CPU Data Representation and Arithmetic Operations

Data Representation on the Central Processing Unit (CPU)

• Data conversion on the CPU involves transforming and manipulating numbers across four primary systems: Decimal, Binary, Octal, and Hexadecimal.
• Arithmetic operations, specifically addition and subtraction, are performed across these different number bases.
• Conversions between these bases are fundamental for processing and understanding data at various levels of computer architecture.

Decimal Number System

• The decimal system is a base-ten system.
• It utilizes ten distinct digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$.
• Weights are assigned based on the position of the digit relative to the decimal point ($.$). The weighting for a decimal number up to three places before and two places after the decimal point is as follows:
  • $10^2 = 100$
  • $10^1 = 10$
  • $10^0 = 1$
  • $.$
  • $10^{-1} = 0.1$
  • $10^{-2} = 0.01$

Binary Number System

• The binary system is a base-two system.
• It utilizes two distinct digits: $0$ and $1$.
• Weights are assigned relative to the binary point ($.$). The weighting for a binary number up to three places before and two places after the binary point is as follows:
  • $2^2 = 4$
  • $2^1 = 2$
  • $2^0 = 1$
  • $.$
  • $2^{-1} = 0.5$
  • $2^{-2} = 0.25$
• The Least Significant Bit (LSB) is the rightmost binary digit, representing the lowest weight in a given number.
• The Most Significant Bit (MSB) is the leftmost binary digit, representing the highest weight in a given number.

Octal Number System

• The octal system is a base-eight system.
• It utilizes eight distinct digits: $0, 1, 2, 3, 4, 5, 6, 7$.
• Weights are assigned relative to the octal point ($.$). The weighting for an octal number up to three places before and two places after the octal point is as follows:
  • $8^2 = 64$
  • $8^1 = 8$
  • $8^0 = 1$
  • $.$
  • $8^{-1} = 0.125$
  • $8^{-2} = 0.015625$

Hexadecimal Number System

• The hexadecimal system is a base-sixteen system.
• It utilizes sixteen distinct digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F$.
• The letters $A$ through $F$ represent decimal values $10$ through $15$.
• Weights are assigned relative to the hexadecimal point ($.$). The weighting for a hexadecimal number up to three places before and two places after the hexadecimal point is as follows:
  • $16^2 = 256$
  • $16^1 = 16$
  • $16^0 = 1$
  • $.$
  • $16^{-1} = 0.0625$
  • $16^{-2} = 0.0039$

One-to-One Comparison of Number Systems

• The following table illustrates the representation of values across the four number systems from decimal $0$ to $16$:

Binary Arithmetic

Binary Addition

• There are four basic rules for adding binary digits:
  • $0 + 0 = 0$
  • $0 + 1 = 1$
  • $1 + 0 = 1$
  • $1 + 1 = 10$ (which is $0$ carry $1$)
• Addition Examples:
  • Example 1 (Carries: $1, 1, 0$): $1010_2 + 1001_2 = 10011_2$
  • Example 2 (Carries: $0, 1, 1$): $101_2 + 011_2 = 1000_2$
  • Example 3 (Carries: $1, 1, 1$): $1110_2 + 1011_2 = 11001_2$
  • Example 4 (Carries: $1, 1, 1, 1$): $10101_2 + 11011_2 = 110000_2$

Binary Subtraction

• Key concept: $10_2 = (1 \times 2^1) + (0 \times 2^0) = 2_{10}$.
• When subtracting a larger digit from a smaller digit (e.g., $0 - 1$), you must borrow from the next higher position. The borrow carries a value of $2$ into the current position.
• Subtraction Examples:
  • Example 1: $110_2 - 101_2$
    • Step: Borrow from the second position, making the first position $10_2$. $2 - 1 = 1$. Result: $001_2$.
    • Verification: $110_2 = 6_{10}$, $101_2 = 5_{10}$. $6 - 5 = 1$.
  • Example 2: $11000_2 - 111_2 = 10001_2$
    • Verification: $11000_2 = (1 \times 2^4) + (1 \times 2^3) = 16 + 8 = 24_{10}$. $111_2 = (1 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 7_{10}$. $24 - 7 = 17_{10}$, which is $10001_2$.

Octal Arithmetic

Octal Addition

• When the sum of two octal digits equals or exceeds $8$, a carry is generated for the next higher position.
• Example 1: $356_8 + 124_8$
  • $6 + 4 = 10_{10}$. Since $10 ≥ 8$, result is $10 - 8 = 2$ with a carry of $1$.
  • $5 + 2 + 1 (carry) = 8_{10}$. Since $8 ≥ 8$, result is $8 - 8 = 0$ with a carry of $1$.
  • $3 + 1 + 1 (carry) = 5$.
  • Therefore, $356_8 + 124_8 = 502_8$.
• Example 2: $1012_8 + 114_8$ could be calculated similarly using base-8 rules.

Octal Subtraction

• If borrowing is required, the borrow from the specific position adds $8$ to the current digit.
• Example 1: $253_8 - 147_8 = 104_8$
  • Units: $3 < 7$, so borrow from the tens position. This becomes $(8 + 3) - 7 = 4$.
  • Tens: Remaining $4 - 4 = 0$.
  • Hundreds: $2 - 1 = 1$.
• Example 2: $1523_8 - 364_8$
  • Units: $3 < 4$, borrow from tens: $(8 + 3) - 4 = 7$.
  • Tens: Remaining $1 < 6$, borrow from hundreds: $(8 + 1) - 6 = 3$.
  • Hundreds: Remaining $4 - 3 = 1$.
  • Thousands: $1 - 0 = 1$.
  • Therefore, $1523_8 - 364_8 = 1137_8$.

