The pigeonhole principle -- Proof Writing 9

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[sillymel]

(7:43) As other people have pointed out, the statement this example asks you to show is actually false. (In general, (a_1, a_2, a_3, …, a_(n+1)) = (0, 1, 2…, n) is a counterexample.) You can fix it by replacing the two instances of “n+1” with “n+2.”

(8:37) “1≤i≤2n” here should be “1≤i≤n+2.”

(12:26) Incorporating a previous correction, the two instances of “n+1” in the narration here should be “n+2.”

(12:33) Continuing to fix parts changed by the first correction, rather than having n or less boxes, you instead need n+1 or fewer boxes.

(12:50) Again, the two instances of “n+1” here should be changed to “n+2.”

(14:28) This is where you got yourself into trouble. You just accidentally introduced an assumption that if there’s one item in that last box, then there must be two items in that box, which is not true.

(17:17) I can see why you like it. It has quite a clever proof (coming up with my own proof before viewing Michael’s proof was the reason it look me so long to post this comment.

(20:41) You should probably specify that k is positive here.

(29:05) It’s actually by the Division Principle, not by the Pigeonhole Principle. The Pigeonhole Principle only lets you say that at least one “box” has more than one object, not that it has 3 or more objects in this case.

(29:52) Narration error: “less than” should be “less than or equal to.”

(30:14) Not an error, but I came up with a different proof that allows you to decrease the number of points necessary. Note that the nine points form a nonagon, which can be partitioned into seven internal triangles. Now, note that the total area of the enclosing space (the original isosceles right triangle) is 1/2. This means that at least one of the seven internal triangles has an area of at most (1/2)/7 = 1/14 (by a continuous analog of the Division Principle), which is less than 1/8, so we’re done. (Note that this proof works for as few as six points, which would result in four internal triangles.)

(35:50) Not just k, but n is a nonnegative integer, not necessarily a natural number, as well.

(36:46) Specifically, that’s by the Subtraction Principle.

(37:04) I assume by “unique pairing, ” you mean that distinct values of A map to distinct values of A-complement, and vice versa. I’m not sure I’d use the word “unique” to describe that pairing, seeing as there are other pairings between k-element sets and (n-k)-element sets that fit that property, but I guess it’s an okay term for now, seeing as you haven’t introduced the concept of one-to-one correspondence yet.

(40:40) That should be {1, …, 2n}, not {1, …, n} (same for the narration).

[noahtaul]

7:43 I’m not sure this is true, if n=2 then among 3 integers we can find two whose sum or difference is a multiple of 4; that’s not the case if we take {0, 1, 2}. You can’t just scrub away that set at 14:23

[TimL_]

I do not understand why a simple and self-explanatory idea such as this, needs a silly name. Nice lesson.

[wannabeactuary]

Is the floor of 2.9 recurring ie
actually
3?

[lexinwonderland]

this feels like a good exercise for the students of this class, credit to J Pi Maths

[Happy_Abe]

The second warm up question should be sun up to n-1, not n