Two ideas power the whole lesson
A multiplier bit decides whether a shifted copy is included. A quotient bit decides whether the divisor fits in the current partial dividend.
Build multiplication from shifted 0-or-1 partial products, choose quotient bits by comparing and subtracting, and keep the ordinary order of operations when several binary operations appear together.
A multiplier bit decides whether a shifted copy is included. A quotient bit decides whether the divisor fits in the current partial dividend.
Moving a binary numeral left by one place multiplies its value by:
Moving it left by k places multiplies by 2k. A multiplier bit of 0 turns the whole shifted row into zero.
Write the three products in binary. The printed questions stay fixed when you change the explorer.
Each left shift multiplies the value by 2.
That 1 is in the fours place, so shift 1101 two places left.
The 1-bits of 101 select the unshifted and twice-shifted copies.
Write a partial row for each multiplier place, then add.
Compare their values: five is smaller than six, so the divisor does not fit.
Bring down a bit, compare, and subtract only when the divisor fits.
Multiply the quotient by the divisor to check an exact answer.
The remainder is smaller than the divisor. If it were not, one more copy of the divisor would still fit, so the quotient would be too small.
Write both quotient and remainder in binary, in separate fields.
The remainder is nonnegative and smaller than the divisor.
Write all three answers in binary. The third division is exact and asks for its quotient only.
Check dividend = divisor × quotient + remainder.
Write the first three results in binary. These printed expressions are independent of the selected explorer example.
Inside the parentheses, division still comes before addition.
Write every answer in this workshop in binary. Correct all eight to complete it.
Answers: 10110₂; 1000001₂; 10111110₂; 101101₂; 110₂; 11₂; 1000100110₂; 111₂. Decimal checks: 11×2=22, 13×5=65, 38×5=190, 225÷5=45, 45=7×6+3, 50×11=550, and (7+105÷5)÷4=7.
Write every answer below in binary. A division with no remainder field asks for its exact quotient.
Answers: 1110101₂; 1001₂; 0; 101101₂; 1011₂. Verify 13×9=117, 117÷13=9, 225÷5=45, and 550÷50=11. Every exact division has remainder zero.
Chapter 26 skill certificate
This certifies that Binary Learner can build shifted partial products, find quotient and remainder, verify D=dq+r, and evaluate mixed binary expressions in the correct order.
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