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Gelfand's "Algebra" prob. 42 - corrections & hints?

  1. Jan 3, 2017 #1
    1. The problem statement, all variables and given/known data

    Fractions ##\frac{a}{b}## and ##\frac{c}{d}## are called neighbor fractions if their difference ##\frac{ad - bc}{bd}## has numerator ##\pm 1##, that is, ##ad - bc = \pm 1##.

    Prove that:
    (a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);
    (b) if ##\frac{a}{b}## and ##\frac{c}{d}## are neighbor fractions, then ##\frac{a + c}{b + d}## is between them and is a neighbor fraction for both ##\frac{a}{b}## and ##\frac{c}{d}##; moreover,
    (c) no fraction ##\frac{e}{f}## with positive integer ##e## and ##f## such that ##f < b + d## is between ##\frac{a}{b}## and ##\frac{c}{d}##

    2. Relevant equations

    n/a

    3. The attempt at a solution

    Two points first:
    1) Please point out flaws in presentation as well as errors in math; I'm studying at home as an older adult without classroom assistance. I've chosen Gelfand's book to help me recover and improve on whatever I've forgotten from high school algebra, now more than 40 years ago.
    2) This problem has been previously posted on the forum - however that thread trailed off in a manner not helpful to me, so I will not refer to it again: Problem 42 on Gelfand's Algebra (on neighbor fractions)

    Back to the problem. I think I've got answers for (a) and (b) - see below - but have no one to review them, so please give corrections or hints. As for (c), I am stumped despite several tries & would appreciate clues or hints so that I can get further.

    (a) Prove that neither fraction can be simplified:

    Assume that for ##\frac{a}{b}##, a common factor f exists such that the fraction is actually ##\frac{af}{bf}## .

    Therefore the test for whether the two fractions are neighbors becomes ##\frac{afd - bfc}{bfd}## and the numerator test for establishing neighbors becomes afd – bfc = ##\pm 1##; we can now express afd – bfc as f (ad – bc ).

    If the neighbor test is already true for ##ad - bc = \pm 1##, there is no f other than 1 for f (ad – bc ) that would not change the neighbor test to produce a result other than 1; therefore there can be no common factor f for a neighbor fraction.

    (b) Prove that if if ##\frac{a}{b}## and ##\frac{c}{d}## are neighbor fractions, then ##\frac{a + c}{b + d}## is between them and is a neighbor fraction for both ##\frac{a}{b}## and ##\frac{c}{d}##:

    Surprisingly (at least to me, if I have got this right) this turns out to be a simple algebraic simplification. Let us check ##\frac{a}{b}## and ##\frac{a+c}{b+d}## to see if they can be shown to be neighbors. We disregard the multiplication of the two denominators and just examine whether the difference between the numerators when the fractions are given a common denominator is either +1 or -1:

    a(b+d) - b(a+c)
    becomes ab + ad - ba - bc

    ab cancels ba and this leaves the original ad - bc which we already know is = ± 1, given that we have already established ##\frac{a}{b}## and ##\frac{c}{d}## as neighbor fractions. Essentially the same proof can be done for ##\frac{a+c}{b+d}## and ##\frac{c}{d}##.

    (c) Prove that no fraction ##\frac{e}{f}## with positive integer ##e## and ##f## such that ##f < b + d## is between ##\frac{a}{b}## and ##\frac{c}{d}##

    Here I haven't gotten very far at all; I don't know how to attack the problem. I don't usually try working out equations until I have a sense of what I am trying to achieve. The way the question is posed, it seems the issue is not whether e/f is itself a neighbor fraction, but rather, is there room for any fraction here that is not a neighbor fraction? Given the stipulation that f < b + d, it seems to me that this also requires e < a + c; if we then attempt to give a/b and e/f a common denominator, the difference between numerators would have to be less than 1, thus zero (a = e); but this seems absurd. Likewise if we attempted to give a/b and e/f a common numerator, then the same thing occurs; the difference between denominators would have to be less than 1, thus zero (f = b); and again this seems absurd.

