Answers to Algebra Practical 3

Question 1 §

Question 1a §

$$(110210{)}_3=((((1\cdot 3+1)\cdot 3+0)\cdot 3+2)\cdot 3+1)\cdot 3+0)=345$$

Question 1b §

$$13539=1934\cdot 7+1$$

$$1934=276\cdot 7+2$$

$$276=39\cdot 7+3$$

$$39=5\cdot 7+4$$

$$5=0\cdot 7+5$$

$$\therefore (54321{)}_7$$

Question 2 §

$$0.6\dot{3}5\dot{1}=\frac{m}{n}:$$

$$\text{let }0.6\dot{3}5\dot{1}=x$$

$$10^{4}x=6351.6\dot{3}5\dot{1}⟹10^{4}x-x=9999x=6351$$

$$\therefore x=\frac{6351}{9999}=\frac{2117}{3333}$$

Question 3 §

Drawing 2023-10-13 10.18.38.excalidraw

$$1+\frac{i}{2}-\frac{1}{4}-\frac{i}{8}+\frac{1}{16}+\frac{i}{32}-\frac{1}{64}-\frac{i}{128}+\dots =\frac{1}{1-\frac{i}{2}}=\frac{4}{5}+\frac{2}{5}i$$

Question 4 §

$$(21.201{)}_3\cdot 3^3=(21201{)}_3$$

$$(21201{)}_3=(((2\cdot 3+1)\cdot 3+2)\cdot 3+0)\cdot 3+1=208$$

$$⟹208\cdot 3^{-3}=\frac{208}{27}$$

$$\therefore 7.\dot{7}0\dot{3}$$

Question 5 §

$$8.57:$$

$$8=1\cdot 6+2,1=0\cdot 6+1\therefore 8=(12{)}_6,$$

$$0.57=57\cdot 10^{-3}⟹57=9\cdot 6+3,9=1\cdot 6+3,1=0\cdot 6+1\therefore 57=(133{)}_6$$

$$57=(133{)}_6⟹0.57=0.6157\dot{4}0\dot{7}$$

$$\therefore 8.57=(12.6157\dot{4}0\dot{7}{)}_6$$

may need correction

Question 6 §

$$(1.\dot{0}01\dot{1}{)}_2=$$