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

[vsgarciaaleman]

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!

 

[fresh_42]

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 A = \begin{bmatrix}a &amp; b\\ c &amp; d\end{bmatrix} then you can define

A \cdot \bar{A} = \begin{bmatrix}a &amp; b\\ c &amp; d\end{bmatrix} \cdot \begin{bmatrix}\bar{a} &amp; \bar{b}\\ \bar{c} &amp; \bar{d}\end{bmatrix} = \begin{bmatrix}a \cdot \bar{a} + b \cdot \bar{c} &amp; a \cdot \bar{b} + b \cdot \bar{d} \\ c \cdot \bar{a} + d \cdot \bar{c} &amp; c \cdot \bar{b} + d \cdot \bar{d} \end{bmatrix} 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 \begin{bmatrix}1 &amp; 3\\ 2 &amp; 7\end{bmatrix} shows.

 

[fresh_42]

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.

 

[haruspex]

the term det(A)=a⋅d−b⋅c is called determinate of A.

Determinant.

 

[fresh_42]

Determinant.

thx, was in a hurry

 

[vsgarciaaleman]

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!

 

[haruspex]

Then $bcf ≤ edb ≤ adf = (1 + bc) f = f + bcf < bcf$ which is a contradiction, i.e $\frac{e}{f}$ cannot exist.

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.

 

[fresh_42]

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.

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