# Thread: Seemingly easy problem.. modern curriculum's fault?

1. ## Seemingly easy problem.. modern curriculum's fault?

Hi I've recently bought a book called "Algebra" by Gelfand & Shen. The scope of the book is to teach students math the "pure way", which to me seems the way it was instructed a long time ago - not like today, e.g. "a collection of formulas to be learned and applicated without knowing the meaning behind them". Sorry my english is not the best... Which leads us to the fact that I'm swedish. Sweden must be one of the places with the most degraded math instruction. Anyway, here's the problem.

Fractions $\frac{a}{b}$ and $\frac{b}{d}$ are called neighbor fractions if their difference $\frac{ad-bc}{bd}$ has the numerator $±1$, that is, $ad-bc=±1$(that's supposed to be plus minus one).
Prove that
(a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);
(b) if a/b and c/d are neighbor fractions, then $\frac{a+b}{c+d}$is between them and is a neighbor for both a/b and c/d; moreover
(c) no fraction e/f with positive integer e and f such that f<b+d is between a/b and c/d.

Thanks for help.. a little discussion would be nice too

2. Here is something you can try.

(a) First assume the contrary that is

the fraction can be simplified (that is, they have common factors in numerator and denominator);

So write $a\ =\ k\ a_1 ,\ \ \ b\ =\ k\ b_1$

and solve using the fact that ad - bc = + or - 1.

Then repeat for c/d using some other constant

You are bound to end up with some contradiction .

3. Hello strunz
Originally Posted by strunz
... Fractions $\frac{a}{b}$ and $\frac{b}{d}$ Grandad says: I think you mean $\color{red}\frac{c}{d}$
are called neighbor fractions if their difference $\frac{ad-bc}{bd}$ has the numerator $±1$, that is, $ad-bc=±1$(that's supposed to be plus minus one).
Prove that
(a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator)
Assume throughout that $\frac{a}{b}>\frac{c}{d}$ and therefore that $ad-bc=+1$.

Then, suppose that $\frac{a}{b}$ can be simplified because $a$ and $b$ have a common factor $k > 1$. Then, for some $p, q: a = kp,\, b = kq$

$\Rightarrow ad-bc = k(pd-qc) \Rightarrow ad - bc > 1$. Contradiction. So $\frac{a}{b}$ cannot be simplified.

Use the same argument assuming that $\frac{c}{d}$ can be simplified.

(b) if a/b and c/d are neighbor fractions, then $\frac{a+b}{c+d}$ Grandad says: I think you mean $\color{red}\frac{a+c}{b+d}$
is between them and is a neighbor for both a/b and c/d
Assuming again that $\frac{a}{b}>\frac{c}{d}$, we must show that $\frac{a}{b} >\frac{a+c}{b+d}>\frac{c}{d}$.

$\frac{a}{b}-\frac{a+c}{b+d} = \frac{a(b+d)-b(a+c)}{b(b+d)}=\frac{ad-bc}{b(b+d)}=\frac{1}{b(b+d)} > 0$. This also proves that $\frac{a}{b}$ and $\frac{a+c}{b+d}$ are neighbours.

Similarly $\frac{a+c}{b+d}- \frac{c}{d}=\frac{1}{d(b+d)}$