• hansolo@lemm.ee
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    40 minutes ago

    …what?

    Horse hair colors vary.

    If you shave a horse, any breeds with dark black or blue-ish grey hair will give you a variety of that pallet, getting dark enough to be stopped and frisked in larger American cities. Sometime blueish-black skin and hair pigmentation matches as well.

    Most other breeds will give you a pink color range.

    https://artpictures.club/autumn-2023.html

    https://www.magonlinelibrary.com/cms/10.12968/ukve.2021.5.1.24/asset/images/medium/ukve.2021.5.1.24_f03.jpg

    Source: was given a saddle for Christmas once.

  • Black616Angel@discuss.tchncs.de
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    3 hours ago

    Actually this just refers to the color of their fur. I’d say that a blonde white man and a white read head are the same color, just their hair has a different color.

    (I’m not totally serious and will not die on this hill)

    • credo@lemmy.world
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      5 hours ago

      Doesn’t “color” (no ‘s’) imply they can’t be more than one? Or… is this theorem further supported? Both of these horses are all the same color.

    • sp3ctr4l@lemmy.zip
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      53 minutes ago

      From that link:

      Assume that n horses always are the same color.

      … I mean… yes, the logic follows… if you… make and hold that assumption… which is ostensibly what you are trying to prove.

      This is otherwise known as circular reasoning.

      Apparently this arose basically as a joke, a way of illustrating that you actually have to prove the induction is valid every step of the way, instead of just asserting it.

      • sp3ctr4l@lemmy.zip
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        16 minutes ago

        Do you mean you went through the proof and verified it, or falsified it?

        As I understand it, it goes something like this:

        You have a set of n horses.

        Assume a set of n horses are the same color.

        Now you also have a set of n+1 horses.

        Set 1: (1, 2, 3, … n)

        Set 2: (2, 3, 4, … n+1)

        Referring back to the assumption, both sets have n horses in them, Set 2 is just incremented forward one, therefore, Set 2’s horses are all one color, and Set 1’s horses are all one color.

        Finally, Set 1 and Set 2 always overlap, therefore that the color of all Set 1 and Set 2’s horses are the same.

        So, if you hold the ‘all horses in a set of size n horses are the same color’ assumption as an actually valid assertion, for the sake of argument…

        This does logically hold for Set 1 and Set 2 … but only in isolation, not compared to each other.

        The problem is that the sets do not actually always overlap.

        If n = 1, and n + 1 = 2, then:

        Set 1 = ( 1 )

        Set 2 = ( 2 )

        No overlap.

        Thus the attempted induction falls apart.

        Set 1’s horse 1 could be brown, Set 2’s horse 2 could be … fucking purple… each set contains only one distinct color, that part is true, but the final assertion that both sets always overlap is false, so when you increment to:

        Set 1 = ( 1, 2 )

        Set 2 = ( 2, 3 )

        We now do not have necessarily have the same colored horse 2 in each set, Set 1’s horse 1 and 2 would be brown, Set 2’s horse 2 and 3 would be purple.

        I may be getting this wrong in some way, it’s been almost 20 years since I last did set theory / mathematical proof type coursework.

      • idunnololz@lemmy.world
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        17 minutes ago

        I took a peek and it is sort of dumb but logically “sound”. Specifically the indictive step.

        In the inductive step you assume the statement is true for some number n and use this statement to prove the statement n + 1 is true. If you can do that then you can prove the induction step.

        So in this example the statement we assume is true is given n horses, all of them are the same color. To prove the statement for n + 1 horses we look at the n + 1 horses. Then we exclude the last horse. By excluding the last horse we have a set of n horses. By the induction statement this set of horses must all be the same color. So now we’ve proven the first n horses are the same color.

        Next we can exclude the first horse. This also gives us a set of n horses. By the induction statement all these horses must also be the same color. Therefore all n + 1 horses must be the same color.

        This sounds really dumb but the proof works in the induction step.

        The logical issue is that the base case is wrong which is necessary for a complete proof by induction.

        • sp3ctr4l@lemmy.zip
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          9 minutes ago

          I think? I worked through how the induction logic actually fails.

          This kind of induction only works if you can actually prove Sets 1 and 2, starting at n and n+1, actually overlap at all stages… and in this case, they don’t.

  • CarbonIceDragon@pawb.social
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    5 hours ago

    But the reasoning given doesnt apply exclusively to horses. Suppose we follow the same chain that gets us “all horses are the same color”, but replace “horses” with “colors”, we would end up with the statement that all colors are the same color. Thus, this is not a counterexample, because black and brown are the same color.

  • vane@lemmy.world
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    4 hours ago

    Maybe author of the sentence was looking at horses in space and wikipedia is completly wrong.