Incommensurable line segments

Question

I have an issue with a proof given in my lecture hopefully someone can help me with. It will be shown that the hypotenuse $c$ of a right-angled and isosceles triangle will be incommensurable to the leg $a$

Two line segments $a,c$ are called commensurable iff there exist a third line segment $d$ and two natural numbers $m,n$ such that $a=m\cdot d, c=n\cdot d$. Otherwise $a,c$ are called incommensurable.

In this context we have discussed the euclidean algorithm. Roughly said, one starts with two line segments ($a,c$) and subtracts in each step the shorter one from the larger one ($c-a$) and replaces thereafter the longer line segment by it (replace $c$ by $c-a$). After each step one compares the new line segments ($a, c-a$) and repeat the whole process till one get two congruent line segments.

Two line segments are commensurable iff the euclidean algorithm stops after finitely many steps.

Now the short proof I do not understand:

1. step: $c>a$. Therefore we get the new line segments $a,r_1$ with $r_1:=c-a$

2. step: $a>r_1$. Therefore we get our new line segments $r_1, r_2$ with $r_2=a-r_1=2a-c$

But now we observe that $\frac{r_2}{c}=\frac{r_1}{a}$. Hence the line segments $a,c$ can not be commensurable.

I understand everything but the last conclusion. Why can not $a,c$ be commensurable because of the equality of the given ratios? Suppose I would get after $N$ steps two line segments $r_N, z_N$ with $\frac{r_N}{c}=\frac{z_N}{a}$. Can I conclude in this situation that $a,c$ will not be incommensurable as well?

I would be very happy if someone can help me.

Best wishes

Answer 1

The justification for this conclusion lies in scaling. The fact that those ratios are equal means that the pair $r_1$, $r_2$ is just a scaled-down version of the pair $a$, $c$. Since the entire process is linear, performing the next two steps will again just scale down the values by the same factor, and so on for every pair of two steps. It follows that the algorithm does not terminate after finitely many steps.

Answer 2

The point is that the pair $a,c$ is supposed to be in lowest terms. If we have $c=14$ and $a=10$, then $c=2\cdot7$ and $a=2\cdot5$. That pair is not in lowest terms. If $14d_1$ is the length of the hypotenuse and $10d_1$ is the length of the leg, then one my let $d_2=2d_1$ and then $7d_1$ is the length of the hypotenuse and $5d_2$ is the length of the leg. The pair $7,5$ cannot be further so reduced, so it's in lowest terms.

Your new pair, $r_1,r_2$, is in lower terms than $a,c$, i.e. we have $0<r_1<a$ and $0<r_2<c$. Thus something that was in lowest terms has been reduced to still lower terms --- a contradiction. That proves the impossibility of having any such pair.