Comment on The science is divided

• Iceman@lemmy.world ⁨3⁩ ⁨hours⁩ ago

  existentialcomics.com/comic/2

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• sp3tr4l@lemmy.zip ⁨18⁩ ⁨hours⁩ 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.

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  • Tiuku@sopuli.xyz ⁨8⁩ ⁨hours⁩ ago

    Yeah, so the following actually is valid: If all pairs of horses are the same color, all horses are the same color. Just starting the induction one step further.

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  • KingRandomGuy@lemmy.world ⁨17⁩ ⁨hours⁩ ago

    Yep this is the exact issue. This problem comes up frequently in a first discrete math or formal mathematics course in universities, as an example of how subtle mistakes can arise in induction.

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• idunnololz@lemmy.world ⁨18⁩ ⁨hours⁩ ago

  I took a peak 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 first 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.

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  • sp3tr4l@lemmy.zip ⁨18⁩ ⁨hours⁩ 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.

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