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Number Base System I | Week 16 Topics2 Quizzes
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Number Base System II | Week 23 Topics
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Number Base System III | Week 32 Topics1 Quiz
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Modular Arithmetic I | Week 42 Topics
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Modular Arithmetic II | Week 53 Topics
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Modular Arithmetic III | Week 64 Topics1 Quiz
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Indices I | Week 73 Topics1 Quiz
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Indices II | Week 81 Topic1 Quiz
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Logarithms I | Week 93 Topics
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Logarithms II | Week 104 Topics1 Quiz
Topic Content:
- Conversion from one Base to Another
It is important to know that a number given in one base other than base ten can be converted to another base via base ten.
Example 2.3.1:
Evaluate the following and express the answer on the bases indicated.
(i) 10401.117 to base eight
(ii) 2013 – 101012 + 2536 to base 6
(i) Convert 10401.117 to base ten
Solution:
i.e \( \scriptsize 10401.11_7 \\ \scriptsize = 1 \: \times \: 7^4 \: + \: 0 \: \times \: 7^3 \: + \: 4 \: \times \: 7^2 \\ \scriptsize \:+\: 0 \: \times \: 7^1 \: + \: 1 \: \times \: 7^0 \: + \: 1 \: \times \: 7^{-1} \: + \: 1 \: \times \: 7^{-2} \)
= \( \scriptsize 20141 \: + \: 0 \: + \: 196 \: + \: 0 \: + \: 1 \: + \: 0.142 \: + \: 0.0204 \)
= \( \scriptsize 2598.1624 \)
Convert to base 8
Whole number part
Decimal part
i.e. \( \scriptsize 0.1624_8 = 0.123_8 \)
Aswer: \( \scriptsize 10401.11_7 = 5046.123_{8} \)
(ii) 2013 – 101012 + 2536 to base 6
Solution:
Convert all numbers to base ten
\(\scriptsize 201_3 \\ \scriptsize = 2 \: \times \: 3^2 \: + \: 0 \: \times \: 3^1 \: + \: 1 \: \times \: 3^0 \\ \scriptsize = 18 \: + \: 0 \: + \: 1 \\ \scriptsize = 19 \) \( \scriptsize 10101_2 \\ \scriptsize = (1 \: \times \: 2^4 )\: + \: (0 \: \times \: 2^3) \: + \: (1 \: \times \: 2^2) \\ \scriptsize \: + \: (0 \: \times \: 2^1) \: + \: (1 \: \times \: 2^0) \\ \scriptsize = 16 \: + \: 0 \: + \: 4 \: + \: 0 \: + \: 1 \\ \scriptsize = 21 \) \(\scriptsize 253_6 \\ \scriptsize = 2 \: \times \: 6^2 \: + \: 5 \: \times \: 6^1 \: + \: 3 \: \times \: 6^0 \\ \scriptsize = 72 \: + \: 30 \: + \: 3 \\ \scriptsize = 105 \)i.e \( \scriptsize 201_3 \: – \: 10101_2 \: + \: 253_6 \\ \scriptsize = 19 \: – \: 21 \: + \: 105 \\ \scriptsize = 124 \: – \: 21 \\ \scriptsize = 103 \)
Answer: \( \scriptsize 201_3 \: – \: 10101_2 \: + \: 253_6 = 251_6 \)
Example 2.3.2:
Find the value of x in the following:
(i) \( \scriptsize 315_x \: -\: 223_x = 72_x \)
(ii) \( \scriptsize (251.25)_{10} = x_7\)
(i) \( \scriptsize 315_x \: -\: 223_x = 72_x \)
Solution:
\( \scriptsize 315_x = 3 \: \times \: x^2 \: + \: 1 \: \times \: x^1 \: + \: 5 \: \times \: x^0 \\ \scriptsize = 3x^2 \: + \: x \: + \: 5 \) \( \scriptsize 223_x = 2 \: \times \: x^2 \: + \: 2 \: \times \: x^1 \: + \: 3 \: \times \: x^0 \\ \scriptsize = 2x^2 \: + \: 2x \: + \: 3 \) \( \scriptsize 72_x = 7 \: \times \: x^1 \: + \: 2 \: \times \: x^0 \\ \scriptsize = 7x \: + \: 2 \)i.e \(\scriptsize 3x^2 \: + \: x \: + \: 5 \: -\: (2x^2 \: + \: 2x \: + \: 3) = 7x \: + \: 2 \\ \scriptsize 3x^2 \: + \: x \: + \: 5 \: – \: 2x^2 \: – \: 2x \: – \: 3 = 7x \: + \: 2 \\ \scriptsize x^2 \: -\: x \: + \: 2 = 7x \: + \: 2\\ \scriptsize x^2 \: -x \: – \: 7x = 2 \: – \: 2 \\ \scriptsize x^2 \: – \: 8x = 0 \)
x = 0 or x = 8
But x cannot be base 0.
