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Why we can't have 0.2

Floating points often get made fun of because they can't compute $0.1 + 0.2$ accurately. This often gets explained as those numbers being inexpressible in a binary format.

We can define a floating point in a binary system with precision $p$ as the set $X$, where $\alpha$ is the mantissa and $\beta$ is the exponent, given natural numbers ($0, 1, 2, 3, ...$) $\mathbb{N}_0$ (where $\mathbb{N}_1$ starts at $1$) and integral numbers ($..., -2, -1, 0, 1, 2, ...$) $\mathbb{Z}$:

$$ X = \bigg\{ \Big(1 + \frac{\alpha}{2^p}\Big) \cdot 2^\beta \mid \alpha\in\mathbb{N}_0\wedge\alpha\lt2^p\wedge p\in\mathbb{N}_1\wedge\beta\in\mathbb{Z}\bigg\} $$

Lets represent an annoying value such as $0.2$. We need to find an $\alpha$ and $\beta$. Since it's smaller than $1$, we know that $\beta\lt0$. Additionally we derive the domain of the $\alpha$-term:

$$ 1 + \frac{\alpha}{2^p} = \begin{cases} 1 & \mid \alpha = 0\\ 2 & \mid \alpha = 2^p \end{cases} $$

Such that:

$$ 1 + \frac{\alpha}{2^p} \in [1, 2\rangle $$

To find $\beta$, first find a power-of-two multiplier that gets it into the range of $[1,2\rangle$:

$$ 0.2 \cdot 2^{-\beta} \in [1,2\rangle $$

For $\beta = -3$ we get:

$$ 1.6 = 0.2 \cdot 2^3 $$

To solve $\alpha$, we look at the fractional part:

$$ \begin{aligned} 1.6 &= 1 + \frac{\alpha}{2^p}\\ 0.6 &= \frac{\alpha}{2^p}\\ \frac{3}{5}\cdot2^p & = \alpha\mid\alpha\in\mathbb{N}_1\wedge p\in\mathbb{N}_1\hspace{2em}?! \end{aligned} $$

The only way to make $\alpha \in \mathbb{N}_1$ is to multiply by a multiple of $5$ (i.e. $0.6 \cdot 5 = 3$) which is impossible since there will be no $5$ (a prime) in the factorization of $2^n$ (power of prime), see fundamental theorem of arithmetic.