The dividend is divided by the divisor, and the answer is the quotient. Compare the divisor to the first digit in the dividend. If the divisor is the larger number, keep adding digits to the dividend until the divisor is the smaller number. (For example, if calculating 172 ÷ 4, we would compare 4 and 1, note that 4 > 1, and compare 4 to 17 instead. ) Write the first digit of the quotient above the last dividend digit you were using in the comparison. Comparing 4 and 17, we see that 4 goes into 17 four times, so we write 4 as the first digit of our quotient, above the 7. Multiply and subtract to find the remainder. Multiply the quotient digit with the divisor, in this case 4 x 4 = 16. Write the 16 underneath the 17, then subtract 17 - 16 to find the remainder, 1. Repeat. Once again, we compare the divisor 4 with the next digit, 1, note that 4 > 1, and “bring down” the next digit of the dividend, to compare 4 with 12 instead. 4 goes into 12 three times with no remainder, so we write 3 as the next digit of the quotient. The answer is 43.
11 > 1, so 11 can’t “go into” 1. Write a 0 as the first digit of the quotient (above the first digit of the dividend).
Bring down the next digit of the dividend. 11 > 10. Write a 0 in the quotient. Bring down the next digit. 11 < 101. Write a 1 in the quotient.
Write the divisor underneath the dividend. Here, we write 11 aligned underneath the first three digits (101) of the dividend. Calculate 101 - 11 to get the remainder, 10. See how to subtract binary numbers if you need a review.
Write 11 underneath the 100 and subtract to get 1. Bring down the final digit of the dividend to make 11. 11 = 11, so write a 1 as the final digit of the quotient (the answer). There is no remainder, so the problem is complete. The answer is 00111, or simply 111.
Binary division problems often end up with repeating fractional portions, more often than they occur in decimal notation. [9] X Research source This is referred to with the more general term “radix point,” which applies in any base, since the “decimal point” is only used in the decimal system. [10] X Research source
26 - 7 = 19 (subtracted 1 time) 19 - 7 = 12 (2) 12 - 7 = 5 (3) 5 - 7 = -2. Negative number, so back up. The answer is 3 with a remainder of 5. Note that this method does not calculate any non-integer portion of the answer.
Find the ones’ complement of the second term, subtracting each digit from 1. This is easily done in binary by switching each 1 to 0 and each 0 to 1. [13] X Research source In our example, 011 becomes 100. Add one to the result: 100 + 1 = 101. This is called the twos complement, and lets us perform subtraction as an addition problem. [14] X Research source Essentially, the result is as though we added a negative number instead of subtracting a positive one, once we finish the process. Add the result to the first term. Write and solve the addition problem: 111 + 101 = 1100. Discard the carry digit. Discard the first digit of your answer to get the final result. 1100 → 100.
Twos’ complement of 000101 = 111010 + 1 = 111011 100011 + 111011 = 1011110 Discard carry bit → 011110
011110 + 111011 = 1011001 → 011001 (quotient 1+1=10) 011001 + 111011 = 1010100 → 010100 (quotient 10+1=11) 010100 + 111011 = 1001111 → 001111 (11+1=100) 001111 + 111011 = 1001010 → 001010 (100+1=101) 001010 + 111011 = 10000101 → 0000101 (101+1=110) 0000101 + 111011 = 1000000 → 000000 (110+1=111) 0 is smaller than 101, so we stop here. The quotient 111 is the answer to the division problem. The remainder is the final result of our subtraction problem, in this case 0 (no remainder).
