Binary number system common mistakes

Common mistakes

• Using the wrong maximum-value formula

  Calculating the maximum unsigned value for n bits as 2^n rather than 2^n - 1.

  Fix itUse 2^n - 1. There are 2^n possible bit patterns, but the values start at 0, so the largest value is one less than 2^n.

• Forgetting positional shifts in multiplication

  Writing every partial product underneath the same starting position.

  Fix itShift each partial product left according to the position of the corresponding 1 in the multiplier, then add the aligned rows.

• Forgetting the final addition of 1

  A student inverts the bits of a positive value and stops, treating the inverted pattern as its negative.

  Fix itAfter inverting every bit, always add 1, using the specified fixed number of bits. For example, +5 as 8 bits is 00000101, so -5 is 11111011, not 11111010.

• Treating fractional bits as whole-number bits

  Reading the bits after the binary point as 1, 2, 4 and 8 instead of 1/2, 1/4, 1/8 and 1/16.

  Fix itWrite the place values above the fractional bits before calculating their contribution.

• Assuming floating point is always exact

  Claiming that floating-point representation removes rounding errors.

  Fix itFloating point can still be inaccurate because a value may not be representable as a binary fraction in the available number of bits.

• Using the stored value as the denominator

  Calculating relative error by dividing the absolute error by the stored value.

  Fix itDivide the absolute error by the actual value: relative error = absolute error / actual value.

• Assuming floating point always has greater precision

  Stating that floating point is always more precise than fixed point.

  Fix itFloating point primarily offers a wider range. Its precision is variable, so it does not automatically provide greater precision for every value.

• Changing the exponent in the wrong direction

  A learner moves the binary point to the right and increases the exponent.

  Fix itMoving the point right increases the mantissa by a power of two, so decrease the exponent by the same number of places. Moving the point left requires the exponent to increase.

• Reversing the two conditions

  Describing overflow as a result that is too small, or underflow as a result that is too large.

  Fix itRemember that overflow is above the maximum representable value, while underflow is below the minimum representable value.