Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 4 - Number Representation and Calculation - 4.3 Computation in Positional Systems - Exercise Set 4.3 - Page 234: 34

Answer

The multiplication of two given numbers in base four is \[{{2122}_{\text{four}}}\]..

Work Step by Step

Since, the computation involves base four, the only digits which are allowed are 0, 1, 2 and 3. The procedure to multiply two numbers in base four is same as in base ten. \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] Hence, first multiply \[{{3}_{\text{four}}}\] in first column with \[{{2}_{\text{four}}}\], which is above it in first column: \[{{3}_{\text{four}}}\times {{2}_{\text{four}}}={{6}_{\text{ten}}}=\left( 1\times 4 \right)+\left( 2\times 1 \right)={{12}_{\text{four}}}\] Now, write 2 in the first column below the horizontal line and carry 1 in the second column: \[\overset{1}{\mathop{3}}\,{{2}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{2}_{\text{four}}}\] Now, multiply \[{{3}_{\text{four}}}\] in first column, with \[{{3}_{\text{four}}}\], which is in the second column and add \[{{1}_{\text{four}}}\] to the product: \[\left( {{3}_{\text{four}}}\times {{3}_{\text{four}}} \right)+{{1}_{\text{four}}}=9+1={{10}_{\text{ten}}}=\left( 2\times 4 \right)+\left( 2\times 1 \right)={{22}_{\text{four}}}\] Write \[{{22}_{\text{four}}}\] in front of \[{{2}_{\text{four}}}\] below the horizontal line: \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] Now, repeat the whole procedure, but with \[{{2}_{\text{four}}}\]. First, place the symbol \[\times \] below \[{{2}_{\text{four}}}\] in \[{{222}_{\text{four}}}\]. \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\times \] Now, multiply \[{{2}_{\text{four}}}\] in second column, with \[{{2}_{\text{four}}}\] in the first column: \[{{2}_{\text{four}}}\times {{2}_{\text{four}}}={{4}_{\text{ten}}}=\left( 1\times 4 \right)+\left( 0\times 1 \right)={{10}_{\text{four}}}\] Write \[{{0}_{\text{four}}}\] below\[{{2}_{\text{four}}}\]in \[{{222}_{\text{four}}}\] and carry 1 in the second column: \[\overset{1}{\mathop{3}}\,{{2}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[0\times \] Now, multiply \[{{2}_{\text{four}}}\] in the second column, with \[{{3}_{\text{four}}}\] in the second column and add \[{{1}_{\text{four}}}\] to the product: \[\left( {{2}_{\text{four}}}\times {{3}_{\text{four}}} \right)+{{1}_{\text{four}}}=6+1={{7}_{\text{ten}}}=\left( 1\times 4 \right)+\left( 3\times 1 \right)={{13}_{\text{four}}}\]. Write\[{{13}_{\text{four}}}\] in front of \[{{0}_{\text{four}}}\]: \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\underline{130\times }\] Now, add \[{{222}_{\text{four}}}\] and \[{{130}_{\text{four}}}\] in the manner shown above: \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\underline{+130\times }\] First, write \[{{2}_{\text{four}}}\] below the symbol \[\times \] as it is: \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\underline{+130\times }\] \[{{2}_{\text{four}}}\] Now, add \[{{2}_{\text{four}}}\] with \[{{0}_{\text{four}}}\]: \[{{2}_{\text{four}}}+{{0}_{\text{four}}}={{2}_{\text{ten}}}=\left( 2\times 1 \right)={{2}_{\text{four}}}\] \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\underline{+130\times }\] \[{{22}_{\text{four}}}\] Now, add \[{{2}_{\text{four}}}\] with \[{{3}_{\text{four}}}\]: \[{{2}_{\text{four}}}+{{3}_{\text{four}}}={{5}_{\text{ten}}}=\left( 1\times 4 \right)+\left( 1\times 1 \right)={{11}_{\text{four}}}\] \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\underline{+\overset{1}{\mathop{1}}\,30\times }\] \[{{122}_{\text{four}}}\] Now, add \[{{1}_{\text{four}}}\] with \[{{1}_{\text{four}}}\]: \[{{1}_{\text{four}}}+{{1}_{\text{four}}}={{2}_{\text{ten}}}=\left( 2\times 1 \right)={{2}_{\text{four}}}\] \[{{32}_{\text{four}}}\] \[\underline{\times {{23}_{\text{four}}}}\] \[{{222}_{\text{four}}}\] \[\underline{+130\times }\] \[{{2122}_{\text{four}}}\] Now, to check whether the above obtained solution is correct, perform the multiplication by converting each number to base ten: \[{{32}_{\text{four}}}=14\], \[{{23}_{\text{four}}}=11\] and \[{{2122}_{\text{four}}}=154\] Since, \[14\times 11\] indeed equals 154, the solution obtained is correct. Hence, the multiplication of two given numbers in base four is \[{{2122}_{\text{four}}}\]..
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