# B "Neighbor fractions" in Gelfand's Algebra

1. Apr 30, 2016

### xwolfhunter

I'm reading Gelfand's algebra, and I encountered some wonky stuff that I can't figure out on my own (hence my need for the book in the first place) in problem 42 of the book. Here is an explicit statement of the problem. I thought parts a. and b. were easy, but when it came to part c., I just kept scratching my head.

To state here the definition of a neighbor fraction given in the book, given some $\frac{a}{b}$ and $\frac{c}{d}$, the two are neighbor fractions when the numerator of $\frac{ad-bc}{bd}$, i.e. $ad-bc$, is equal to $\pm 1$. (He explicitly states that $ad-bc=\pm1$.)

Now part c. is about proving that, given some $\frac{e}{f}$ such that $f<b+d$, there is no $e$ where $\frac{a}{b}>\frac{e}{f}>\frac{c}{d}$ (assuming $\frac{a}{b}>\frac{c}{d}$).

First, I went the common sense route and just thought of a neighbor fraction. Why not $\frac{3}{18}$ and $\frac{4}{18}$? I even worked out $\frac{a+c}{b+d}$ to see if it would neighbor the outsides, like I proved in part two, and reduced it down to $\frac{5}{30}$,$\frac{6}{30}$ and $\frac{9}{45}$,$\frac{10}{45}$. So far so good, right? So if part c. was about proving that there is no $\frac{e}{f}$ between the two, I thought I'd try to find one, to see if I could learn anything about the structure by doing so, and be led to the solution. So, let's see, $b+d=15$, and $f<b+d$, so I'll say $f=14$. Now let's try to find . . . wait.

$\frac{1}{6}=.1\bar{6}$.

$\frac{2}{9}=.\bar{2}$.

$\frac{3}{14}=.2142857$(a pattern I recognize from one of the earlier problems! Good old integer 7.)

So much for that.

Then the following occurred to me:

Let's just suppose for example that there exists some $abcd$ such that the numerator of $\frac{ad-bc}{bd}=\pm1$. Now I give $a'=2a\\b'=2b\\c'=2c\\d'=2d$and checking with the definition, we see that the numerator of $\frac{a'd'-b'c'}{b'd'}=\pm2$, which means that by definition $\frac{a'}{b'}$ and $\frac{c'}{d'}$ are not neighbor fractions, but $\frac{a'}{b'}$=$\frac{a}{b}$ and $\frac{c'}{d'}$=$\frac{c}{d}$, and by supposition $\frac{a}{b}$ and $\frac{c}{d}$ are neighbor fractions, so contradiction. So the definition doesn't seem to work.

Clearly I am in desperate need of foundation work, so if somebody could maybe explain what my misconception is, I would be very grateful. Thanks in advance, I always find this place very helpful!

Edit: Okay, with the last part, part a. addresses that . . . so never mind. Maybe he's just imposing restrictions on what fractions are called neighbor fractions, since, again, $\frac{1}{6}$ and $\frac{2}{9}$ have $\frac{3}{14}$ between them, but also don't satisfy the definition . . . I'm just not sure what's really going on here mathematically.

Edit edit: Ahhhhhh, I completely missed the point of part a. of the problem, upon rereading it. I'm going to revisit that and try again, but responses are still very much welcome.

Last edited: Apr 30, 2016
2. May 5, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. May 5, 2016

### Staff: Mentor

@xwolfhunter, not sure what you're doing, but if you want to "take the common sense route" you need to start with two fractions that actually are neighber fractions according to the definition.

Your first examples of 3/18 and 4/18 don't work, because $ad - bc \ne \pm1$. Note that the reduced forms of these are 1/6 and 2/9, and these aren't neighbor fractions, either (1 * 9 - 2 * 6 = -3).

Here are a couple that actually are neighbor fractions: $\frac 1 2$ and $\frac 2 3$. Here ad - bc = 3 - 4 = -1.