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 $y_{n-1} - y_0 = DE - \frac1{n}DE = \frac{n-1}{n}DE > \frac1{2}AB > x_{n-1} - x_0$. 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, $$y_{n-2} - y_0 = (y_{n-1} - y_0) - \frac1{n-1}(y_{n-1} - y_0) = \frac{n-2}{n-1}(y_{n-1} - y_0) > \frac{n-1}{n-2}(x_{n-1} - x_0) \geq \frac1{2}(x_{n-1} - x_0) \geq x_{n-2} - x_0$$

where the penultimate 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

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

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

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

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 $x_i - x_0 < \frac1{2}(x_{i+1} - x_0)$ 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 $y_{n-1} - y_0 = DE - \frac1{n}DE = \frac{n-1}{n}DE > \frac1{2}AB > x_{n-1} - x_0$. 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, $$y_{n-2} - y_0 = (y_{n-1} - y_0) - \frac1{n-1}(y_{n-1} - y_0) = \frac{n-2}{n-1}(y_{n-1} - y_0) > \frac{n-1}{n-2}(x_{n-1} - x_0) \geq \frac1{2}(x_{n-1} - x_0) > x_{n-2} - x_0$$

where the penultimate 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.

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 $y_{n-1} - y_0 = DE - \frac1{n}DE = \frac{n-1}{n}DE > \frac1{2}AB > x_{n-1} - x_0$. 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, $$y_{n-2} - y_0 = (y_{n-1} - y_0) - \frac1{n-1}(y_{n-1} - y_0) = \frac{n-2}{n-1}(y_{n-1} - y_0) > \frac{n-1}{n-2}(x_{n-1} - x_0) \geq \frac1{2}(x_{n-1} - x_0) \geq x_{n-2} - x_0$$

where the penultimate 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

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