\( \scriptsize \therefore x = 8 \)(ii) \( \scriptsize (251.25)_{10} = x_7\)
Solution:
Convert \( \scriptsize (251.25)_{10} \) to base 7
Whole number part:
⇒ \( \scriptsize 251_{10} = 506_7 \)
Decimal part:
⇒ \( \scriptsize 0.25_{10} = 151_7 \)
\( \scriptsize \therefore 251.25_{10} = \underset{whole\: number \: part}{506}.\underset{decimal\: part}{151_7} \)⇒ \( \scriptsize 506.151_7 = x \)
⇒ \( \scriptsize x = 506.151_7 \)
Example 2.3.3
Find the number bases p and q in the following simultaneous equations
(i)
\(\scriptsize 26p \: -\: 34q = 1000_2 \)
\(\scriptsize 38p \: -\: 21q = 111_5 \)
(ii)
\( \scriptsize A2p \: -\: 80q = 90_{10} \)
\( \scriptsize B4p \: +\: 77q = 250_{10} \)
(i)
\(\scriptsize 26p \: -\: 34q = 1000_2 \)
\(\scriptsize 38p \: -\: 21q = 111_5 \)
Solution
Convert to base 10
\(\scriptsize 26p \: -\: 34q = 1000_2 \)⇒ \(\scriptsize 2 \: \times \: p^1 \: + \: 6 \: \times \: p^0 \: – \: (3 \: \times \: q^1 \: + \: 4 \: \times \: q^o) \\ \scriptsize = 1 \: \times \: 2^3 \: + \: 0 \: \times \: 2^2 \: + \: 0 \: \times \: 2^1 \: + \: 0 \: \times \: 2^0 \)
\( \scriptsize \therefore 2p \: + \: 6 \: – \: 3q \: -\: 4 = 8 \: + \: 0 \\ \scriptsize 2p\: – \: 3q = 8 \: – \: 2 \\ \scriptsize 2p \: -\: 3q = 6 \: …..(1) \)Also, Convert to base 10,
\(\scriptsize 38p \: -\: 21q = 111_5 \)⇒ \(\scriptsize 3 \: \times \: p^1 \: + \: 8 \: \times \: p^0 \: – \: (2 \: \times \: q^1 \: + \: 1 \: \times \: q^o) \\ \scriptsize = 1 \: \times \: 5^2 \: + \: 1 \: \times \: 5^1 \: + \: 1 \: \times \: 5^0 \)
\( \scriptsize \therefore 3p \: + \: 8 \: – \: 2q \: -\: 1 = 25 \: + \: 5 \: + \: 1 \\ \scriptsize 3p \: – \: 2q = 31 \: – \: 8 \: + \: 1 \\ \scriptsize 3p\: – \: 2q = 31 \: – \: 7 \\ \scriptsize 3p \: -\: 2q = 24 \: …..(2) \)Solving simultaneously and eliminating p by multiplying equation (1) by 3 and equation (2) by 2, we have:
\( \scriptsize 2p \: -\:3q = 6 \) ………….(1) × 3
\( \scriptsize 3p \: – \: 2q = 24 \) ……………(2) × 2
\( \scriptsize 6p \: -\:9q = 18 \) …………(3)
\( \scriptsize 6p \: – \: 4q = 48 \) ………….(4)
Subtract equation (3) from (4)
\( \scriptsize -\: 4q \: – \: (- \: 9q) = 48 \: -\: 18 \) \( \scriptsize -\: 4q \: + \: 9q = 48 \: -\: 18 \) \( \scriptsize 9q \: – \: 4q = 48 \: -\: 18 \) \( \scriptsize 5q = 30 \)divide both sides by 5
