Which of these lists correctly orders the binary
numbers from smallest to largest?
Choose 1 answer:
0011, 0110, 0101, 1010
1010, 0101, 0110, 0011
0011, 0101, 0110, 1010
1010, 0110, 0101,0011

The smallest computing data unit is the binary digit or bit. It is represented by 0 or 1, respectively. Binary numbers consist of binary digits (bits). To calculate the smallest to largest binary digit first convert the binary to decimal.

Binary conversion

Let

[tex]\to 2^0=1[/tex]

Converting Binary to decimal:

For point a:

[tex](0011)_2 = (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)=0+0+2+1 = (3)_{10}\\\\(0110)_2 = (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=0+4+2+0 = (6)_{10}\\\\(0101)_2 = (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)=0+4+0+1 = (5)_{10}\\\\(1010)_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=8+0+2+0 = (10)_{10}\\\\[/tex]

For point b:

[tex](1010)_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=8+0+2+0 = (10)_{10}\\\\(0101)_2 = (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)=0+4+0+1 = (5)_{10}\\\\(0110)_2 = (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=0+4+2+0 = (6)_{10}\\\\(0011)_2 = (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)=0+0+2+1 = (3)_{10}\\\\[/tex]

For point c:

[tex](0011)_2 = (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)=0+0+2+1 = (3)_{10}\\\\(0101)_2 = (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)=0+4+0+1 = (5)_{10}\\\\(0110)_2 = (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=0+4+2+0 = (6)_{10}\\\\(1010)_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=8+0+2+0 = (10)_{10}\\\\[/tex]

For point d:

[tex](1010)_2 = (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=8+0+2+0 = (10)_{10}\\\\(0110)_2 = (0 \times 2^3) + (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0)=0+4+2+0 = (6)_{10}\\\\(0101)_2 = (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)=0+4+0+1 = (5)_{10}\\\\(0011)_2 = (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0)=0+0+2+1 = (3)_{10}\\\\[/tex]

Therefore, the final answer is "Option c"

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