What does Spivak mean in chapter on derivatives when he writes that Leibnizian notation is impossible to reconcile with 13 properties of real numbers?

Question

In chapter 9 of Spivak's Calculus, on derivatives, he mentions the "Leibnizian Notation" for the derivative of a function $f$, $\frac{df(x)}{dx}$. In a footnote on page 155, he writes

Leibniz was led to this symbol by his intuitive notion of the derivative, which he considered to be, not the limit of quotients $\frac{f(x+h)-f(x)}{h}$, but the "value" of this quotient when $h$ is an "infinitely small" number. This "infinitely small" quantity was denoted $dx$ and the corresponding "infinitely small" difference $f(x+dx)-f(x)$ by $df(x)$. Although this point of view is impossible to reconcile with properties (P1)-(P13) of the real numbers, some people find this notion of the derivative congenial.

The bold section has been highlighted by me. What does he mean with that?

Answer 1

Within that axiomatic, there are no infinitesimal numbers, that is, there is no number $\mu>0$ such that $(\forall n\in\Bbb N):\mu<\frac1n$. That's so because the Archimedean property follows from those axioms. And that property states that $\Bbb N$ has no upper bound. But if such a number $\mu$ existed, we would have $(\forall n\in\Bbb N):n<\frac1\mu$.

Answer 2

The precise meaning of "infinitesimal" depends on context, but it common to define an infinitesimal number as a positive number $x$ such that $x<1/n$ for every natural number $n$; similarly, an infinite number satisfies $x>n$ for every natural number $n$. With this definition, it can be proven that $\mathbb R$ does not contain infinite or infinitesimal numbers.

To prove that infinite numbers do not exist in $\mathbb R$, assume for the sake of contradiction that there is an infinite number $x\in\mathbb R$. Then, $x$ is an upper bound of $\mathbb N$, and so by the axiom of completeness, $\mathbb N$ has a least upper bound $\alpha$. Since $\alpha-1<\alpha$, it follows that $\alpha-1$ is not an upper bound of $\mathbb N$; in particular, there is an $n_0\in\mathbb N$ such that $n_0>\alpha_0-1$. But then $n_0+1>\alpha$ and $n_0+1\in\mathbb N$, contradicting the fact that $\alpha$ is an upper bound of $\mathbb N$. Hence, $\mathbb N$ is not bounded above in $\mathbb R$.

To prove that infinitesimal numbers do not exist in $\mathbb R$, assume for the sake of contradiction that there is an infinitesimal $y\in\mathbb R$. Then, $n<1/y$ for every natural number $n$, and so $1/y$ is an upper bound of $\mathbb N$, contradicting the fact that $\mathbb N$ is not bounded above in $\mathbb R$.