\( \frac{5q}{5} = \frac{30}{5} \) \( \frac{\not{5}q}{\not{5}} = \frac{30}{5} \) \( \scriptsize q = 6 \)Substitute for q in equation (1) to obtain p
\( \scriptsize 2p \: -\:3q = 6 \) ………….(1)
\( \scriptsize 2p \: -\:3 \: \times \: 6 = 6 \) \( \scriptsize 2p \: -\:18 = 6 \) \( \scriptsize 2p = 6 \: + \: 18 \) \( \scriptsize 2p = 24 \)divide both sides by 2
\( \frac{2p}{2} = \frac{24}{2} \) \( \frac{\not{2}p}{\not{2}} = \frac{24}{2} \) \( \scriptsize p = 12 \)(ii)
\( \scriptsize A2p \: -\: 80q = 90_{10} \)
\( \scriptsize B4p \: +\: 77q = 250_{10} \)
Solution
Convert to base 10
\( \scriptsize A2p \: -\: 80q = 90_{10} \)A = 10
= \( \scriptsize 10 \: \times \: p^1 \: + \: 2 \: \times \: p^0 \: -\: 8 \: \times \: q^1 \: + \: 0 \: \times \: q^0 = 90 \)
\( \scriptsize \therefore 10p \: + \: 2 \: – \: 8q = 90 \) \( \scriptsize = 10p \: – \: 8q = 90 \: -\: 2\) \( \scriptsize = 10p \: – \: 8q = 88\)simplify by dividing through by 2
\( \scriptsize = 5p \: – \: 4q = 44\: ……………(1)\)Also, convert to base 10,
\( \scriptsize B4p \: +\: 77q = 250_{10} \)B = 11
= \( \scriptsize 11 \: \times \: p^1 \: + \: 4 \: \times \: p^0 \: +\: 7 \: \times \: q^1 \: + \: 7 \: \times \: q^0 = 250 \)
\( \scriptsize \therefore 11p \: + \: 4 \: + \: 7q \: + \: 7 = 250 \) \( \scriptsize = 11p \: + \: 7q \: + \: 11 = 250\) \( \scriptsize = 11p \: + \: 7q = 250 \: – \: 11\) \( \scriptsize = 11p \: + \: 7q = 239\:…………..(2)\)Eliminating q, by multiplying equation(1) by 7 and equation (2) by 4, we have:
\( \scriptsize = 5p \: – \: 4q = 44\: ……………(1) \: \: \times \: 7\)
\( \scriptsize = 11p \: + \: 7q = 239\:…………..(2)\: \: \times \: 4\)
\( \scriptsize = 35p \: – \: 28q = 308\: ……………(3) \)
\( \scriptsize = 44p \: + \: 28q = 956\:…………..(4)\)
Add equation (4) to (3)
\( \scriptsize (44p \: + \: 25p) + (28 \: + (- 28) = (956 \: + \: 308) \) \( \scriptsize (79p) + (0) = (1264) \) \( \scriptsize 79p = 1264 \)divide both sides by 79
\( \frac{79p}{79} = \frac{1264}{79} \) \( \scriptsize p = 16 \)Substitute for the value of p in equation (i) to obtain q
\( \scriptsize = 5p \: – \: 4q = 44\: ……………(1) \)
\( \scriptsize = 5 \: \times \: 16 \: – \: 4q = 44 \)
\( \scriptsize 80 \: – \: 4q = 44 \) \( \scriptsize \: – \: 4q = 44 \:-\:80 \) \( \scriptsize \: – \: 4q =\: -36 \) \( \scriptsize q = \normalsize \frac{-36}{-4} \) \( \scriptsize q = 9 \)