Let S(n) be the statement “every herd of n sheep is monochromatic” - let’s work out what the one color is later. Without some careful thought, each statement S(n) may be true or may equally well be false - but we’ll convince ourselves here that S(n) is true regardless of what number n represents.
It’s easy to prove that the statement S(1) is true: all the sheep in a herd of size 1 are all the same color.
Next, let’s show that S(k) implies S(k+1) - i.e. we’ll temporarily decide to believe S(k) for some given integer k, and then notice, as a consequence, that we now inescapably believe S(k+1).
To this end, consider an arbitrary herd H which contains k+1 sheep. Assume, without loss of generality, that H contains two special sheep, named Archibald and Beatrice (renaming some sheep if necessary).
Let A be the sub-herd of H consisting of all sheep in H except Archibald.
Let B be the sub-herd of H consisting of all sheep in H except Beatrice.
Both A and B are herds of size k, and S(k) would imply that herd A is monochromatic. Likewise, S(k) says herd B is monochromatic. Lastly, every sheep aside from Archibald and Beatrice is in BOTH herds, so the color of herd A has to be the SAME as the color of herd B.
We have thus shown that all sheep in H are the same color, assuming only that we believe S(k). But herd H was to represent an arbitrary herd of k+1 sheep! That is, we have used S(k) to convince ourselves that every herd of k+1 sheep is monochromatic. Thus we have shown that S(k) implies S(k+1).
The consequence is that we now believe S(n) for any n. For instance, to prove that a herd of 10 sheep is all the same color, we convince ourselves of the truth of S(10). But we’ve shown that S(10) follows from S(9), which follows from S(8), and so on… and since we believe S(1), we’re good.
There are surely finitely many sheep in the world. Suppose there are N of them. The statement S(N), which we now believe, asserts that they’re all the same color, so we just have to figure out what that color is. Luckily, that’s easy, because I happen to have a cute little plush sheep which is pink. The argument did not distinguish between adorable little plush sheep and living meat-sheep. Thus every sheep is the same color as my plush sheep: pink.
There you go!
Obviously there is a flaw in the above argument somewhere, and the exercise is to find it. We give this exercise to undergraduate math students when they are learning a proof technique called “induction”. The mistake I have (deliberately) made in the above argument is slightly subtle and students tend to make it all the time.
This is why it took four hours.
goddamn it i’ve been reading this out loud for the last however many minutes and i’m mad about it all over again
you think you’re mad now, wait till you hear about how every positive integer is fascinating.
Let me make a guess, the mistake is the assumption of S(A) and S(B) being equal, there is a distinction for a reason
professor @birdthatlookslikeastick, what is your response to this bold conjecture?
i don’t think that’s quite correct - A and B are indeed distinct herds, as you say, but the size of each is k. The statement S(k) does apply to both - as well as any other herd of size k.
But you’re very close, and you’re right to be suspicious of this part of the argument. Let’s inspect the reasoning about herd A and herd B more carefully. It goes like this:
1) Herd A is monochromatic,
2) Herd B is also monochromatic, possibly a different color than herd A,
3) BUT there’s some sheep common to both herd A and herd B, so in fact all of herd H would have to be monochromatic.
Steps 1) and 2) are legit, but there is a problem with step 3…
