Problem Understanding Euclid Book 10 Proposition 1 [closed]

Question

this is embarrassing, but I am having trouble reading through Proposition 1 of Book 10 of Euclid's elements. I'm struggling with Euclid's terminology and don't have a clear picture of what divisions he's making in the lines involved, so not clear what the proof says. Here's the text of the proof with some comments/questions by me embedded in square brackets (I've also numbered the sentences):

Theorem: Let AB and C be two unequal magnitudes of which AB is the greater. I say that, if from AB there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude which is less than the magnitude C.

1. Some multiple DE of C is greater than AB. [So let's say that DE is 31 times C and is greater than AB for example.]

2. Divide DE into the parts DF, FG, and GE equal to C. [Euclid can't mean that he's divided DE into three equal parts each of length C, and in Step 7 he assumes that FD equals C, but is GE also equal to C?]

3. From AB subtract BH greater than its half, and from AH subtract HK greater than its half, and repeat this process continually until the divisions in AB are equal in multitude with the divisions in DE. [Does this mean divide AB into 31 different parts where BH > half AB, and HK > half AH, and KL > half AK, and so on 31 times?]

4. Let, then, AK, KH, and HB be divisions equal in multitude with DF, FG, and GE. [Is Euclid just saying that he's considering each line as having been split into three parts?]

5. Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half [Why is EG less than half of DE? Is this where we have to assume that EG is C? I can't picture what assumptions we're making about the division of DE in Step 2], and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.

6. And, since GD is greater than HA, and there has been subtracted from GD the half GF [Is GF supposed to be half of GD?], and from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.

7. But DF equals C, therefore C is also greater than AK. Therefore AK is less than C.

8. Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C.

Answer 1

I'm reading in David Joyce's transcription. All that follows is what I think he's driving at, but expressed in somewhat more modern terms.

So in the first place, Euclid means suppose that $n$ times the length $C$ (which he'll denote as some length $DE$) is greater than the length $AB$. Presumably the truth of that supposition (for some $n$) is either "self-evident" or was considered earlier. He more or less wants to prove that $2^k C$ exceeds $AB$ for some $k$, and in fact he (more or less) shows $2^n C$ exceeds $AB$. In order to avoid cumbersome notation and to make clear how the general proof would go, he indicates just the argument for $n = 3$. We his students are supposed to supply the ellipsis ... for general $n$ which nowadays of course would be formalized by a proof by induction. But to avoid confusion, I'll explain what (I think) he means in more modern language.

So he assumes, without saying so but without loss of generality, that $n > 2$.

Subdivide $AB$ into $n$ subintervals $[x_i, x_{i+1}]$ for $i = 0, \ldots, n-1$ such that $x_0 = A, x_n = B$, and $x_i - x_0 \leq \frac1{2}(x_{i+1} - x_0)$ for $i = 1, \ldots, n-1$. Subdivide $DE$ into $n$ subintervals $[y_i, y_{i+1}]$ of equal length $C$, where $y_0 = D$ and $y_n = E$.

Since $n > 2$, we have that $y_{n-1} - y_0 = DE - \frac1{n}DE > DE - \frac1{2}DE = \frac1{2}DE > \frac1{2}AB \geq x_{n-1} - x_0$. We could have replaced DE by the expression $y_n - y_0$.

Now he's going to repeat the previous step, replacing $n$ by $n-1$. So similarly, $$y_{n-2} - y_0 = (y_{n-1} - y_0) - \frac1{n-1}(y_{n-1} - y_0) \geq \frac1{2}(y_{n-1} - y_0) > \frac1{2}(x_{n-1} - x_0) \geq x_{n-2} - x_0$$

where the last strict inequality was from the previous paragraph.

He'll keep descending likewise through fractions $\frac{n-k}{n-k-1}$ until he reaches the fraction $\frac1{2}$, which is where the proof ends.

The proof would be clearer to me if he simply made $x_{i-1} - x_0 = \frac1{2}(x_i - x_0)$. The proof seems to boil down to the fact that

$$\frac1{n} = \frac{n-1}{n} \cdot \frac{n-2}{n-1} \cdot \ldots \cdot \frac1{2} > \frac1{2}\cdot \frac1{2} \cdot \ldots \cdot \frac1{2} = \frac1{2^n}.$$