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I am trying to solve the following problem:

Prove that the following fractions are irreducible for any n (n is a natural number and it cannot be null).

  1. $\frac{n}{n+1}$
  2. $\frac{n+1}{2n+3}$
  3. $\frac{3n-5}{4n-7}$

I don't know if my logic is right!

Now for the first one, i used the property which states that $n$ and $n+1$ are always coprime. For the second one i relied on the following property,

  1. if $n|a$ and $n|b$ then $n|(a*k - b*p)$.

So lets suppose $\frac{n+1}{2n+3}$ is reducible, then there is a whole number $d$ which divides both the numerator and denominator. By property 1, $d|[2n+3 - 2(n+1)]$ this means $d|1$ so $d=1\ldots$.

So the fraction is not reducible since every number is divisible by 1.

Is my logic right? how would someone go about solving this problem?

  • 2
    Yes, perfect. Go on.2012-11-17

3 Answers 3