Brushing up on my Algebra skills with the book "Algebra" by I.M. Gelfand and reached a problem I am unable to solve.

Fractions $a/b$ and $c/d$ are called neighboring fractions if their difference $\frac{ad-bc}{bd}$ has numerator $\pm1$, that is, $ad-bc = \pm1$.

Prove (a) In this case neither fraction can be simplified (that is, neither has any common factors in numerator or denominator).

What is a correct answer to this problem? Thank you.

  • $\begingroup$ Welcome to MSE. What is your question exactly? Also, please type out the problem rather than linking to a picture. $\endgroup$ – Théophile Sep 16 '18 at 1:39
  • $\begingroup$ Hello. Hopefully you are able to see the image I linked. The problem is contained within. $\endgroup$ – Jack Sep 16 '18 at 1:42
  • $\begingroup$ Yes, I can see the image; please type it out. But what is your question? $\endgroup$ – Théophile Sep 16 '18 at 1:44
  • $\begingroup$ Ok. I am asking for the solution to the problem - a proof $\endgroup$ – Jack Sep 16 '18 at 1:48
  • $\begingroup$ Thanks for writing it out; I've edited it to typeset the math. Have a look at the edits: in general, you can enclose a statement in dollar signs to typeset it (e.g. $x+y$ to give $x+y$). $\endgroup$ – Théophile Sep 16 '18 at 2:03

Assume one can be simplified, so for example $a=pk,b=qk$ with $k \gt 1$. Plug that in and conclude you cannot have $ad-bc=\pm 1$


Thank you Ross Millikan. This is a proof by contrapostive as I understand it. Is the following correct?

Assume $a = pk$ and $ b = qk$ with $k > 1$. Then,

$$ {a\over b} - {c\over d} = {pk\over qk} - {c\over d} = {{dpk - qkc} \over qkd} = {k({dp-cq}) \over bd} $$ ${k({dp-cq})} \ne \pm1$

  • 1
    $\begingroup$ This is effectively correct, but it would be good to justify the last line. $\endgroup$ – Théophile Sep 16 '18 at 3:06

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