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I feel like I'm too ignorant of mathematics to be confused by this "proof". When you set out to prove "for any set of n horses, every horse in that set has the

I feel like I'm too ignorant of mathematics to be confused by this "proof". When you set out to prove "for any set of n horses, every horse in that set has the same color" and you then "suppose for all sets of n horses, every horse in the set has the same color" as a first step, you're messing with a tautology, aren't you? I mean if you've supposed that it's true for all, it's definitely true for any, right? What am I missing that makes the problem subtle and interesting, rather than just blatantly circular from the start?

mannerheim 5y ago
The inductive step assumes the hypothesis is true for n, and proves it for the n+1 case. The overall proof is then for all positive integers, n.
- phkahler 5y ago
  >> The inductive step assumes the hypothesis is true for n You don't get to assume the thing you're trying to prove is true as a part of the proof that it's true. The only time we assume something like that is in a proof by contradiction, where the contradiction implies either an error in our logic, or an error in the assumptions.
  - deleted 5y ago
    [deleted]
  - mannerheim 5y ago
    An inductive proof has two steps: 1) Prove H(0) 2) Prove that if H(n), then H(n+1) Then, by the axiom of induction, this is proven for all positive integer values of n. Because if those two conditions hold, we would have H(0) is true H(1) is true because H(0) -> H(1) ((2) with n=0) and H(0) H(2) is true because H(1) -> H(2) ((2) with n=1) and H(1) and so on. The assumption isn't what's being proven, it's proving the inductive hypothesis for the next integer.
    - phkahler 5y ago
      Yes, I was being dumb. As the article says there is a problem talking about equality among members of a set with size 1. So saying "all the horses in a set of size 1 are the same color" begs the question "same as what?" and that's where the real problem is, and the difficulty in pinning down that kind of thing is why I'm not a mathematician ;-)
      - mannerheim 5y ago
        There are several 'mathematical' ways you could phrase it (glossing over how to define the colour of a horse): 1) In a set of horses with size n, if horses A and B being both members of this set implies horse A has the same colour as horse B, then all horses in the set have the same colour. 2) In a set of horses with size n, and let f be a function on this set that produces each horse's colour. Then all horses in this set have the same colour if and only if the codomain of f on this set has size 1. You could probably prove these are equivalent definitions.
- na85 5y ago
  But why would anyone assume that n horses, for arbitrary n, are the same color? That's absurd. Anyone who's ever seen a field with horses in it would trivially know that this is a faulty assumption to make. You might as well say "We can prove that coconuts can migrate. First, assume that coconuts can migrate. Thus, coconuts migrate". It's equally valid logic.
  - mannerheim 5y ago
    So it's worth noting that the inductive hypothesis isn't false for all values of n - namely, it is true for n = 0,1. But beyond that, in mathematical logic, for p -> q, you can still prove this to be true, even if p is false. In fact, if you can prove p is always false, p -> q is 'vacuously true' - it's true because you can never come up with an example of 'p and not q'. So mathematically, there's no problem with assuming an induction hypothesis that's actually false here. The real problem occurs is that the result doesn't follow from the assumption because the induction step of the 'proof' sneakily assumes that n >= 3, i.e., the induction step simply doesn't work for the n = 2 case.
  - ajuc 5y ago
    > But why would anyone assume that n horses, for arbitrary n, are the same color? because that's how inductive proofs work. You assume something is true for N and show that this implies that it's also true for N+1. This combined with base case (for example for N=1) proves that it's true for all N >= 1. The assumption is just a tool used to check if you can prove the implication, the real "meat" of the proof is in the implication. Are you aware of proofs by contradiction? A: "assume N is the largest natural number". If that's true then there shouldn't be any natural numbers larger than N, but we can create N+1 and show it's larger AND natural, so assumption A leads to contradiction, so there is no such thing as the largest natural number. We used false assumption in our proof, but the proof was correct. It's a similar idea with inductive proofs - you make an assumption and see where it leads you. You don't use the assumption for the proof, you use the implication A(N)=> A(N+1) for the proof, the assumption A is just a way to see if you can prove the implication. > You might as well say "We can prove that coconuts can migrate. First, assume that coconuts can migrate. Thus, coconuts migrate". It's equally valid logic. The logic isn't actually circular, your coconut analogy is wrong. Correct analogy would be: "when we assume N coconuts can migrate we can formally show that it implies N+1 coconuts can migrate" + "we can show that 1 coconut can migrate". You only need these 2 facts to prove all coconuts can migrate.
shkkmo 5y ago
There are two steps of an inductive proof one is proving the inductive rule that: p(n) implies p(n+1) the other is proving that there exists some number i for which p(i) is true. This, combined with the proof of the inductive rule, allows you to prove that p(n) is true for all n >= i.
- LanceH 5y ago
  In simpler terms the two steps for this problem are: 1. There exists some number of horses where all horses are of the same color. This is obviously true when you have 1 horse. 2. The second step is that you have to prove that if you have some number of horses n, and they are all of the same color, then if you add another horse all the horses are of the same color. He's saying there is some set of n+1 horses, n of which are of the same color. Then he makes the (false) claim that removing one horse at random leaves you with n horses, which therefore must all be of the same color. This is incorrect because the assumption is that there is a set of n horses, not "if you have n number of horses they are all the same color. The base case of n=1 is just fine. It's adding a randomly colored horse to a set of horses that doesn't always result in an n+1 set of like-colored horses. He doesn't even explain the illogical step correctly, so the whole thing is a mess.
  - shkkmo 5y ago
    > He's saying there is some set of n+1 horses, n of which are of the same color. > This is incorrect because the assumption is that there is a set of n horses, not "if you have n number of horses they are all the same color. You have this exactly backwards. The assumption you say is not being made is exactly the one that is being made (and is a step that occurs similarly in every inductive proof.)
ajuc 5y ago
It's not circular, because you aren't using the assumption A in the final proof, you are just using the implication A(N) => A(N+1) in the final proof. the final proof is: A(1) is true AND A(N) implies A(N+1) means that A(N) is true for all N >= 1 the assumption that A is true for N was just used as a condition in the implication If A(N) is false in general - you would realize that despite assuming it - because the implication or the base case couldn't be proven.