    To me I am going around in circles here: I have developed a rather sketchy verbal argument but don't know how to translate it into equations, and without equations I can't determine its validity. So as I say, hints would be welcome. But also let me know if you think I should demonstrate more of an attempt to solve the problem; or if mistakes made in my attempts at (a) and (b) may be preventing me from understanding (c).

    P.S. One of the issues with Gelfand is that solutions are provided for only about half the problems, which makes it harder for someone doing self-study - this problem #42 is a good example of that. On the other hand, it's a very enjoyable text and I have heard it praised. Should I try & find another text where all the problems have solutions provided? I've tried Googling to learn more about "neighbor fractions", but the only hits I get are links to the Gelfand book; so it seems to be a concept that he & Shen developed, at least by that name. But Gelfand usually seems to give problems for sound pedagogical reasons, so obviously there is something the student is intended to learn here about either fractions or problem-solving.
     
    Last edited: Jan 3, 2017
  2. jcsd
  3. Jan 3, 2017 #2

    Ray Vickson

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    For (b), you have not presented (here, at least) an argument/proof that ##\frac{a+c}{b+d}## lies between ##\frac{a}{b}## and ##\frac{c}{d}##.

    As for (c): of course there are fractions ##e/f## that lie between ##a/b## and ##c/d##, because between any two rational numbers there lie infinitely many other rational numbers. However, the claim is that such fractions must have a pretty large denominator ##f##: ##f## must be at least as large as ##b+d##.
     
  4. Jan 3, 2017 #3
    Thanks, I'll look at (b) and (c) from those perspectives.
     
    Last edited: Jan 3, 2017
  5. Jan 3, 2017 #4
    Meanwhile I am wondering whether the "neighbor fractions" presented in Gelfand are in fact "Farey neighbors." I got the idea to Google for "fractions denominators series" and that led me to this page which referenced "Farey sequences" with a link to an article on that topic in Wikipedia; midway through the article we come to this:

    Etc. I won't read any further for now, but will look through the article later.
     
  6. Jan 4, 2017 #5

    Delta²

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    Just to emphasize something, that all of the numbers appearing in the numerators and denominators (like a,b,c,d,e,f) are integers, cause much of the reasoning presented here does not work if we allow them to take non integer values.

    In a), all we know is that the test is true for ##\frac{af}{bf}## cause we assume that's our original fraction(we don't know yet if it is true for ##\frac{a}{b}##). So we have ##afd-bfc=f(ad-bc)=1##. A product of integers is 1 if and only if both integers are 1 or -1 so f has to be either 1 or -1.

    My idea for (c) would be that if we take ##f<b+d## and e to be the smallest integer for which ##\frac{e-1}{f}<\frac{a}{b}<\frac{e}{f}## then we have to prove that ##\frac{c}{d}<\frac{e}{f}##. (I assumed that ##\frac{a}{b}<\frac{c}{d}##.)
     
    Last edited: Jan 4, 2017
  7. Jan 4, 2017 #6
    Agreed; and I think this is strongly implied by the presentation of this problem and nearby problems in the Gelfand book. But just so I understand, doesn't the term "fraction" by convention imply integers for numerator & denominator - at the very least, unless explicitly stated otherwise? I.e. if I go to Mathworld.Wolfram.com, which I assume is held to be reasonably reliable, and look up "Fraction", the definition begins as follows:
    And if I follow the link for "rational number", that definition begins rather similarly:
     
    Last edited by a moderator: May 8, 2017
  8. Jan 4, 2017 #7

    Delta²

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    So if I understand correctly what you doing in a), you consider that both ##\frac{a}{b}## and ##\frac{af}{bf}## satisfy the neighbour test with ##\frac{c}{d}##. But this is not given by a) all that a) gives us is that ##\frac{a}{b}## satisfies the neighbour test.

    The way you doing it, is proving that given that ##\frac{a}{b}## satisfies the neighbour test with ##\frac{c}{d}## then any fraction of the form ##\frac{af}{bf}## that satisfies the neighbour test has f=1. Ok that is correct but this is not exactly what a) is asking for.
     
