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Question Number 146979 by ajfour last updated on 16/Jul/21

If there just were 0,±1,±2,±3,±4,±i in place of the decimal or binary equivalets..

$${If}\:{there}\:{just}\:{were} \\ $$$$\:\:\mathrm{0},\pm\mathrm{1},\pm\mathrm{2},\pm\mathrm{3},\pm\mathrm{4},\pm{i} \\ $$$${in}\:{place}\:{of}\:{the}\:{decimal}\:{or} \\ $$$${binary}\:{equivalets}.. \\ $$

Commented by Rasheed.Sindhi last updated on 17/Jul/21

Sir, you mean 5-Base system?

$$\mathcal{S}{ir},\:{you}\:{mean}\:\mathrm{5}-{Base}\:{system}? \\ $$

Answered by ajfour last updated on 23/Jul/21

4+1=10 multiplication tables 1×1=1 2×1=2 1×2=2 2×2=4 1×3=3 2×3=11 1×4=4 2×4=13 1×10=10 2×10=20 3×1=3 4×1=4 3×2=11 4×2=13 3×3=14 4×3=22 3×4=22 4×4=31 3×10=30 4×10=40 (1/(10))=0.1 , 0−1=−1 2−21=−14, .. so on ......

$$\:\:\mathrm{4}+\mathrm{1}=\mathrm{10} \\ $$$$\:{multiplication}\:{tables} \\ $$$$\:\:\mathrm{1}×\mathrm{1}=\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$$$\:\:\mathrm{1}×\mathrm{2}=\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{2}×\mathrm{2}=\mathrm{4} \\ $$$$\:\:\mathrm{1}×\mathrm{3}=\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{2}×\mathrm{3}=\mathrm{11} \\ $$$$\:\:\mathrm{1}×\mathrm{4}=\mathrm{4}\:\:\:\:\:\:\:\:\mathrm{2}×\mathrm{4}=\mathrm{13} \\ $$$$\:\:\mathrm{1}×\mathrm{10}=\mathrm{10}\:\:\:\mathrm{2}×\mathrm{10}=\mathrm{20} \\ $$$$ \\ $$$$\:\:\mathrm{3}×\mathrm{1}=\mathrm{3}\:\:\:\:\:\:\:\:\mathrm{4}×\mathrm{1}=\mathrm{4} \\ $$$$\:\:\mathrm{3}×\mathrm{2}=\mathrm{11}\:\:\:\:\:\mathrm{4}×\mathrm{2}=\mathrm{13} \\ $$$$\:\:\mathrm{3}×\mathrm{3}=\mathrm{14}\:\:\:\:\:\mathrm{4}×\mathrm{3}=\mathrm{22} \\ $$$$\:\:\mathrm{3}×\mathrm{4}=\mathrm{22}\:\:\:\:\:\mathrm{4}×\mathrm{4}=\mathrm{31} \\ $$$$\:\:\mathrm{3}×\mathrm{10}=\mathrm{30}\:\:\:\mathrm{4}×\mathrm{10}=\mathrm{40} \\ $$$$ \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{10}}=\mathrm{0}.\mathrm{1}\:\:\:,\:\mathrm{0}−\mathrm{1}=−\mathrm{1} \\ $$$$\:\:\mathrm{2}−\mathrm{21}=−\mathrm{14},\:..\:\:{so}\:{on} \\ $$$$\:\:...... \\ $$

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