- #1
vsgarciaaleman
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Homework Statement
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}##
Homework 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##...).
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!