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Numerical Methods

Lecture 3 – Base representation
Notes

Lecturer: Stephan Juricke
Date of lecture: February 11th, 2022
Author: Lirik Maxhuni

Base representation
Let 𝑏 ∈ 𝑁\{1}
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𝑛

𝑥 = 𝑎0 𝑏0 + 𝑎1𝑏1 + ⋯ 𝑎𝑛 𝑏𝑛 = ∑ 𝑎𝑖 𝑏𝑖
𝑖=0

𝑎𝑖 ∈ 𝑁0 , 𝑎𝑖 < 𝑏, 𝑎𝑖 ∈ {0, ⋯ , 𝑏 − 1}
where 𝑏 is the base, 𝑎𝑖 are the digits
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:
Base 𝑏 = 10: 37294 = 4 ∙ 100 + 9 ∙ 101 + 2 ∙ 102 + 7 ∙ 103 + 3 ∙ 104
Base 𝑏 = 2: 1011 = 1 ∙ 20 + 1 ∙ 21 + 0 ∙ 12 + 1 ∙ 23 = (1)10 + (2)10 +
(0)10 + (8)10 = (11)10

These calculations can be done via algorithms, that can convert number
systems
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1 𝑖 = 𝑛 = 3:
𝑎3 = 13 𝑑𝑖𝑣 23 = 13 𝑑𝑖𝑣 8 = 1 (𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑜𝑛)
𝑥 = 13 𝑚𝑜𝑑 23 = 13 𝑚𝑜𝑑 8 = 5 (𝑟𝑒𝑠𝑡)
Step 3
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3 𝑖 = 1
𝑎1 = 1 𝑑𝑖𝑣 21 = 1 𝑑𝑖𝑣 2 = 0
𝑥 = 1 𝑚𝑜𝑑 21 = 1 𝑚𝑜𝑑 2 = 1
Step 3
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e
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, (0
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0001100110011
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e
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,
(551
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110 010 100)2
Sets of 3 in binary are 1 digit in octal
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- Horne’s Scheme algorithm does not need division by large numbers
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Computer precision and type of numbers
On computers we have only finite precision (number of digit bits)
Def: Normalized floating point representation with respect to base b, stores a
number x as 𝑥 = 0
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The fact that 𝑎1 ≠ 0 is called a
normalization, it makes representation unique
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:
base 𝑏 = 10: 32
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32213 ∙ 102
base 𝑏 = 2: 𝑥 = ± 0
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e
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e

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