  9. Jan 4, 2017 #8
    No, I don't consider that af/bf passes the neighbor test. What I am trying to set up is a "proof by contradiction" - i.e., attempting to insert a common factor f demonstrates that if the test is to be passed, f can only equal 1. I am not sure of the precise definition for "common factor" except that I'm pretty sure 1 doesn't qualify as such, at least not in this context; thus this shows that neighbor fractions such as ##\frac{a}{b}## don't contain common factors.

    Would it help if I used different letters to denote the complication of presuming an additional factor inside a/b? I'm not sure how I would tackle it otherwise; can you suggest a first step?
     
  10. Jan 4, 2017 #9

    Delta²

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    Yes I believe you should assume that ##\frac{a}{b}=\frac{a'f}{b'f}## as first step and given that ##\frac{a}{b}## satisfies the neighbour test (but this doesn't mean that ##\frac{a'}{b'}## satisfies the neighbour test) to prove that f=1.
     
  11. Jan 6, 2017 #10

    TheBlackAdder

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    Apparently there is a typo in (b), taken from an unofficial solutions pdf:

     
  12. Jan 6, 2017 #11
    Yes, others have mentioned the typo & I have already corrected for it, thanks.

    I am hoping to get my brain uploaded again in the next few days so I can tackle these problems again. I have severe sleep deprivation issues due to chronic pain & also a circadian rhythm disorder ("Advanced Sleep Phase Disorder). I'm working with a cognitive behavioral psychologist on slowly getting my sleep cycle more towards normal, but it's a long slog; and in the meantime I have many days where I press the "Think" key on my brain keyboard & get a blue screen, so to speak. Today seems to be another of those days.

    Using a phrase from the part you excerpted, I have found that "solutions" document & downloaded it for future use on any problems I get stuck on - won't be looking at it just yet, though. What web search term did you use to find it??
     
    Last edited: Jan 6, 2017
  13. Jan 6, 2017 #12
    Ha ha ha, my brain seems to have come up to at least a minimal level of function. Despite my morning stupor, I think I've solved this first part of (b). Rather obvious, I suppose, as it just involves doing out the algebraic comparison to see if it works out or not. If someone can check my work & see if I'm right? Here we go:

    Problem 42(b): Prove that if ##\frac { a }{ b }## and ##\frac { c }{ d }## are neighbor fractions (such that their difference ##\frac { ad-bc }{ bd }## has numerator ##\pm 1##, that is, ##ad-bc=\pm 1##), then ##\frac { a+c }{ b+d }## is between them and is a neighbor fraction for both ##\frac { a }{ b }## and ##\frac { c }{ d }##.

    My solution to the first part of the above, showing that ##\frac { a+c }{ b+d }## lies between ##\frac { a }{ b }## and ##\frac { c }{ d }##:

    Let us assume that the particular order of our original two neighbor fractions is ##\frac { a }{ b } <\frac { c }{ d }## (FYI, not that it matters, but this standard convention of left-to-right = smaller-to-bigger also applies to neighboring fractions in "Farey sequences", the phenomenon that Gelfand & Shen borrowed from to create this problem). We need to show that both ##\frac { a }{ b } <\frac { a+c }{ b+d }## and ##\frac { a+c }{ b+d } <\frac { c }{ d }## are true. Let’s start with the former:

    Transform to have a common denominator:
    ##\frac { a(b+d) }{ b^{ 2 }+bd } <\frac { b(a+c) }{ b^{ 2 }+bd }##

    Apply the distributive law to the numerators:
    ##(ab+ad)<(ba+bc)##

    Cancel the common terms to wind up with:
    ##ad < bc##

    That this statement is true is evidenced by its being identical to the relationship obtained for the previously defined neighbor fractions ##\frac { a }{ b }## and ##\frac { c }{ d }##, in which by stipulation ##\frac { a }{ b } < \frac { c }{ d }##, with the resulting transformation to ##\frac { ad }{ bd } <\frac { bc }{ bd }## where the numerator comparison is again ##ad < bc##. I won’t do the second step here for ##\frac { a+c }{ b+d } <\frac { b }{ d }##, but suffice to say it produces the identical result.

    For my solution to the second part of (b), see post #1. It is essentially done along the same lines, just doing out the implied algebra to see if it is true.
     
    Last edited: Jan 6, 2017
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