number systems 既問題(2條)

1.convert 110110(2) into a denary number.

110110(2)
=1x2^5+1x2^4+0x2^3+1x2^2+1x2^1+0x2^0
=32+16+0+4+2+0
=54(10)

2.convert 37(10) into a binary number.

*要用短除法來轉為二進制
(...後為餘數)
37/2=18...1←6
18/2= 9...0←5
09/2= 4...1←4
04/2= 2...0←3
02/2= 1...0
↑ ↑
1 2 (箭咀為答案次序)
Ans由最底的"1"向上抄
=100101(2)

2008-07-30 23:05:38 補充：
格式問題:
2.convert 37(10) into a binary number.

*要用短除法來轉為二進制
(...後為餘數)
37/2=18...1←6
18/2= 9...0←5
09/2= 4...1←4
04/2= 2...0←3
02/2= 1...0←2
------↑
------ 1 (箭咀為答案次序)
Ans由最底的"1"向上抄
=100101(2)

1.
110110(2)
= 1x2^5 + 1x2^4 + 0x2^3 + 1x2^2+ 1x2^1 + 0x2^0
= 32 + 16 + 0 + 4 + 2 + 0
= 54(10)
2.
37(10)
= 32 + 0 + 0 + 4+ 0 + 1
= 1x2^5 + 0x2^4 + 0x2^3 + 1x2^2 + 0x2^1 + 1x2^0
= 100101(2)

2008-07-31 09:12:26 補充：
By definition,
Denary number abcd(10)
= ax10^3 + bx10^2 + cx10^1 + dx10^0
= 1000a + 100b + 10c + d
Similarily,
Binary number abcd(2)
= ax2^3+ bx2^2 + cx2^1 + dx2^0
= 8a + 4b + 2c + d

To change 37(10) to binary number,
37(10)
= 1x32 + 1x4 + 1x1
= 1x2^5 + 0x2^4 + 0x2^3 + 1x2^2 + 0x2^1 + 1x2^0
=100101(2)

1. 110110(2) = 0 x 1 + 1 x 2 + 1 x 2^2 + 0 x 2^3 + 1 x 2^4 + 1 x 2^5 = 0 + 2 + 4 + 0 + 16 + 32 = 54(10).
2. 37/2 = 18........remainder 1
18/2 = 9............. rem. 0
9/2 = 4.............. rem. 1
4/2 = 2.............. rem. 0
2/2= 1.............. rem. 0
1/2 = 0............ rem. 1
So 37(10) = 100101(2).