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Hello guys I have this problem which has been really bugging me. And it goes as follows:

Using induction, we want to prove that all human beings have the same hair colour. Let S(n) be the statement that “any group of n human beings has the same hair colour”.

Clearly S(1) is true: in any group of just one, everybody has the same hair colour.

Now assume S(k), that in any group of k everybody has the same hair colour. If we replace any one in the group with someone else, they still make a total of k and hence have the same hair colour. This works for any initial group of people, meaning that any group of k + 1 also has the same hair colour. Therefore S(k + 1) is true.

I cant seem to figure out where the problem lies in this proof. I have tried a few things and I have concluded the base case is correct. But other than that I can't seem to disprove this proof.

Martin
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  • Well, what is the smallest $n$ for which you suspect that $S(n)$ is false? – Hagen von Eitzen Oct 27 '13 at 19:14
  • Argument fails at $k=1$. – André Nicolas Oct 27 '13 at 19:14
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    Think about the inductive step going from $k=1$ to $k=2$. Then you can't connect the two disjoint groups into a larger group having the same color hair. – coffeemath Oct 27 '13 at 19:14
  • The main problem is that we can pick examples of both people who do have the same hair color and those who don't, so that the statement can be shown to be true up to the limit of the number of people in the world with that color of hair. This requires that we are able to pick from a pool of people where there is at least one not yet picked who has the same color of hair as those already picked. – abiessu Oct 27 '13 at 19:14
  • @coffeemath, Thanks for the quick response. I still can't seem to see what you are pointing to. – Martin Oct 27 '13 at 19:19
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    Indeed the inductive argument $S(k)\Rightarrow S(k+1)$ is correct for $k\ge2$. – Carsten S Oct 27 '13 at 19:55

2 Answers2

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The base case is correct.

Inductive step:

Assume that the result is true for $n = k$, which is to say that everybody in a group of $k$ people has the same hair colour.

For the proof to work, you now have to prove that $P(k)$ implies $P(k+1)$, but it doesn't. Just because it's true that for any group of $k$ people, everyone in the group has the same hair colour, it doesn't follow that for any group of $k+1$ people, everyone in the group has the same hair colour. One counterexample suffices to disprove the claim: take a group with one person in, who has red hair. Now add someone with blonde hair.

George Tomlinson
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  • oh I see now, we can say that k = n = 1 where it works because there is only 1 person in the group... but it fails to work when we have 2 people in the group.

    But could it be said, that if there are zero people in a group K=0, then everyone has the same hair colour (No hair colour). That means k+1 is correct because of we add a blond hair person to the group 0+1 then it is true.

     – Martin Oct 27 '13 at 19:25
  • That would only show that $P(0)$ implies $P(1)$, at best. This isn't the same as showing that $P(k)$ implies $P(k+1)$. The latter implication is irrespective of the size of $k$. It's saying that for a group of $k$ people, where $k$ can be any natural number at all, if $P(k)$ is true, then $P(k+1)$ is true. – George Tomlinson Oct 27 '13 at 19:35
  • okay one last question would this also work for the statement "any group of k - 1 also has the same hair colour. Therefore S(k - 1) is true" – Martin Oct 27 '13 at 19:51
  • There' no point in considering $k-1$. That can only help us prove that $0$ and negative numbers of people have the same hair colour, which isn't helping us to prove the statement that “any group of n human beings has the same hair colour”. – George Tomlinson Oct 27 '13 at 20:12
  • ahh yes u are correct. And would we be able to write this in some mathmatical notation? – Martin Oct 27 '13 at 20:15
  • The best I can do is $k-1<k$. – George Tomlinson Oct 27 '13 at 20:18
  • no I mean the proof.. that when k is bigger than 1 it fails to work – Martin Oct 27 '13 at 20:20
  • Oh sorry. I don't see that you need to and I'm not sure you can anyway. It's not that it fails for k bigger than 1. It's that a group that's one bigger than k can fail. k could be 2 or 2000. the point is that if k is 2 a third person could have a different hair colour. To prove the claim, you'd have to show that if a group of 2 people always has 2 people with the same hair colour, then a group of 3 always has 3 people with the same hair colour, but I think you already got that. – George Tomlinson Oct 27 '13 at 20:27
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The problem is when you go from 1 to 2 persons.

Assuming S(K) valid for k=1: that in any group of 1 human being, everybody has the same hair colour

Now you should be able to prove the validity of S(K+1): the equivalent to any group of 2 human beings.

Ok. Let's try. If you have any group of 2 human beings in a room, then you ask one of them to go out. The one who stayed is blond, for example. When the first person returns, then you ask to the blond one (who stayed) to go out now, and apply the hipothesis: being a group of 1 human being, "all" should have the same hair colour, ANY COLOUR, for example, brown hair.

user103074
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