a.
101010
= (0x20) + (1x21) + (0x22)
+ (1x23) + (0x24) + (1x25)
= (0) + (2) + (0) + (8) + (0) + (32)
= 42
1010102 = 4210
b.
100001
= (1x20) + (0x21) + (0x22)
+ (0x23) + (0x24) + (1x25)
= (1) + (0) + (0) + (0) + (0) + (32)
= 33
1000012 = 3310
c.
10111
= (1x20) + (1x21) + (1x22)
+ (0x23) + (1x24)
= (1) + (2) + (4) + (0) + (16)
= 23
101112 = 2310
d.
0110
= (0x20) + (1x21) + (1x22)
+ (0x23)
= (0) + (2) + (4) + (0)
= 6
01102 = 610
e.
11111
= (1x20) + (1x21) + (1x22)
+ (1x23) + (1x24)
= (1) + (2) + (4) + (8) + (16)
= 31
111112 = 3110
II.
Convert
each of the following base ten representations to its equivalent binary form:
a.
32
32/2=16 (0)
16/2=8 (0)
8/2=4 (0)
4/2=2 (0)
2/2=1 (0)
1/2=0 (1)
3210 = 1000002
b.
64
64/2=32 (0)
32/2=16 (0)
16/2=8 (0)
8/2=4 (0)
4/2=2 (0)
2/2=1 (0)
1/2=0 (1)
6410 = 10000002
c.
96
96/2=48 (0)
48/2=24 (0)
24/2=12 (0)
12/2=6 (0)
6/2=3 (0)
3/2=1 (1)
1/2=0 (1)
9610 = 11000002
d.
15
15/2=7 (1)
7/2=3 (1)
3/2=1 (1)
1/2=1 (1)
1510 = 11112
e.
27
27/2=13 (1)
13/2=6 (1)
6/2=3 (0)
3/2=1 (1)
1/2=0 (1)
9610 = 110112
III.
Perform
the following additions in binary notation:
a.
11011
1100 +
100111
b.
1010.001
1.101 +
1011.110
c.
11111
0001 +
100000
d.
111.11
00.01 +
1000.00
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