Wikipedia talk:Articles for deletion/Three forms of mathematical induction

message: In mathematics, many standard functions, including operations such as + and relations such as =, are binary, meaning that they take two arguments. Often these functions possess properties that implicitly extend them to more than two arguments. For example, once addition a + b is defined and is known to satisfy the associativity property (a + b) + c = a + (b + c), then the trinary addition a + b + c makes sense, either as (a + b) + c or as a + (b + c). Similarly, many axioms and theorems in mathematics are stated only for the binary versions of mathematical operations and relations, and implicitly extend to higher arity versions.
Suppose that we wish to prove a statement about an n-arity operation implicitly defined from a binary operation, using mathematical induction on n. Then it should come as no surprise that the n = 2 case carries special weight. Here are some examples.
====== Product rule ======
In this example, the binary operation in question is multiplication (of functions). The usual product rule taught in calculus says
:(fg)' = f'g + g'f.\,
For a product of n functions, one has
:(f_1 f_2 f_3 \cdots f_n)' \,
::= (f_1' f_2 f_3 \cdots f_n)
+ (f_1 f_2' f_3 \cdots  f_n)
+ (f_1 f_2 f_3'\cdots  f_n)+\cdots +(f_1 f_2 \cdots f_{n-1} f_n').
In each of the n terms, just one of the factors is a derivative; the others are not.
The n = 1 case is trivial (f' = f'), and the n >= 3 cases are easy to prove from the preceding n − 1 cases. The real difficulty lies in the n = 2 case, which is why this is the one stated in the standard product rule.
====== Pólya's proof that there is no "horse of a different color" ======
In this example, the binary relation in question is equality, =, applied to color of horses.
In the middle of the 20th century, a commonplace colloquial locution to express the idea that something is unexpectedly different from the usual was "That's a horse of a different color!".  George Pólya posed the following exercise: Find the error in the following argument, which purports to prove by mathematical induction that all horses are of the same color:
:If there's only one horse, there's only one color.
:Suppose within any set of n horses, there is only one color.  Now look at any set of n + 1 horses.  Number them: 1, 2, 3, ..., n, n + 1.  Consider the sets {1, 2, 3, ..., n} and {2, 3, 4, ..., n + 1}.  Each is a set of only n horses, therefore with each there is only one color.  But the two sets overlap, so there must be only one color among all n + 1 horses.
Again the n = 1 case is trivial (any horse is the same color as itself), and the inductive step is correct in all cases n >= 3 cases. However, the logic of the inductive step is incorrect when n = 2, because the statement that "the two sets overlap" is false. Indeed, the n = 2 case is clearly the crux of the matter; if one could prove the n = 2 case, then all higher cases would follow from the transitive property of equality.
====== Triangle inequality ======
This example is somewhat more complicated, because it involves a binary operation, +, and a binary function, the distance function d in a metric space.
The usual triangle inequality in metric spaces says
:d(x_1,x_3) \le d(x_1,x_2) + d(x_2,x_3).\,
For a sequence of n points, one has
:d(x_1,x_n) \le d(x_1,x_2)+\cdots+d(x_{n-1},x_n).\,
Again the n = 1 case is trivial, since
:d(x_1, x_1) \leq 0. 
In this example the n = 2 case is also trivial:
:d(x_1, x_2) \leq d(x_1, x_2).
Because two binary functions are involved, with the "middle argument" x_2 matching, the first nontrivial case occurs when n = 3. This case is the crux of the matter, which is why the triangle inequality deals with it. The cases n >= 4 then follow inductively from the corresponding n − 1 cases.
:::Well, I think that in the triangle inequality one should let n be the number of terms in the sum on the right and that that should be recognized not to be a mere matter of convention.  Yes you can trivially reindex, but what I'm saying is that the special property of the number 2 applies to the case indexed as n = 2 precisely when one does it that way. Michael Hardy 02:41, 8 March 2006 (UTC)
replies: text: But it is a mere matter of convention.  — Arthur Rubin | (talk) 02:43, 8 March 2006 (UTC)
replies: text: I agree with Michael.  It is, abstractly speaking, a matter of convention, but in the context of binary operators it is perfectly natural to let n be the number of terms in the sum.  Also, this is the number of legs in the polygonal path whose vertices are the points appearing in the inequality, which makes it likewise natural to index on this quantity.  Although I suppose you could argue that n should be the number of sides in the polygon.  Still, it doesn't help the example to insist on an indexing which fails to bring out the essential feature it's trying to exemplify. Ryan Reich 02:58, 8 March 2006 (UTC)
replies: text: Reindexing is fine; go ahead. This restores n = 2 to the crucial case, and makes the arity of the + operation the key consideration. What I'm really concerned about is whether the people interested in preserving this article's content would be satisfied by merging an explanation similar to this into mathematical induction. We can polish it later. Joshuardavis 03:33, 8 March 2006 (UTC)

