Inductive proof that every person supports the same team

Question

I can't find what's wrong with this.

Proof:

Every person supports the same team.

We can observe that in a set with only one person, all people within it support the same team. Now suppose that the statement is true for every set with cardinality $≤ n$. Then if there are $n + 1$ persons in a set, we take one of them and, by the hypothesis, all $n$ persons support the same team. Now put this person back to the initial set and remove a different one. Again, all the $n$ persons left support the same team. Therefore all $n + 1$ persons support the same team, and for every $k$ in $\mathbb N$, those $k$ people support the same team.

Notice that in the inductive step it is being assumed n-1 > 0. You haven't demonstrated in initial step where n -1 > 0. — fleablood, Aug 15 '16 at 01:29

score 13 · Answer 1 · answered Aug 15 '16 at 00:14

13

Hint: The base case needs to be solved for $n = 1$ AND $n = 2$ in this problem. Do you see why?

answered Aug 15 '16 at 00:14

Sandeep Silwal

8,135

2

+1. Alternatively, the inductive step implicitly assumes that $n>1$, and hence is not valid. – Noah Schweber Aug 15 '16 at 00:18
I suspected that the problem was there, but I can't see why it is problematic to solve it only for $n=1$. – izzorts Aug 15 '16 at 00:22
@izzorts Try running the inductive step for $n=1$ and see what goes wrong. – f'' Aug 15 '16 at 00:31
1

@f'': well, nothing goes wrong at that step if you mechanically treat "supports the same team" as a single-valued predicate, like "belongs to a set" or "is red". And therefore read "all n support the same team" as if it meant "each one of them supports the same team and therefore they all support the same team". It's just a question of exactly where you want the proof to be wrong ;-) – Steve Jessop Aug 15 '16 at 00:46
@izzorts: the problem at $n = 1$ is that of your $n+1$ (i.e. 2) people, first you take a set of one and say that one supports the same team as himself. Then you take another set of one and say that one supports the same team as himself. But because the two sets don't intersect, they might support different teams. The erroneous "proof" skips over the fact that two sets $A$, $B$ need a non-empty intersection in order for us to prove "all of $A$ support the same team and all of $B$ support the same team implies all of $A \cup B$ support the same team". – Steve Jessop Aug 15 '16 at 00:50
... although to see why that is, you might need a more formal expression of what "supports the same team" actually means. The "proof" uses supposed properties of this phrase without ever formally defining it, in the hope you won't notice that it doesn't really have the properties claimed. – Steve Jessop Aug 15 '16 at 00:55
Sandeep answered best. To be able to find "another" person to remove in the argument above, one has to have at least 3 people. So the induction gets blocked at $n=2.$ – Behnam Esmayli Aug 15 '16 at 01:29
Go through the inductive step slowly. You have n +1 people. Remove 1. n people support the same team (call i) Al. Removed person unkown. Remove one person. One removed person supports A. One removed person is unknown. Remaining n-1 people IF THERE ARE ANY support A. Add the unknown person. You have n people so they must support the same team. IF THERE WERE ANY n-1 people supporting A, these n people including the unknown support A. Bring first person back. BUT WHAT IF n-1 WAS 0? – fleablood Aug 15 '16 at 02:26

score 7 · Answer 2 · edited Aug 15 '16 at 06:37

Now put the person back and remove another. Again those n people support the same team.

Yes, but the $n$ people might support a different team than the previous $n$ people.

You are assuming:

$ set_1 = \{$ $n$ people supporting team $A$ $\}$

$ set_2 = \{$ $n$ people supporting team $B$ $\}$

$ set_{phantom} = set_1 \cap set_2 = \{$ $n-1$ people who support both team $A$ and team $B$ $\}$

(NEVER mentioned but assumed to exist)

We assume $set_{phantom}$ isn't empty and as people can't support different teams, we conclude $A$ and $B$ are the same team.

Then we have:

$set_{magic} = set_1 \cup set_2 = \{ n + 1$ people supporting either team $A$ or team $B$ but they are the same team $\}$

But notice! Our base case was $n=1$. For $n=1$ then $set_{pantom}$ with $n-1 = 0$ members IS empty. So team $A$ and team $B$ may be different. And $set_{magic}$ could be $n+1 =2$ people supporting different teams.

So to do this proof right we need to prove it for a case where $set_{phantom}$ will have at least $1$ member. Thus we need to show this for a base case of $n >= 2$.

So we need to show all sets of two people support the same team.

It is impossible for us to show this.

Inductive proof that every person supports the same team

2 Answers2