# Thread: Proof of neighbor fractions

1. ## Proof of neighbor fractions

*If I turn out to have a wrong answer, please no hints or showing an valid proof. I want to do it on my own !

ad/bd-bc/bd=+-1/bd is neighbor fraction

Now, reduce the common numbers :

a/b-c/d=+-1/bd

We must now prove that the left hand side has irreductible fractions. Lets see what would happen if these fraction were reductible.

let a=z*y , b=z*l , c=p*m d=p*n

z*y/z*l-p*m/p*n equality to be determined +-1/(z*l)*(p*n)

Reduce:

*ln (y/l-m/n) equality to be determined (+-1/(z*l)*(p*n))*ln

yn-lm equality to be determined +-1/z*p

yn-lm not= +-1/z*p

We have considered the initial fractions to be reductible and have arrived at a false result. An integer cannot be equal to (+-1/z*p)

So, the initial fractions must be irreductible.

*Of course, I'm considering the variables to represents integers.

2. ## Re: Proof of neighbor fractions

What, exactly, are you trying to prove? You say you want to show that the fractions on the left side, $\frac{a}{b}$ and $\frac{c}{d}$, are reduced to lowest terms. But that is given, isn't it? The definition says that two fractions are "neighbor fractions" if and only if $\frac{a}{b}- \frac{c}{d}= \frac{z}{bd}$ where $z= \pm 1$ which assumes that the fractions are in lowest terms.

3. ## Re: Proof of neighbor fractions

What I'm trying to prove is that these neighbor fraction are irreductible. This is exactly what the first question asks. We don't know that these fractions are irreductible or not. We just know that they're neighbor.(Which doesn't imply that they're irreductible.That's how I understood it atleast.)

Hope I was clear!

4. ## Re: Proof of neighbor fractions

So "neighbor fractions" is NOT a property of the numbers, just how they happen to be written? That is, 1/2 and 1/3 are "neighbor fractions" but 2/4 and 1/3 are not? Prove that if a and b, in the fraction a/b, have common factor, n, then ad- bc is a multiple of n?