Thinking mathematically, response + problem

Response:

My first stop(s) would be all of the little bits of advice which are given for helping get through moments of being stuck on a problem. They mostly seem somewhat obvious, but I suppose the point is that it's not always obvious to students on how to handle being stuck. Just as a side note, it bothers me somewhat that so many words are capitalized. That might be my own fussy little thing, and I need to think more on why that bothers me... But I wonder if I feel like it's a little reductive to imply that there is like this formula of key-worded steps to properly think mathematically. I would however, counter those negative thoughts of mine by saying I think it's actually a great thing to celebrate getting stuck on a problem. It usually doesn't feel too good getting stuck, but we should all instead strive to be excited when faced with a problem requiring some careful and creative thought. Ironically, another stop for me was the "Threaded Pins" problem, which I found frustrating and didn't spend more than a couple minutes looking at because I thought it was a poorly formulated question. Now, it might just be me - I know I can be ignorant of very obvious things - but it was not at all obvious (to me at least) what was meant by a "clockwise gap." So flying in the face of celebrating getting stuck, I decided that spending any more time on that problem was a waste. The other problem I looked at was much easier for me to understand.

The Problem: Rational Divisors
Given that 14/15 divides into 28/3 a whole number of times (10 times), we
might say that 14/15 is a rational divisor of 28/3.
● Find all the rational divisors of 28/3.
● Find all the rational divisors of 1/2, and then find all the numbers that
are rational divisors of both 28/3 and 1/2.
Does it make sense to talk about the greatest common rational divisor of
two fractions, and the lowest common rational multiple of two fractions?

- I started off by working through the example division they gave. The idea was to get a sense of how dividing with fractions results in a whole number.

- Using the "keep change flip" method of dividing fractions, combined with "cross-cancelling," I realized that to get a whole number, the denominators (after the flip) would need to be factors of the numerators (after the flip)

- here I start to try and generalize that idea a little bit, to hopefully find a method to obtain all rational dividers of 28/3.

- I try to be a little bit more clear about the conditions required for a rational number to divide 28/3

- realizing that 3 has infinitely many multiples, it was clear that you could have as many rational divisors as you like. for example: 28/3, 28/6, 28/9, and so on, would all divide 28/3 nicely.

- now trying to answer the question, I set to expressing it as as set of numbers. 

- I saw that it would easy to show the infinitely many answers using a variable, and that it might be nice to include positive and negatives as well.

- I thought that I might be able express the numerators in a more compact manner, maybe with a variable as well, but I felt that this was more clear to simply list all of the possible numerators. especially since, unlike the denominator, there are not infinitely many choices.

- while I had technically solved the first part of the problem, I was still interested in rewriting this idea more generally, mostly for fun.

- I think this could still be refined, probably stated a little bit more cleanly. and in general I don't really like using the "divisible by" symbol. 

- It seems like it would be more useful to do something like "there exists a factor k for which m = a*k" but I think this gets the point across.

- using the method I had developed earlier, it was relatively simple to apply it to find the set of rational divisors of 1/2

- now looking to find the greatest common divisor, I started by thinking how to find the set of all common divisors.

- referring back to my method, I had a bit of an "Aha" moment, realizing that my conditions work just as well when considering multiple numbers.

- from there I just needed to find the common factors of the numerators, and the common multiples of the denominators. thankfully these were relatively simple.

 

- a key to this for me was knowing that all common multiples of 3 and 2 would be multiples of their least common multiple 6, which made it easy for me to write my answer in a similar looking set. I was enjoying the consistency of that format of answer at this point

- from there, it was straightforward to find the GCD, the only thing I stumbled over for a minute was that a greater denominator would actually correspond to a "lesser" rational number overall.

- this allowed me to realize in general, that to have the greatest possible divisor, we would want to minimize the denominator, but maximize the numerator.

- for fun, I generalized this for any given rational number.

- I also had the thought that this would work for any number of rational numbers, so I decided to use ellipses, m_i and n_i to indicate this idea

- I finished by answering the question that "yes" this does make sense, to me at least. I couldn't see any reason why this result wouldn't be applicable for any rational numbers. 

- I did however exclude 0 from the discussion. So I suppose in the interest of completeness I would add that it doesn't make too much sense to consider the rational divisors of 0, only because it is simply all numbers other than 0.

- here, I began to consider the LCM of rational numbers. and was assuming the traditional view point of multiples, which is they would be the result of multiplying by the positive integers

- I decided to work through some examples just to get some insight into the problem.

- here I saw that if the denominators were mutually prime, you would essentially need to find the LCM of the numerators.

- I did not formalize these ideas here, but I do think it would have been fun to play with that some more

- another example... it seems that if the numerators are the same, and one denominator is a multiple of the other, then the LCM would simply be the greater fraction

- what if the numerators were mutually prime but one of the denominators was a multiple of the other? (reverse of the first example)

- here I started to think about this problem as depending on cases of whether or not denominators/numerators were mutually prime

- at this point, I've reached the conclusion that it makes plenty of sense to consider the LCM of rational numbers, though some cases are simpler than others.

- I did also consider: what if weren't using the "traditional" definition of multiples, and instead, were thinking of multiples of rational numbers as being products of any two rational numbers? however, I had the thought that you could always choose to multiply by a rational number to obtain any rational number you like, so in theory the LC"M" of any two rational numbers is arbitrarily small. example 1/2 and 1/3... multiply 1/2 by 2/99999999... and multiply 1/3 by 3/99999999...