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fix: Modify some formulas in number_encoding.md to enhance readability

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8
docs/chapter_data_structure/number_encoding.md
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@ -105,7 +105,7 @@ $$
二进制数 `float` 对应值的计算方法为：
$$
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 + b_{22} b_{21} \ldots b_0\right)_2
$$
转化到十进制下的计算公式为：
@ -119,7 +119,7 @@ $$
$$
\begin{aligned}
\mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}]
(1 + \mathrm{N}) = & (1 + \sum_{i=0}^{22} b_{i} 2^{i-23}) \subset [1, 2 - 2^{-23}]
\end{aligned}
$$
@ -141,8 +141,8 @@ $$
| 指数位 E | 分数位 $\mathrm{N} = 0$ | 分数位 $\mathrm{N} \ne 0$ | 计算公式 |
| ------------------ | ----------------------- | ------------------------- | ---------------------------------------------------------------------- |
| $0$ | $\pm 0$ | 次正规数 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ |
| $1, 2, \dots, 254$ | 正规数 | 正规数 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ |
| $0$ | $\pm 0$ | 次正规数 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0+\mathrm{N})$ |
| $1, 2, \dots, 254$ | 正规数 | 正规数 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1+\mathrm{N})$ |
| $255$ | $\pm \infty$ | $\mathrm{NaN}$ | |
值得说明的是，次正规数显著提升了浮点数的精度。最小正正规数为 $2^{-126}$ ，最小正次正规数为 $2^{-126} \times 2^{-23}$ 。

8
en/docs/chapter_data_structure/number_encoding.md
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@ -105,7 +105,7 @@ According to the IEEE 754 standard, a 32-bit `float` consists of the following t
The value of a binary `float` number is calculated as:
$$
\text{val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2 - 127} \times \left(1 . b_{22} b_{21} \ldots b_0\right)_2
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 + b_{22} b_{21} \ldots b_0\right)_2
$$
Converted to a decimal formula, this becomes:
@ -119,7 +119,7 @@ The range of each component is:
$$
\begin{aligned}
\mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} \times 2^{-i}) \subset [1, 2 - 2^{-23}]
(1 + \mathrm{N}) = & (1 + \sum_{i=0}^{22} b_{i} 2^{i-23}) \subset [1, 2 - 2^{-23}]
\end{aligned}
$$
@ -141,8 +141,8 @@ As shown in the table below, exponent bits $\mathrm{E} = 0$ and $\mathrm{E} = 25
| Exponent Bit E | Fraction Bit $\mathrm{N} = 0$ | Fraction Bit $\mathrm{N} \ne 0$ | Calculation Formula |
| ------------------ | ----------------------------- | ------------------------------- | ---------------------------------------------------------------------- |
| $0$ | $\pm 0$ | Subnormal Numbers | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ |
| $1, 2, \dots, 254$ | Normal Numbers | Normal Numbers | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ |
| $0$ | $\pm 0$ | Subnormal Numbers | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0+\mathrm{N})$ |
| $1, 2, \dots, 254$ | Normal Numbers | Normal Numbers | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1+\mathrm{N})$ |
| $255$ | $\pm \infty$ | $\mathrm{NaN}$ | |
It's worth noting that subnormal numbers significantly improve the precision of floating-point numbers. The smallest positive normal number is $2^{-126}$, and the smallest positive subnormal number is $2^{-126} \times 2^{-23}$.

8
zh-hant/docs/chapter_data_structure/number_encoding.md
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@ -105,7 +105,7 @@ $$
二進位制數 `float` 對應值的計算方法為：
$$
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 . b_{22} b_{21} \ldots b_0\right)_2
\text {val} = (-1)^{b_{31}} \times 2^{\left(b_{30} b_{29} \ldots b_{23}\right)_2-127} \times\left(1 + b_{22} b_{21} \ldots b_0\right)_2
$$
轉化到十進位制下的計算公式為：
@ -119,7 +119,7 @@ $$
$$
\begin{aligned}
\mathrm{S} \in & \{ 0, 1\}, \quad \mathrm{E} \in \{ 1, 2, \dots, 254 \} \newline
(1 + \mathrm{N}) = & (1 + \sum_{i=1}^{23} b_{23-i} 2^{-i}) \subset [1, 2 - 2^{-23}]
(1 + \mathrm{N}) = & (1 + \sum_{i=0}^{22} b_{i} 2^{i-23}) \subset [1, 2 - 2^{-23}]
\end{aligned}
$$
@ -141,8 +141,8 @@ $$
| 指數位 E | 分數位 $\mathrm{N} = 0$ | 分數位 $\mathrm{N} \ne 0$ | 計算公式 |
| ------------------ | ----------------------- | ------------------------- | ---------------------------------------------------------------------- |
| $0$ | $\pm 0$ | 次正規數 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0.\mathrm{N})$ |
| $1, 2, \dots, 254$ | 正規數 | 正規數 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1.\mathrm{N})$ |
| $0$ | $\pm 0$ | 次正規數 | $(-1)^{\mathrm{S}} \times 2^{-126} \times (0+\mathrm{N})$ |
| $1, 2, \dots, 254$ | 正規數 | 正規數 | $(-1)^{\mathrm{S}} \times 2^{(\mathrm{E} -127)} \times (1+\mathrm{N})$ |
| $255$ | $\pm \infty$ | $\mathrm{NaN}$ | |
值得說明的是，次正規數顯著提升了浮點數的精度。最小正正規數為 $2^{-126}$ ，最小正次正規數為 $2^{-126} \times 2^{-23}$ 。