Hi folks, Mark Tapley wrote: Actually this isn't *quite* true. An n x n bit unsigned multiplier producing a 2n bit result also produces the correct signed results for the bottom n bits. Thus the 6809's MUL instruction produces the correct answer to A = -20 * B = -3 as it will store -60 in the B register. The reason is because with the n x n bit multiplier multiplying a * b, where b is -ve => a* ((2^n) - (-b) ) = a*(2^n) + a*b = a*b since a*(2^n) doesn't contribute to to the lower n bits. Consider a 2-bit multiplier dealing with the -ve * -ve and -ve * +ve cases: 01 * 10 = 0010, = -2 correct. 01 * 11 = 0011, = -1 correct. 10 * 10 = 0100, = 0 correct. 10 * 11 = 0110, = 2 (or -2) correct. 11 * 11 = 1001, = 1 correct. -cheers from julz