Given a room
with any (finite
in it, if 1
of them is a redhead
, then all of them are.
This is patently false, of course. Nonetheless we have the following "proof" by induction.
We'll denote the number of people in the room by n.
- base case: n=1.
This case is clear: it says that for a room with one person in it, if one of them (and there's only one) is a redhead, then all of them are redheads.
- induction case: we shall suppose the untheorem true for n, and prove it for n+1.
Consider a room with n+1 people in it, one of them a redhead. Ask one of the non-redheads to step outside the room for a moment. Then we are left with a room with n people in it, and one of them is (still) a redhead. By the induction hypothesis, all n people are redheads!
Now ask the (only) non-redhead who stepped out to return to the room; ask some other person to step out, and again we're left with n people in a room, one of them (in fact, all but one of them!) a redhead. By the hypothesis, we deduce that all n are redheads, in particular the remaining person. Returning the redhead who stepped outside to the room, we see that all n+1 people are redheads.
By the principle
of mathematical induction
, the theorem
is proved (or rather, the untheorem
What's wrong here??
Notes about comments:
Both prole and DaVinciLe0 are giving wrong arguments. prole is mistaken about the proof: every time we ask someone who is not a redhead to step out, thus fulfilling the induction hypothesis. DaVinciLe0 says something about not being able to prove by induction anything regarding n people. Apparently he does not believe (a finite number of) people can be counted, since the primary way of telling if a set can be placed in a bijection with an initial subset of the naturals is to count it!
This is a common problem with this unproof: it's false, but for much simpler reasons than what people try to make up!