What is the definition of $(dx)?$ I often see in calculus that we treat the symbols $dx$ and $dy$ as infinitesimal numbers but I don't really accept this definition because in the real number line there's no such thing as infinitesimals.
I read in Calculus by Michael Spivak that the $dx$ has no meaning [Page $264$].

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In MathJax
The symbol $dx$ has no meaning in isolation, any more that the symbol $x \rightarrow$ has any meaning, except in the context $\lim \limits_{x \to a} f(x).$

So Can someone explain it in explicit way and the true definition for it ?
 A: the exact definition is this:

if $\mathbf{x}$ denotes the general point of $\mathbb{R^n}$, we denote
the $i^{th}$ projection function mapping $\mathbb{R^n}$ to
$\mathbb{R}$ by the symbol $x_i$. Then $dx_i$ equals the elementary
1-form $\tilde{\phi_i}$ in $\mathbb{R^n}$.

A: Let a differentiable real function $f:U\subset \Bbb R\to \Bbb R$, a $x\in U$ and a real number $Δx$. We define the differential of $f$ as the real function of two real variables $df$ via $df(x,Δx):=f'(x)Δx$. If $f$ is the identity map$f(x)=id(x)=x$, $x\in U$, then $df(x,Δx)=dx(x,Δx)=(x)'Δx=Δx$, so $dx=Δx$, thus $dx$ means the differential of the identity map. (It is useful a geometric approach by plotting a graph of an arbitrary differentiable function.)
A: I'll take a page off of actual differential geometry here because in one dimension it's actually pretty easy.
Consider the set of all smooth (or even differentiable) functions $\Bbb{R}\to\Bbb{R}$. Call this set $C$. Then at every point $p\in \Bbb{R}$ we can define the following operator:
$$\frac{d}{dx}|_p : C \to \Bbb{R}$$
And it is defined by:
$$\frac{d}{dx}|_p(f) = f'(p)$$
This operator is linear, in the sense that:
$$\frac{d}{dx}|_p(\alpha f + \beta g) = \alpha \frac{d}{dx}|_p (f) + \beta \frac{d}{dx}|_p (g)$$
where $\alpha, \beta$ are real numbers and $f, g \in C$.\
We are interested in operators with these properties because they have amazing algebraic properties and structure.
Now let us define another operator $df$, this time it takes an operator of the kind discussed above and it feeds a pre-determined function $f \in C$ to it.
$$ df(\frac{d}{dx}|_p) = f'(p)$$
This might look like fancy notation for the same thing as before, but as a matter of fact this allows you to "fix the derivative and choose the function to derive". So for example $dg$ will take a derivative and apply it to $g$. This operator is also linear in the sense that:
$$df(\alpha \frac{d}{dx}|_p) = \alpha df(\frac{d}{dx}|_p)$$
with $\alpha$ real number.
In particular, $dx$ applies the operator to the function $x$, which is the identity function, and evaluates it at $p$. So $dx$, as a function of operators, is the function such that $dx(\frac{d}{dx}|_p) = 1$ for all $p$.
This has nothing to do with infinitesimals, and in fact the relationship between this definition of $dx$ and the integral is more convoluted than just assuming $dx$ to mean tiny steps.\
To conclude, in standard analysis (and differential geometry) infinitesimals do not exist in the sense of tiny steps.
For those who know what I'm talking about already: I know I'm being sloppy, but I think it's best to give an intuitive idea of what is going on.
