I was dissatisfied with my conclusion last time, because, while there are lots of peculiar situations when dealing with infinity, we can explain those peculiarities, whereas in the endless fraction situation, it seems like we are just making an arbitrary declaration to make the contradiction disappear. Example: There are the same number of positive even integers as there are positive integers. This seems like a contradiction, because all the positive even integers are positive integers *and* there are lots of other positive integers that are not even, namely, all the positive odd integers. We explain this by analyzing what we mean when we say there are a particular number of some named “thing” (chairs, people, fractions, etc.). When I say there are 5 fingers on my right hand, I am saying that it is possible to set up a one-to-one correspondence between the fingers of my right hand and the members of the set {1, 2, 3, 4, 5}, that is, the positive integers from 1 to 5. (At a previous meeting of the Math-mavens Club, we agreed on the symbols for the positive integers and the order in which we would use them. Sorry you missed that one.)

If a one-to-one correspondence exists between a set of named “things” and the positive integers from 1 to *n*, we say there are *n* of these “things”. If there is at least one of these “things”, yet no one-to-one correspondence can be set up between these “things’ and the positive integers from 1 to *n*, no matter what *n* is, we say there are an infinity of these “things”. There are an infinity of positive integers. There are the same infinity of positive even integers, because a one-to-one correspondence can be set up between the set of positive integers and the set of positive even integers. In symbols: *I* = {*m* / *m *= 2*n*, *n* ∈ }.* * Translation: The set of positive even integers is the set of all numbers* m *such that *m* is two times a positive integer. This sets up the one-to-one correspondence: 1→2, 2→4, 3→6, 4→8, etc. Hence, there are the same “number” of positive even integers as there are positive integers. (Actually, we say they has the same cardinality, because infinity is not a number. It just indicates that there are more of whatever you’re talking about than any positive integer.)

In order for our conclusion last time not to be an arbitrary fiat, we must have an explanation for rejecting both solutions that is consistent with what we already know, so I played around with a few examples, and here is what I have so far.

If we set up the sequence *P*, , , . , , etc., then the limit as *n* approaches infinity of the *nth* term of this sequence should be the value of the endless fraction.

For example, in Puzzle 42, *P* = 1 and *S* = 2, so the sequence is 1, , 1, , 1, , 1, , 1, , etc., which can be separated into two sequences, 1, 1, 1, 1, 1, etc., and , , , , , etc., each of which converges to 1, so the sequence formed by alternating members of these two sequences also converges to the common limit of these two sequences. This limit is the same as the solution obtained for Puzzle 42 using the easier algebraic method.

If *P* = 2 and *S* = 3, the sequence we get is 2, , 2, , 2, , 2, , 2, , etc., which separates into 2, 2, 2, 2, 2, etc. and , , , , , etc. Since these two sequences converge to 2 and to 1 respectively, the combined sequence does not converge, so there is no limit and neither 1 nor 2 is the value of the endless fraction. Using our algebraic method to evaluate the endless fraction, we get *x* = 1 and *x *= 2, so we must reject both of those solutions.

In fact, every time I did this with an endless fraction where *S* = *P* + 1, I obtained two sequences that converged to *P* and to 1 respectively, which led me to the following:

Prove: Considering the endless fraction, , if *P* and *S* are positive integers such that *S* = *P* + 1, then the sequence of fractions formed by ending the fraction at successive appearances of the variable “*S*” converges to 1, and the sequence of fractions formed by ending the fraction at successive appearances of the variable “*P*” converges to *P*.

Proof: The first sequence is , . , etc. Substituting *P* + 1 for *S* and simplifying all the complex fractions, we get the sequence , , , etc., where all the terms are fractions in which the numerator and denominator are consecutive positive integers, a sequence that converges to 1. The second sequence is *P*, , , etc. Substituting *P* + 1 for *S* and simplifying all the complex fractions, we get the sequence *P*, *P*, *P*, etc., a sequence that converges to *P*. Both of these convergences can be confirmed using mathematical induction.

There are some other findings about endless fractions that I will make explicit next time.

*TO BE CONTINUED*