mannerheim 5y ago

The inductive step assumes the hypothesis is true for n, and proves it for the n+1 case. The overall proof is then for all positive integers, n.

phkahler 5y ago
>> The inductive step assumes the hypothesis is true for n You don't get to assume the thing you're trying to prove is true as a part of the proof that it's true. The only time we assume something like that is in a proof by contradiction, where the contradiction implies either an error in our logic, or an error in the assumptions.
- deleted 5y ago
  [deleted]
- mannerheim 5y ago
  An inductive proof has two steps: 1) Prove H(0) 2) Prove that if H(n), then H(n+1) Then, by the axiom of induction, this is proven for all positive integer values of n. Because if those two conditions hold, we would have H(0) is true H(1) is true because H(0) -> H(1) ((2) with n=0) and H(0) H(2) is true because H(1) -> H(2) ((2) with n=1) and H(1) and so on. The assumption isn't what's being proven, it's proving the inductive hypothesis for the next integer.
  - phkahler 5y ago
    Yes, I was being dumb. As the article says there is a problem talking about equality among members of a set with size 1. So saying "all the horses in a set of size 1 are the same color" begs the question "same as what?" and that's where the real problem is, and the difficulty in pinning down that kind of thing is why I'm not a mathematician ;-)
    - mannerheim 5y ago
      There are several 'mathematical' ways you could phrase it (glossing over how to define the colour of a horse): 1) In a set of horses with size n, if horses A and B being both members of this set implies horse A has the same colour as horse B, then all horses in the set have the same colour. 2) In a set of horses with size n, and let f be a function on this set that produces each horse's colour. Then all horses in this set have the same colour if and only if the codomain of f on this set has size 1. You could probably prove these are equivalent definitions.
na85 5y ago
But why would anyone assume that n horses, for arbitrary n, are the same color? That's absurd. Anyone who's ever seen a field with horses in it would trivially know that this is a faulty assumption to make. You might as well say "We can prove that coconuts can migrate. First, assume that coconuts can migrate. Thus, coconuts migrate". It's equally valid logic.
- mannerheim 5y ago
  So it's worth noting that the inductive hypothesis isn't false for all values of n - namely, it is true for n = 0,1. But beyond that, in mathematical logic, for p -> q, you can still prove this to be true, even if p is false. In fact, if you can prove p is always false, p -> q is 'vacuously true' - it's true because you can never come up with an example of 'p and not q'. So mathematically, there's no problem with assuming an induction hypothesis that's actually false here. The real problem occurs is that the result doesn't follow from the assumption because the induction step of the 'proof' sneakily assumes that n >= 3, i.e., the induction step simply doesn't work for the n = 2 case.
- ajuc 5y ago
  > But why would anyone assume that n horses, for arbitrary n, are the same color? because that's how inductive proofs work. You assume something is true for N and show that this implies that it's also true for N+1. This combined with base case (for example for N=1) proves that it's true for all N >= 1. The assumption is just a tool used to check if you can prove the implication, the real "meat" of the proof is in the implication. Are you aware of proofs by contradiction? A: "assume N is the largest natural number". If that's true then there shouldn't be any natural numbers larger than N, but we can create N+1 and show it's larger AND natural, so assumption A leads to contradiction, so there is no such thing as the largest natural number. We used false assumption in our proof, but the proof was correct. It's a similar idea with inductive proofs - you make an assumption and see where it leads you. You don't use the assumption for the proof, you use the implication A(N)=> A(N+1) for the proof, the assumption A is just a way to see if you can prove the implication. > You might as well say "We can prove that coconuts can migrate. First, assume that coconuts can migrate. Thus, coconuts migrate". It's equally valid logic. The logic isn't actually circular, your coconut analogy is wrong. Correct analogy would be: "when we assume N coconuts can migrate we can formally show that it implies N+1 coconuts can migrate" + "we can show that 1 coconut can migrate". You only need these 2 facts to prove all coconuts can migrate.

shkkmo 5y ago

There are two steps of an inductive proof one is proving the inductive rule that: p(n) implies p(n+1) the other is proving that there exists some number i for which p(i) is true. This, combined with the proof of the inductive rule, allows you to prove that p(n) is true for all n >= i.

LanceH 5y ago
In simpler terms the two steps for this problem are: 1. There exists some number of horses where all horses are of the same color. This is obviously true when you have 1 horse. 2. The second step is that you have to prove that if you have some number of horses n, and they are all of the same color, then if you add another horse all the horses are of the same color. He's saying there is some set of n+1 horses, n of which are of the same color. Then he makes the (false) claim that removing one horse at random leaves you with n horses, which therefore must all be of the same color. This is incorrect because the assumption is that there is a set of n horses, not "if you have n number of horses they are all the same color. The base case of n=1 is just fine. It's adding a randomly colored horse to a set of horses that doesn't always result in an n+1 set of like-colored horses. He doesn't even explain the illogical step correctly, so the whole thing is a mess.
- shkkmo 5y ago
  > He's saying there is some set of n+1 horses, n of which are of the same color. > This is incorrect because the assumption is that there is a set of n horses, not "if you have n number of horses they are all the same color. You have this exactly backwards. The assumption you say is not being made is exactly the one that is being made (and is a step that occurs similarly in every inductive proof.)

ajuc 5y ago

It's not circular, because you aren't using the assumption A in the final proof, you are just using the implication A(N) => A(N+1) in the final proof. the final proof is: A(1) is true AND A(N) implies A(N+1) means that A(N) is true for all N >= 1 the assumption that A is true for N was just used as a condition in the implication If A(N) is false in general - you would realize that despite assuming it - because the implication or the base case couldn't be proven.