Convert the following binary numbers to decimal equivalents: a. 001100 b. 000011 c. 011100 d. 111100 e. 111111 2. Convert the following binary numbers to their decimal equivalents a. 11100.001 b. 110011.10011 c. 101010101010.1 Show all the workings and not just the answers.
The Correct Answer and Explanation is:
1. Converting Binary Numbers to Decimal Equivalents
Each binary digit (bit) represents a power of 2. Starting from the right, the first digit is 202^020, the second is 212^121, and so on. For numbers with fractional parts, the first digit to the right of the binary point is 2−12^{-1}2−1, then 2−22^{-2}2−2, and so forth.
a. 001100
(0×25)+(0×24)+(1×23)+(1×22)+(0×21)+(0×20)=0+0+8+4+0+0=12(0 \times 2^5) + (0 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (0 \times 2^0) \\ = 0 + 0 + 8 + 4 + 0 + 0 = 12(0×25)+(0×24)+(1×23)+(1×22)+(0×21)+(0×20)=0+0+8+4+0+0=12
b. 000011
(0×25)+(0×24)+(0×23)+(0×22)+(1×21)+(1×20)=0+0+0+0+2+1=3(0 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) \\ = 0 + 0 + 0 + 0 + 2 + 1 = 3(0×25)+(0×24)+(0×23)+(0×22)+(1×21)+(1×20)=0+0+0+0+2+1=3
c. 011100
(0×25)+(1×24)+(1×23)+(1×22)+(0×21)+(0×20)=0+16+8+4+0+0=28(0 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (0 \times 2^0) \\ = 0 + 16 + 8 + 4 + 0 + 0 = 28(0×25)+(1×24)+(1×23)+(1×22)+(0×21)+(0×20)=0+16+8+4+0+0=28
d. 111100
(1×25)+(1×24)+(1×23)+(1×22)+(0×21)+(0×20)=32+16+8+4+0+0=60(1 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (0 \times 2^0) \\ = 32 + 16 + 8 + 4 + 0 + 0 = 60(1×25)+(1×24)+(1×23)+(1×22)+(0×21)+(0×20)=32+16+8+4+0+0=60
e. 111111
(1×25)+(1×24)+(1×23)+(1×22)+(1×21)+(1×20)=32+16+8+4+2+1=63(1 \times 2^5) + (1 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) \\ = 32 + 16 + 8 + 4 + 2 + 1 = 63(1×25)+(1×24)+(1×23)+(1×22)+(1×21)+(1×20)=32+16+8+4+2+1=63
2. Binary Numbers with Fractional Parts
a. 11100.001
Integer part:(1×24)+(1×23)+(1×22)+(0×21)+(0×20)=16+8+4+0+0=28(1 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (0 \times 2^0) = 16 + 8 + 4 + 0 + 0 = 28(1×24)+(1×23)+(1×22)+(0×21)+(0×20)=16+8+4+0+0=28
Fractional part:(0×2−1)+(0×2−2)+(1×2−3)=0+0+0.125(0 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3}) = 0 + 0 + 0.125(0×2−1)+(0×2−2)+(1×2−3)=0+0+0.125
Total:28+0.125=28.12528 + 0.125 = 28.12528+0.125=28.125
b. 110011.10011
Integer part:(1×25)+(1×24)+(0×23)+(0×22)+(1×21)+(1×20)=32+16+0+0+2+1=51(1 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 32 + 16 + 0 + 0 + 2 + 1 = 51(1×25)+(1×24)+(0×23)+(0×22)+(1×21)+(1×20)=32+16+0+0+2+1=51
Fractional part:(1×2−1)+(0×2−2)+(0×2−3)+(1×2−4)+(1×2−5)=0.5+0+0+0.0625+0.03125=0.59375(1 \times 2^{-1}) + (0 \times 2^{-2}) + (0 \times 2^{-3}) + (1 \times 2^{-4}) + (1 \times 2^{-5}) = 0.5 + 0 + 0 + 0.0625 + 0.03125 = 0.59375(1×2−1)+(0×2−2)+(0×2−3)+(1×2−4)+(1×2−5)=0.5+0+0+0.0625+0.03125=0.59375
Total:51+0.59375=51.5937551 + 0.59375 = 51.5937551+0.59375=51.59375
c. 101010101010.1
Integer part:
From left to right:(1×211)+(0×210)+(1×29)+(0×28)+(1×27)+(0×26)+(1×25)+(0×24)+(1×23)+(0×22)+(1×21)+(0×20)(1 \times 2^{11}) + (0 \times 2^{10}) + (1 \times 2^9) + (0 \times 2^8) + (1 \times 2^7) + (0 \times 2^6) + (1 \times 2^5) + (0 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)(1×211)+(0×210)+(1×29)+(0×28)+(1×27)+(0×26)+(1×25)+(0×24)+(1×23)+(0×22)+(1×21)+(0×20)=2048+0+512+0+128+0+32+0+8+0+2+0=2730= 2048 + 0 + 512 + 0 + 128 + 0 + 32 + 0 + 8 + 0 + 2 + 0 = 2730=2048+0+512+0+128+0+32+0+8+0+2+0=2730
Fractional part:(1×2−1)=0.5(1 \times 2^{-1}) = 0.5(1×2−1)=0.5
Total:2730+0.5=2730.52730 + 0.5 = 2730.52730+0.5=2730.5
Explanation (300 words)
Binary to decimal conversion involves evaluating the positional values of digits based on powers of two. The rightmost digit always represents 20=12^0 = 120=1, the next one is 21=22^1 = 221=2, then 22=42^2 = 422=4, and this continues as powers of two increase leftward. If a binary number contains a fractional part, its digits represent negative powers of two starting from 2−12^{-1}2−1, then 2−22^{-2}2−2, and so on.
For example, in the binary number 001100, only the third and fourth bits from the right are ones. These correspond to 23=82^3 = 823=8 and 22=42^2 = 422=4, resulting in a decimal value of 12.
When the binary number includes a fractional part, such as 11100.001, the process remains the same, but the binary point separates the integer and fractional positions. Here, the digits after the point are interpreted using negative exponents of two. In 11100.001, the fractional part 001 equates to 2−3=0.1252^{-3} = 0.1252−3=0.125, so the complete decimal equivalent becomes 28.125.
This positional system allows binary, the language of computers, to map precisely to the decimal system used in daily life. Each binary digit’s value depends entirely on its position and whether it is a 1 or 0. This system is foundational to digital electronics, data storage, and programming, where numeric precision often determines performance and accuracy.
