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Problem 42 on Gelfand's Algebra (on neighbor fractions)

  1. Nov 12, 2015 #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 + b}{c + 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

    We saw in page 23 that “it is a horrible error (which, of course, you avoid) to add numerators and denominators separately:

    $$\frac{2}{3} + \frac{5}{7} \rightarrow \frac{2 + 5}{3 + 7} = \frac{7}{10}$$

    Instead of the sum this operation gives you something in between the two fractions you started with (##7/10 = 0.7## is between ##2/3 = 0.666##... and ##5/7 = 0.714285##...).


    3. The attempt at a solution

    The first part is proved as follows:


    We prove that if a fraction has common factors in numerator and denominator, we cannot get a possible solution:


    ##\frac{xi}{xj} - \frac{c}{d} = \frac{\pm 1}{xjd}##

    ##\frac{xid - xjc}{xjd} = \frac{\pm 1}{xjd}##

    ##\frac{x\times(id - jc)}{xjd} = \frac{\pm 1}{xjd}##

    ##\frac{id - jc}{jd} = \frac{\pm 1}{xjd}##

    ##\frac{id - jc}{jd}\times jd = \frac{\pm 1}{xjd}\times jd##

    ##id - jc \neq \frac{\pm 1}{x}##

    Except in the case that ##x = \pm 1##


    I think it's ok that way. But now in the following part of the problem I think that it has a mistake, and where he says '##\frac{a + b}{c + d}## is between them and...' it should say '##\frac{a + c}{b + d}## is between them and...'due to what we saw in page 23. If that were the case, then the solution should be:

    ##\frac{a}{b} - \frac{a + c}{b + d} = \frac{\pm 1}{b (b + d)}##

    ##\frac{a (b + d) - b (a + c)}{b (b + d)} = \frac{\pm 1}{b (b + d)}##

    ##\frac{ab + ad - ba - bc}{b (b + d)} = \frac{\pm 1}{b (b + d)}##

    ##\frac{ad - bc}{b (b + d)} = \frac{\pm 1}{b (b + d)}##

    ##\frac{ad - bc}{b^{2} + bd} \times b^{2} = \frac{\pm 1}{b^{2} + bd} \times b^{2}##

    ##\frac{ad - bc}{bd} = \frac{\pm 1}{bd}##

    And that's what we had as a neighbor fraction. The last question is driving me crazy. I can't see how to solve that, and would appreciate a hint or something.

    Thanks for your attention and excuse me for my bad English!
     
  2. jcsd
  3. Nov 12, 2015 #2

    fresh_42

    Staff: Mentor

    Let me give you some common tools at hand so that you can learn something more out of it than just getting a solution.

    If you arrange your fractions as a matrix [tex] A = \begin{bmatrix}a & b\\ c & d\end{bmatrix}[/tex] then you can define

    [tex]A \cdot \bar{A} = \begin{bmatrix}a & b\\ c & d\end{bmatrix} \cdot \begin{bmatrix}\bar{a} & \bar{b}\\ \bar{c} & \bar{d}\end{bmatrix} = \begin{bmatrix}a \cdot \bar{a} + b \cdot \bar{c} & a \cdot \bar{b} + b \cdot \bar{d} \\ c \cdot \bar{a} + d \cdot \bar{c} & c \cdot \bar{b} + d \cdot \bar{d} \end{bmatrix}[/tex] which is called matrix multiplication.

    Further the term ##det(A) = a \cdot d - b \cdot c## is called determinate of ##A##. You can show that ##det(A \cdot \bar{A}) = det(A) \cdot det(\bar{A})##.

    Applying this to your problems (a) and the last part of (b) the solutions can be stated in a more elegant way, although it does just that what you did. For the second part of (b) consider multiplying ##A## with \begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix} and \begin{bmatrix} 1& 0\\ 1 & 1\end{bmatrix}.

    And you are right: ##\frac{a+b}{c+d}## is not in between ##\frac{a}{b}## and ##\frac{c}{d}## as [tex]\begin{bmatrix}1 & 3\\ 2 & 7\end{bmatrix}[/tex] shows.
     
  4. Nov 12, 2015 #3

    fresh_42

    Staff: Mentor

    Hint on (c): Let us assume all integers are positiv, ##a d - b c = 1## and ##0 ≤ \frac {c}{d} ≤ \frac {e}{f} ≤ \frac {a}{b}##
    Then ##bcf ≤ edb ≤ adf = (1 + bc) f = f + bcf < bcf## which is a contradiction, i.e ##\frac{e}{f}## cannot exist.
     
  5. Nov 12, 2015 #4

    haruspex

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    Determinant.
     
  6. Nov 12, 2015 #5

    fresh_42

    Staff: Mentor

    thx, was in a hurry :wink:
     
  7. Nov 16, 2015 #6
    Thank you very much for your answers! Since I haven't studied matrices yet I can't apply what you told me @fresh_42 but I've put a mark on the problem to be back when I begin with matrices!
     
  8. Nov 16, 2015 #7

    haruspex

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    I didn't follow the last step, where you write "< bcf". Where does that come from? You don't seem to have used f<b+d, so maybe it's from that somehow.
     
  9. Nov 16, 2015 #8

    fresh_42

    Staff: Mentor

    You are right. I made a fat mistake. Embarrassing :sorry: I have to restart it.
     
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