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xwolfhunter

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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

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}##

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.

*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.

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