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

**Fractions $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{c}{d}$ are called neighbor fractions if their difference $\displaystyle \frac{ad-bc}{bd}$ 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!