Back to Course

0% Complete
0/0 Steps

• ## Do you like this content?

Lesson 2, Topic 3
In Progress

# Conversion from one Base to Another

Lesson Progress
0% Complete

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 1

Evaluate the following and express the answer in the bases indicated.

(i) 10401.117 to base eight

(ii) 2013 – 101012 + 2536 to base 6

Solution:

(i) Convert 10401.117 to base ten

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: $$\scriptsize 10401.11_7 = 5046.123_{8}$$

Solution:

(ii) 2013 – 101012 + 2536 to base 6

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

Find the value of x in the following:

(i) $$\scriptsize 315_x \: -\: 223_x = 72_x$$

(ii) $$\scriptsize (251.25)_{10} = x_7$$

Solution (i):

$$\scriptsize 315_x = 3 \: \times \: x^2 \: + \: 1 \: \times \: x^1 \: + \: 5 \: \times \: x^0 \\ \scriptsize = 3x^2 \: + \: x \: + \: 5$$

$$\scriptsize 223_x = 2 \: \times \: x^2 \: + \: 2 \: \times \: x^1 \: + \: 3 \: \times \: x^0 \\ \scriptsize = 2x^2 \: + \: 2x \: + \: 3$$

$$\scriptsize 72_x = 7 \: \times \: x^1 \: + \: 2 \: \times \: x^0 \\ \scriptsize = 7x \: + \: 2$$

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.

$$\scriptsize \therefore x = 8$$

Solution (ii):

Convert $$\scriptsize (251.25)_{10}$$ to base 7

i.e. whole number part

i.e $$\scriptsize 251_{10} = 506_7$$

Decimal part

i.e $$\scriptsize 0.25_{10} = 151_7$$

$$\scriptsize \therefore 251.25_{10} = \underset{whole\: number \: part}{506}.\underset{decimal\: part}{151_7}$$

$$\scriptsize 506.151_7 = x$$

$$\scriptsize x = 506.151_7$$

Example 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)

$$\scriptsize A2p \: -\: 80q = 90_{10}$$

$$\scriptsize B4p \: +\: 77q = 250_{10}$$

Solution (i)

$$\scriptsize 26p \: -\: 34q = 1000_2$$

ï»¿

$$\scriptsize 38p \: -\: 21q = 111_5$$

Convert to base 10

$$\scriptsize 26p \: -\: 34q = 1000_2$$

= $$\scriptsize 2 \: \times \: p^1 \: + \: 6 \: \times \: p^0 \: – \: (3 \: \times \: q^1 \: + \: 4 \: \times \: q^o) \\ \scriptsize = 1 \: \times \: 2^3 \: + \: 0 \: \times \: 2^2 \: + \: 0 \: \times \: 2^1 \: + \: 0 \: \times \: 2^0 \\ \scriptsize \therefore 2p \: + \: 6 \: – \: 3q \: -\: 4 = 8 \: + \: 0 \\ \scriptsize = 2p\: – \: 3q = 8 \: – \: 2 \\ \scriptsize 2p \: -\: 3q = 6 \: …..(1)$$

AlsoConvert to base 10, $$\scriptsize 38p \: -\: 21q = 111_5$$

= $$\scriptsize 3 \: \times \: p^1 \: + \: 8 \: \times \: p^0 \: – \: (2 \: \times \: q^1 \: + \: 1 \: \times \: q^o) \\ \scriptsize = 1 \: \times \: 5^2 \: + \: 1 \: \times \: 5^1 \: + \: 1 \: \times \: 5^0 \\ \scriptsize \therefore 3p \: + \: 8 \: – \: 2q \: -\: 1 = 25 \: + \: 5 \: + \: 1 \\ \scriptsize = 3p\: – \: 2q = 31 \: – \: 7 \\ \scriptsize 3p \: -\: 2q = 24 \: …..(2)$$

Solving simultaneously and eliminate p by multiplying equation (1) by 3 and equation (2) by 2, we have:

$$\scriptsize 2p \: -\:3q = 6$$ ………….(1) x 3

$$\scriptsize 3p \: – \: 2q = 24$$ ……………(2) x 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

$$\frac{2p}{2} = \frac{24}{2}$$

$$\frac{\not{2}p}{\not{2}} = \frac{24}{2}$$

$$\scriptsize p = 12$$

Solution (ii)

$$\scriptsize A2p \: -\: 80q = 90_{10}$$

$$\scriptsize B4p \: +\: 77q = 250_{10}$$

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:

$$\scriptsize = 5p \: – \: 4q = 44\: ……………(1) \: \: \times \: 7$$

$$\scriptsize = 11p \: + \: 7q = 239\:…………..(2)\: \: \times \: 4$$

$$\scriptsize = 35p \: – \: 28q = 308\: ……………(3)$$

$$\scriptsize = 44p \: + \: 28q = 956\:…………..(4)$$

$$\scriptsize (44p \: + \: 25p) + (28 \: + (- 28) = (956 \: + \: 308)$$

$$\scriptsize (79p) + (0) = (1264)$$

$$\scriptsize 79p = 1264$$

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

$$\scriptsize = 5p \: – \: 4q = 44\: ……………(1)$$

$$\scriptsize = 5 \: \times \: 16 \: – \: 4q = 44$$

$$\scriptsize 80 \: – \: 4q = 44$$

$$\scriptsize \: – \: 4q = 44 \:-\:80$$

$$\scriptsize \: – \: 4q =\: -36$$

$$\scriptsize q = \normalsize \frac{-36}{-4}$$

$$\scriptsize q = 9$$

error: