Floating Points

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 ($...,-2,-1,0,1,2,...$) $\mathbb{Z}$:

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

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 $\beta <0$. Additionally we derive the domain of the $\alpha$-term:

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

Such that:

$$1+\frac{\alpha }{2^p}\in [1,2⟩$$

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

$$0.2\cdot 2^{-\beta }\in [1,2⟩$$

For $\beta =-3$ we get:

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

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

$$\begin{array}{rl}1.6& =1+\frac{\alpha }{2^p}\\ 0.6& =\frac{\alpha }{2^p}\\ \frac{3}{5}\cdot 2^p& =\alpha \mid \alpha \in {\mathbb{N}}_1\wedge p\in {\mathbb{N}}_1?!\end{array}$$

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.