Prove that no cancellation is possible for $$\frac{a_1 + a_2}{b_1 + b_2}$$ if $a_1 b_2-a_2 b_1=\pm 1$.

I'm new at number theory so if you can be simple it would be great.

Here is what I think:

For there to be no cancellation the gcd of them should be $\pm 1$

$$\gcd((a_1+a_2), (b_1+b_2))=d$$

So we have to prove that $d=\pm 1$. So suppose that $d=\pm 1$.

$$\pm 1|a_1+a_2$$

$$\pm 1|b_1+b_2$$

Then $\pm 1=a_1 b_2-a_2 b_1$ ... Nothing from this


Hint: If $d$ divides $a_1+a_2$ and $b_1+b_2$, then $d$ divides $(a_1+a_2)b_2-(b_1+b_2)a_2$. Expand.

  • $\begingroup$ Great. Thanks bro. $\endgroup$ – Shtyllat Dec 28 '15 at 19:29
  • 1
    $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Dec 28 '15 at 19:32
  • $\begingroup$ I'm a bit confused, I understand that $d$ divides $a_1 b_2-a_2 b_1$ but I'm not sure about $a_1 b_2-a_2 b_1=\pm 1$, many thanks $\endgroup$ – aibohphobia Dec 29 '15 at 22:00
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    $\begingroup$ We are told that $a_1b_2-a_2b_1=\pm 1$. We are asked to deduce that $a_1+a_2$ and $b_1+b_2$ have no factor $\gt 1$ in common. Suppose to the contrary that $d\ge 2$ divides both $a_1+a_2$ and $b_1+b_2$. Then by the answer above, $d$ divides $a_1b_2-a_2b_1$. But this is impossible, since $a_1b_2-a_2b_1=\pm 1$, and no $d\ge 2$ divides $1$ or $-1$. $\endgroup$ – André Nicolas Dec 29 '15 at 22:05

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