# Arithmetic on signed 12-bit octal number stored in sign magnitude form

What is 4365 − 3412 when these values represent signed 12-bit octal numbers stored in sign-magnitude format? The result should be written in octal. Show your work.

Octal to binary:

• 4365: 100 011 110 101
• 3412: 011 100 001 010

By recognising the role of the sign bit, we can represent positive and negative 64-bit numbers in terms of the bit value times a power of 2. The binary number x, where xi means the ith bit, represents the number:

(x11 * -2^11) + (x10 * 2^10) + (x9 * 2^9) + ... + (x1 * 2^1) + (x0 * 2^0)

I have used the formula given above to convert the value of octal 4365 in decimal:

(1 * -2^11) + (1 * 2^7) + 2^6 + 2^5 + 2^4 + 2^2 + 2^0 = -2048 + 245 = -1803

Similarly, the value of octal 3412 in decimal is 1802.

Having obtained the two values in decimal, I subtract (-1803 - 1802), obtaining the result -3605.

Binary representation of 3605 is 111 000 010 101.

Converting it back to octal gives 7025.

This answer is wrong. It should be octal 7777 or decimal -3777.

The concept is not clear to me. Where am I going wrong?

$$100 011 110 101$$ if is taken as SM in 12bit is $$-245$$ decimal and $$011 100 001 010$$ is $$1802$$ decimal, so we obtain $$-245-1802=-2047$$. $$2047$$ in binary is $$11111111111$$ with $$11$$ bits so for minus we add one more $$1$$ to left = $$111111111111$$. It is octal 7777.