text: "Mere convention": Only a few months ago I was surprised to find out that some respectable mathematicians think the fact that the empty product is 1 is merely convention.  And that that topic is not worth discussing.  Maybe it would not be worth discussing if it were mere convention.  But it's beginning to look to me like a big gap in most mathematicians' understanding of what seems like trivial matters, and perhaps so is this.  More shortly.... Michael Hardy 22:31, 8 March 2006 (UTC)
replies: text: Is your point about conventions that they are chosen to preserve larger patterns in mathematics (and therefore, as a practical benefit, simplify notation)? Because I think that people who say "it's just a convention" know this. They know that the convention is not fully arbitrary, that in fact it's the only one that's "right". What they are really saying is that the conventional case (such as the empty product being 1) does not have the same obvious intuitive meaning as the typical case. Joshuardavis 02:58, 9 March 2006 (UTC)
thread_title:  Special significance of the ''n'' = 2 case 

message: This example is somewhat more complicated, because it involves a binary operation, +, and a binary function, the distance function d in a metric space.
The usual triangle inequality in metric spaces says
:d(x_1,x_3) \le d(x_1,x_2) + d(x_2,x_3).\,
For a sequence of n points, one has
:d(x_1,x_n) \le d(x_1,x_2)+\cdots+d(x_{n-1},x_n).\,
Again the n = 1 case is trivial, since
:d(x_1, x_1) \leq 0. 
In this example the n = 2 case is also trivial:
:d(x_1, x_2) \leq d(x_1, x_2).
Because two binary functions are involved, with the "middle argument" x_2 matching, the first nontrivial case occurs when n = 3. This case is the crux of the matter, which is why the triangle inequality deals with it. The cases n >= 4 then follow inductively from the corresponding n − 1 cases.
:::Well, I think that in the triangle inequality one should let n be the number of terms in the sum on the right and that that should be recognized not to be a mere matter of convention.  Yes you can trivially reindex, but what I'm saying is that the special property of the number 2 applies to the case indexed as n = 2 precisely when one does it that way. Michael Hardy 02:41, 8 March 2006 (UTC)
replies: text: But it is a mere matter of convention.  — Arthur Rubin | (talk) 02:43, 8 March 2006 (UTC)
replies: text: I agree with Michael.  It is, abstractly speaking, a matter of convention, but in the context of binary operators it is perfectly natural to let n be the number of terms in the sum.  Also, this is the number of legs in the polygonal path whose vertices are the points appearing in the inequality, which makes it likewise natural to index on this quantity.  Although I suppose you could argue that n should be the number of sides in the polygon.  Still, it doesn't help the example to insist on an indexing which fails to bring out the essential feature it's trying to exemplify. Ryan Reich 02:58, 8 March 2006 (UTC)
replies: text: Reindexing is fine; go ahead. This restores n = 2 to the crucial case, and makes the arity of the + operation the key consideration. What I'm really concerned about is whether the people interested in preserving this article's content would be satisfied by merging an explanation similar to this into mathematical induction. We can polish it later. Joshuardavis 03:33, 8 March 2006 (UTC)

text: "Mere convention": Only a few months ago I was surprised to find out that some respectable mathematicians think the fact that the empty product is 1 is merely convention.  And that that topic is not worth discussing.  Maybe it would not be worth discussing if it were mere convention.  But it's beginning to look to me like a big gap in most mathematicians' understanding of what seems like trivial matters, and perhaps so is this.  More shortly.... Michael Hardy 22:31, 8 March 2006 (UTC)
replies: text: Is your point about conventions that they are chosen to preserve larger patterns in mathematics (and therefore, as a practical benefit, simplify notation)? Because I think that people who say "it's just a convention" know this. They know that the convention is not fully arbitrary, that in fact it's the only one that's "right". What they are really saying is that the conventional case (such as the empty product being 1) does not have the same obvious intuitive meaning as the typical case. Joshuardavis 02:58, 9 March 2006 (UTC)
thread_title:  Triangle inequality