082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

1.3 Number bases 13
giving
83=64+16+3
We now focus on the 3 and again usingTable 1.1 we may write
83=64+16+2+1
=2 6 +2 4 +2 1 +2 0
= 1(2 6 ) +0(2 5 ) +1(2 4 ) +0(2 3 ) +0(2 2 ) +1(2 1 ) +1(2 0 )
= 1010011 2
Example1.14 Express 200 10
as a binary number.
Solution From Table 1.1 we note that 128 is the highest number that does not exceed 200 so we
write
200 = 128 +72
UsingTable 1.1 repeatedly we may write
200 = 128 +72
=128+64+8
=2 7 +2 6 +2 3
= 1(2 7 ) +1(2 6 ) +0(2 5 ) +0(2 4 ) +1(2 3 ) +0(2 2 ) +0(2 1 ) +0(2 0 )
= 11001000 2
Anotherwaytoconvertdecimalnumberstobinarynumbersistodivideby2repeatedly
and note the remainder. We rework the previous two examples using thismethod.
Example1.15 Convert the following decimal numbers tobinary: (a)83 (b) 200
Solution (a) We divide by 2 repeatedly and note the remainder.
Remainder
83÷2=41r1 1
41÷2=20r1 1
20÷2=10r0 0
10÷2= 5r0 0
5÷2= 2r1 1
2÷2= 1r0 0
1÷2= 0r1 1
To obtain the binary number we write out the remainder, working from the bottom
one tothe topone. This gives
as before.
83 10
= 1010011 2