initially read it; thanks for the clarification. I just wish there wouldn’t be so many over-the-top insults thrown before the two sides even understand each other.

>>14699120
is anybody else just completely uninterested in this topic? There's nothing I could possibly care less about than discrete math

>>14699120
This stupid shit is constructed on the assumption you won't question why horseness is somehow intrinsically linked to a color, when that is nowhere embedded in the definition of "horse". This is unlike how it works with integers and induction. This is also why induction generally is only useful to things reducible to integers.

>>14699361
>I could prove that if all groups of 3 horses have the same color, then all groups of more than 3 horses do as well
indeed. so? what is the problem?

>>14702913
lol what are you even trying to say? That colour is not an intrinsic property of a horse? If you don't think horses and colour are "intrinsically linked", just replace it with something more autistic. Here, let's prove all vector spaces have the same dimension:
For a set containing 1 vector space, all of the vector spaces in the set have the same cardinality.
Now assume every set containing k vector spaces has the property that every vector space in it has the same dimension.
Let S be a set with k+1 vector spaces, label the vector spaces from [math]V_1[/math] to [math]V_{k+1}[/math]. Then the sets [math]A=\{V_1,..., V_k \}[/math] and [math]B=\{V_2,...,V_k+1\}[/math] have k elements. By the induction hypothesis, all of the vector spaces in A and B have k elements. It follows that the vector spaces from [math]V_1[/math] through to [math]V_{k+1}[/math] have the same dimension. Therefore, every vector space has the same cardinality.

Are vector spaces "intrinsically linked" enough to their own dimension for you?

>>14703010
Had a brain fart, just replace every instance of "cardinality" to "dimension" (or the other way around, they both work!)

>>14701418
but you don't matter, as you just came to the thread with the sophisticated plan of insulting everyone smarter than you, i.e. everyone for short.

Horse color is a social construct. Who is he to judge which color the horse identifies with?

STOP PLAYNG WITH WORDS
^N^ is not a number

>>14699120
So are there actually 2 flaws? The one on wikipedia, that the reasoning used in the proof of the induction step only works for 3 or more horses. But also, even if we were to accept the induction step (and take 3 horses for the base case or something), it says "ANY set of N horses are the same colour". Simply pointing at a particular set of N horses of the same colour does not prove that ALL sets of N horses are necessarily the same colour. That seems like a second flaw.

>>14705176
I don't see a second flaw. The induction hypothesis can be "all sets of n horses have the same color". Then you consider an arbitrary set of n+1 horses and show they all have the same color. Therefore all sets of n+1 horses have same color.

>>14699120
When N = 2 this argument doesn't work, yes all subsets of 1 horse will have the same color, but that doesn't have to be the same.
Can't belive most of /sci/ is this filled with brainlets, I'll stick to /mg/.

>>14705205
Fuck you, I'm going to shit up /mg/ because you said that

>>14705226
Pls don't, I was only joking ;_;

>>14699120
>Well, given any set of N+1 horses, if you exclude the last horse
You start from N and get to N+1, not start from N+1, ignore one to bring it down to N, then add that one back. That's just circular logic

>>14699120
Counterclaim:
It is well known there are 6 horses who are very good friends with each other, their colors are purple, white, blue, orange, pink, and yellow.
Conclusion:
You have been coaxed into a snafu. And the theorem is false, obviously.