Question about arclength (smooth infinitesimal analysis) In the original version of this question, I asked about the value of $\sqrt{\epsilon^2}$ in smooth infinitesimal analysis. A helpful hint from Andreas Blass led me to check the definition of root, and I found that my original question was misguided since in SIA roots are only guaranteed for positive quantities, excluding $0$ and all infinitesimals.
My question was motivated by trying to understand the derivation of the formula for arclength; in particular the justification for the equation
$$\sqrt{\epsilon^2 + \epsilon f'(x)^2} = \epsilon(\sqrt{1 + f'(x)^2})$$
I know this question came up here, but it wasn't (so far as I can see) addressed head on the context of SIA. And it seems (in light of the above) that there are issues here. Don't we need to know that $\epsilon^2 + \epsilon f'(x)^2 \not=0$ to even know that root is defined? How else to we get that an infinitesimal increment in length is in fact $\epsilon(\sqrt{1 + f'(x)^2})$?
 A: OK, so I managed to resolve my confusion completely in the end. The main points are:
First, square root is only for $> 0$ quantities in SIA.
Second, the arc-length formula for $f$ can therefore not be derived in the intuitive way using the pythagorean theorem on an infinitesimal increase in $f$. What you'd like to do is to observe that the increase in the length along $f$ from $f(x)$ to $f(x + \epsilon)$ is $\epsilon \cdot \sqrt{1 + f'(x)}$ on the way to proving the standard formula. The natural approach is to consider the right triangle whose hypotenuse is the increment in $f$ and whose legs are $\epsilon$ and $\epsilon f'(x)$; the length of the increment would then be given by the pythagorean formula as $\sqrt{\epsilon^2 + \epsilon f'(x)^2}$. But we don't have this square root, since the term occurring in it is plainly $0$. This was what confused me, back in the days when I thought square root was defined everywhere.
How then is the result attained? The answer is there in the Bell, I just got really confused and didn't see it. $1 + f'(x)^2$ is always $>0$, so the root $\sqrt{1 + f'(x)^2}$ is always defined and represents the length of hypotenuse of the right triangle with legs 1 and $f'(x)$. If we divide the length of the infinitesimal increment in $f$ by this quantity we get $\epsilon$ (using some simple trig); one can then use this result to get the standard formula for arc-length (exactly as Bell does in his "primer" on p44).
Sad to see the intuitive argument not vindicated here. :'( :'(
