You can continue your place value chart. Each place value is determined by a base 2 exponent. [4] X Research source For example:20=ones place{\displaystyle 2^{0}={\text{ones place}}}21=twos place{\displaystyle 2^{1}={\text{twos place}}}22=fours place{\displaystyle 2^{2}={\text{fours place}}}23=eights place{\displaystyle 2^{3}={\text{eights place}}}24=sixteens place{\displaystyle 2^{4}={\text{sixteens place}}}25=thirty-twos place{\displaystyle 2^{5}={\text{thirty-twos place}}}
For example, you might write 1 in the eights place, 1 in the fours place, 0 in the twos place, and 1 in the ones place: 1101.
For example, in the binary number 1101, there is a 1 in the ones place, so the value is 1. So the binary number 1 is equal to the decimal number 1.
For example, in the binary number 1101, there is a 0 in the twos place, so the value is 0. So the binary number 01 is equal to the decimal number 1, because there are 0 twos and 1 one: 0 + 1 = 1.
For example, in the binary number 1101, there is a 1 in the fours place, so the value is 4. So the binary number 101 is equal to the decimal number 5, because there are 1 four, 0 twos, and 1 one: 4 + 0 + 1 = 5.
For example, in the binary number 1101, there is a 1 in the eights place, so the value is 8. So the binary number 1101 is equal to the decimal number 13, because there are 1 eight, 1 four, 0 twos, and 1 one: 8 + 4 + 0 + 1 = 13.
For example, if adding 0111 and 1110, for the ones column you would add 1 one plus 0 ones = 1, so place a 1 in the answer’s ones column.
For example, if adding 0111 and 1110, for the twos column you would add 1 two, plus 1 two = 2 twos = 4, so place a 0 in the answer’s twos column and carry a 1 into the fours column.
For example, if adding 0111 and 1110, for the fours column you would add 1 four, plus 1 four, plus 1 four = 3 fours= 12, so place a 1 in the answer’s fours column and carry a 1 into the eights column.
For example, if adding 0111 and 1110, for the eights column you would add up 2 eights, since you carried from the fours column. For a sum of 2, place a 0 in the eights column and carry a 1 into the sixteens column. Since there are no other numbers in the sixteens column, just bring that 1 down to the final answer. So 0111 + 1110 = 10101.
For example, if adding 1010 + 1111 + 1011 + 1110, you should circle 1 pair of 1s.
For example, since you circled one pair of 1s, you would carry a 1 into the twos column, and drop a 0 in the answer’s ones place.
For example, if adding 1010 + 1111 + 1011 + 1110, you should circle 2 pairs of 1s, and have one 1 left over.
For example, since you circled 2 pairs of 1s and had one 1 leftover, you would carry a 1 into the fours column twice, and drop a 1 in the answer’s twos place.
For example, if adding 1010 + 1111 + 1011 + 1110, you should circle 2 pairs of 1s, since you carried 1 over twice from the twos column.
For example, since you circled 2 pairs of 1s with none left over, you would carry a 1 into the eights column twice, and place a 0 in the answer’s fours column.
For example, if adding 1010 + 1111 + 1011 + 1110, for the eights place you should circle 3 pairs of 1s, since you carried 1 over twice from the fours column. So you would place a 0 in the eights place of your answer and carry three 1s into the sixteens column. In the sixteens column you have one pair of 1s with one 1 left over, so you would place a 1 in the sixteens place of your answer, and a 1 in the thirty-twos column of your answer. So 1010 + 1111 + 1011 + 1110 = 110010.
An unsigned integer has no sign. The values that can be represented by an n{\displaystyle n} bit string is between 0{\displaystyle 0} and 2n−1{\displaystyle 2^{n-1}}. To convert the result back to decimal, perform 2n−1∗in+2n−2∗in−1+. . . +21∗i2+20∗i1{\displaystyle 2^{n-1}*i_{n}+2^{n-2}*i_{n-1}+. . . +2^{1}*i_{2}+2^{0}*i_{1}}, where in{\displaystyle i_{n}} is the n{\displaystyle n}th bit. A two’s compliment integer has both negative and positive results. The values that can be represented by an n{\displaystyle n} bit string is between −2n−2{\displaystyle -2^{n-2}} and 2n−2{\displaystyle 2^{n-2}}. To convert the result back to decimal, first toggle all the bits then add 1 to the first bit if the last bit is on. Then use the same formula for computing the decimal result 2n−1∗in+2n−2∗in−1+. . . +21∗i2+20∗i1{\displaystyle 2^{n-1}*i_{n}+2^{n-2}*i_{n-1}+. . . +2^{1}*i_{2}+2^{0}*i_{1}}, where in{\displaystyle i_{n}} is the n{\displaystyle n}th bit. If you had to invert the bits, make sure to prepend a − sign before the answer.
