The set of positive integers starts {1, 2, 3, …} and there are no positive integers smaller than 1. On the other hand the set of positive real numbers has no smallest element, since for any positive real number x, the number x/2 is smaller and still a positive real. It’s not that the reals are unorderable (they are most certainly ordered), it’s that certain subsets of the reals do not necessarily contain a minimum/maximum.

It's not just about ordering, but well-ordering. Which says not just that you can put them in order, but that you can put them in order with a first element (one that's always less than all of the others).

So you can have a set like "all real numbers greater than 0". It doesn't have an obvious first element, even though you can always compare any two numbers and say "this one is less than that one".

There may be a non-obvious first element, i.e. some strategy for picking a first element that applies to all sets. That's the Well-Ordering Principle. Which feels false to most mathematicians, but it's equivalent to the Axiom of Choice, which feels true.

How does it contradict that? The set of negative integers does not have a least element, and neither does the set of negative reals. (But the former does have a greatest element (–1) while the latter doesn’t have that, either.)

Oh dear, I dropped the critical word "positive" from this in my reading. My bad.

> why real numbers are somehow unorderable

They are orderable ... For all x, y in R, either x >= y or x < y.

But R is not bounded (no inf, sup) and therefore has no minimum and no maximum.

What's the consequence of extending the reals with infinities, as my textbook for real analysis does?

That every closed set has a maximum and a minimum.

Relatedly, every set gets a supremum and an infumum. (The first two must be members of the set. The second two can lie outside the set).

Without the extension, you need to make an exception for unbounded sets with the above rules.