Even though $$\varphi = \frac{1 + \sqrt{5}}{2}, \quad \psi = \frac{1 - \sqrt{5}}{2}$$ are irrational numbers, I can calculate the number \$F_n\$,

$$F_n = \frac{\varphi^n - \psi^n}{\varphi - \psi} = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$

for any positive integer \$n\$ without ever calculating any kind of an approximation for \$\varphi\$ or \$\psi\$.

I bet you can, too, because the sequence \$F_0=0, F_1=1, F_2=1, F_3=2, F_4=5, \dots, F_n = F_{n-1} + F_{n-2}\$ is the sequence of

Fibonacci numbers,

OEIS A000045.

Thus, while you cannot specify the value of irrational numbers exactly, you can specify their behaviour and use, which makes them exceedingly useful as numbers. Many other irrational numbers (like \$\pi\$, \$e\$ for example) have similar use cases.

Let's say we have a continuous function \$f(x)\$.

The slope of this function is \$f^\prime(x)\$:

$$f^\prime(x) = \frac{d ~ f(x)}{d ~ x} = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

The right side does not mean that \$\frac{d f(x)}{d x} = \frac{f(x + h) - f(x)}{h}\$; you can't just ignore the

limit operator \$\lim\$ like that!

What that right side actually means (being the definition of a

derivative), is that the value of the expression at \$x\$ is determined by the

*behaviour* of the ratio between the change in \$f\$ and the change in \$x\$, as the change in \$x\$ tends towards zero.

That is, we never, ever try to divide anything by zero there. Instead, we can think of the process as looking at how the ratio changes as \$h\$ gets closer to zero, and

extrapolate the value the ratio would be if we could calculate it at \$h = 0\$.

The power of

calculus is that instead of having to do that rigmarole for every value of \$x\$ we're interested in – especially considering how careful one would need to be and how much work it would be to obtain the correct ratio! –, calculus tells us directly how we can obtain the slope as a function, \$f^\prime(x)\$, from the mathematical expression of \$f(x)\$ itself, with very little work.

For example, if we have $$f(x) = 3 \sin(x) + 2 \cos(3 x^2 - x + 2)$$calculus tells us that $$f^\prime(x) = \frac{d f(x)}{d x} = 3 \cos(x) + (2 - 12 x)\sin\left(3 x^2 - x + 2\right)$$

In short, because the limit operator tells us the behaviour of a function at a value where the function itself is not defined, there is absolutely no "division by zero". And it is a powerful operator, because even in cases where we have something like \$\lim_{x \to c} \frac{f(x)}{g(x)}\$ with \$f(c) = g(c) = 0\$ (or \$\pm\infty\$) and both \$f(x)\$ and \$g(x)\$ continuous (or differentiable) around \$x = c\$ but perhaps not at exactly \$x = c\$, it can still yield the correct answer (via

l'Hôpital's rule).

Simplifying all this hard work done by mathematicians over the centuries by claiming

*my number line is different* is, well, pretty arrogant. Does your number line work as well as the one we use right now does?