# Thread: Proof concerning "neighbor fractions"

1. ## Proof concerning "neighbor fractions"

Hello, I am working with Israel Gelfand's book "Algebra" and have another question about one of the exercises!

Fractions $\frac{a}{b}$ and $\frac{c}{d}$ are called neighbor fractions if their difference $\frac{ad-bc}{bd}$ has numerator ±1, that is, ad–bc = ±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 $\frac{c}{d}$.

I understand that this is a three-part proof and so I'm not expecting any complete answers, but just some comments or tips in the right direction would help immensely.

For (a), I understand how it is true and I can sort of describe why ("multiples of factors greater than one will either be identical or will be separated by a distance greater than one, so if x is the factor then a(x)d–b(x)c will either equal 0 or have an absolute value greater than the absolute value of one"), but I'm not sure how to write this up or if it's totally accurate.

For (b) and (c) I am confused because from the examples I tried it seems like $\frac{c+d}{a+b}$ should be the one in between $\frac{a}{b}$ and $\frac{c}{d}$ and not $\frac{a+b}{c+d}$. Also it seems like (c) should be a stipulation for (b) (and not just an addition to it), since there are pairs like $\frac{4}{9}$ and $\frac{3}{7}$ which are neighbor fractions but for which neither (b) nor (c) are true.

Any help is greatly appreciated!

2. Originally Posted by Ragnarok
Hello, I am working with Israel Gelfand's book "Algebra" and have another question about one of the exercises!

Fractions $\frac{a}{b}$ and $\frac{c}{d}$ are called neighbor fractions if their difference $\frac{ad-bc}{bd}$ has numerator ±1, that is, ad–bc = ±1. Prove that
(a) in this case neither fraction can be simplified (that is, neither has any common factors in numerator and denominator);

I understand that this is a three-part proof and so I'm not expecting any complete answers, but just some comments or tips in the right direction would help immensely.

For (a), I understand how it is true and I can sort of describe why ("multiples of factors greater than one will either be identical or will be separated by a distance greater than one, so if x is the factor then a(x)d–b(x)c will either equal 0 or have an absolute value greater than the absolute value of one"), but I'm not sure how to write this up or if it's totally accurate.
Any help is greatly appreciated!
Hello,

I will comment on part (a) for now, and then when I have more time, I will comment on the remaining parts (unless someone else beats me to it).

Your thinking is correct. Suppose that a and b have a common factor x>1. Then a = mx and b=nx for some integers m and n. But then:

ad-bc=mxd-nxc = x(md-nc). Here both x and md-nc are integers, and furthermore x>1. Hence their product cannot possibly be 1.

(Why? If md-nc>0, then x(md-nc) is greater than or equal to x, which is strictly greater than 1. If md-nc=0, then the product is 0. If md-nc<0, then x(md-nc) < md-nc <0, which thus can't be 1)

3. Thank you so much for taking the trouble! I know it is a complicated problem (at least, it was for me). I will keep working on parts (b) and (c) in the meantime – I haven't had a chance to until now.

4. I realize that in my other post, I forgot to argue why ad-bc cannot be -1, but the argument of this would be similar. The easy argument is of course that if x>1, then x(ad-bc) cannot be 1 or -1, because if ab-bc>0, then $x(ad-bc)\geq x>1$, and if ad-bc<0, then $x(ad-bc)\leq -x < -1$, and if ad-bc=0, then x(ad-bc)=0.

As for the fractions $\frac{4}{9}$ and $\frac{3}{7}$, I get that $\frac{a+c}{b+d} = \frac{7}{16}$ is a fraction between the two.

Edit: I realize that the text says (a+b)/(c+d) and not (a+c)/(b+d) like I wrote, which is probably the cause of the confusion. But it seems that what I thought it said is also what it intended to say.

For part b, suppose that $\frac{a}{b}$ and $\frac{c}{d}$ are neighbor fractions, and assume that $\frac{a}{b} < \frac{c}{d}$. We may assume that both b and d are positive integers, thus letting a or c be equipped with the potential sign. So by multiplying out the denominators of this inequality, we get ad < bc, or ad-bc <0. Prove first that:

$\frac{a}{b}<\frac{a+c}{b+d}.$

Since both denominators are positive, this is equivalent to

a(b+d) < b(a+c)

or ad < bc

or ad-bc < 0.

But this last inequality is true, and hence so was the desired inequality. SImilarly you can prove that

$\frac{a+c}{b+d}<\frac{c}{d}$.

Note: For this point, we didn't even use that the fractions were neighbors. In conclusion, the fraction $\frac{a+c}{b+d}$ is between $\frac{a}{c}$ and $\frac{b}{d}$ no matter what the fractions are.

To prove that when $\frac{a}{b}$ and $\frac{c}{d}$ are neighbors, then so are (for example) $\frac{a}{b}$ and $\frac{a+c}{b+d}$, we need to check that the number

$a(b+d) - b(a+c)$

is either +1 or -1.

But this expression is identical to:

ab + ad - ba - bc = ad-bc,

which is either +1 or -1 by assumption.

Similarly one can prove that (a+b)/(c+d) and c/d are neighbors.

Hope this helps. I will probably look at part c at some later time.

5. Thank you so much! I'm sorry I haven't gotten back to this earlier; I've been super busy the last few days. I believe I can do part (c) on my own. Again, thank you – I really appreciate it.