Hexadecimal Arithmetic

Hexadecimal Addition

• Addition follows standard rules, where values exceeding $15$ result in a carry-out to the next place value.
• Example 1: $23_{16} + 16_{16} = 39_{16}$.
• Example 2: $2B_{16} + 84_{16}$
  • $B (11) + 4 = 15 = F$.
  • $2 + 8 = 10 = A$.
  • Result: $AF_{16}$.
• Example 3: $DF_{16} + AC_{16}$
  • $F (15) + C (12) = 27_{10}$. $27 - 16 = 11 (B)$ with carry $1$.
  • $D (13) + A (10) + 1 (carry) = 24_{10}$. $24 - 16 = 8$ with carry $1$.
  • Result: $18B_{16}$.

Hexadecimal Subtraction

• Example 1: $75_{16} - 21_{16} = 54_{16}$.
• Example 2: $84_{16} - 2A_{16}$
  • $4 < A (10)$, borrow $16$: $(16 + 4) - 10 = 10 = A$.
  • Remaining $7 - 2 = 5$.
  • Result: $5A_{16}$.

Number Base Conversions

Binary to Decimal

• Multiply each binary digit by its weighted position value and sum the results.
• Example 1: Convert $110101_2$ to decimal.
  • $(1 \times 2^5) + (1 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) = 32 + 16 + 0 + 4 + 0 + 1 = 53_{10}$.
• Example 2: Convert $0.1011_2$ to decimal.
  • $(1 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3}) + (1 \times 2^{-4}) = 0.5 + 0 + 0.125 + 0.0625 = 0.6875_{10}$.

Decimal to Binary

• For integers: Repeatedly divide by $2$ and collect the remainders from bottom to top.
• Example 1: Convert $18_{10}$ to binary.
  • $18 / 2 = 9$ Remainder $0$
  • $9 / 2 = 4$ Remainder $1$
  • $4 / 2 = 2$ Remainder $0$
  • $2 / 2 = 1$ Remainder $0$
  • $1 / 2 = 0$ Remainder $1$
  • Result: $10010_2$.
• For fractions: Repeatedly multiply by $2$. If the result exceeds or equals $1$, the carry is $1$. Continue with the fractional part until it becomes zero.
• Example 2: Convert $0.3125_{10}$ to binary.
  • $0.3125 \times 2 = 0.625$ (Carry $0$)
  • $0.625 \times 2 = 1.250$ (Carry $1$)
  • $0.25 \times 2 = 0.500$ (Carry $0$)
  • $0.50 \times 2 = 1.000$ (Carry $1$)
  • Result: $0.0101_2$.

Octal to Decimal

• Multiply each digit by its weighted position value ($8^n$) and sum.
• Example 1: Convert $237.04_8$ to decimal.
  • $(2 \times 8^2) + (3 \times 8^1) + (7 \times 8^0) + (0 \times 8^{-1}) + (4 \times 8^{-2}) = 128 + 24 + 7 + 0 + 0.0625 = 159.0625_{10}$.

Decimal to Octal

• For integers: Repeatedly divide by $8$ and collect remainders.
• Example 1: Convert $359_{10}$ to octal.
  • $359 / 8 = 44$ Remainder $7$
  • $44 / 8 = 5$ Remainder $4$
  • $5 / 8 = 0$ Remainder $5$
  • Result: $547_8$.
• For fractions: Repeatedly multiply by $8$ and collect the carry integers.
• Example 2: Convert $0.3125_{10}$ to octal.
  • $0.3125 \times 8 = 2.50$ (Carry $2$)
  • $0.5 \times 8 = 4.00$ (Carry $4$)
  • Result: $0.24_8$.

Binary and Octal Inter-conversion

• Binary to Octal: Break the binary digits into groups of three starting from the LSB. Add leading zeros as MSBs if necessary.
• Octal to Binary: Each octal digit is represented by exactly three binary digits.
  • $0: 000, 1: 001, 2: 010, 3: 011, 4: 100, 5: 101, 6: 110, 7: 111$.
  • Example 1: Convert $13_8$ to binary. $1 \rightarrow 001$, $3 \rightarrow 011$. Total: $001011_2$.
  • Example 2: Convert $37.12_8$ to binary. $3 \rightarrow 011$, $7 \rightarrow 111$, $1 \rightarrow 001$, $2 \rightarrow 010$. Total: $011111.001010_2$.

Hexadecimal to Decimal

• Multiply each hexadecimal digit by its weighted position ($16^n$).
• Example 1: Convert $356_{16}$ to decimal.
  • $(3 \times 16^2) + (5 \times 16^1) + (6 \times 16^0) = 768 + 80 + 6 = 854_{10}$.
• Example 2: Convert $AF2_{16}$ to decimal.
  • $(10 \times 16^2) + (15 \times 16^1) + (2 \times 16^0) = 2560 + 240 + 2 = 2802_{10}$.

Decimal to Hexadecimal

• Repeatedly divide by $16$ and record remainders (convert remainders $10–15$ to letters $A–F$).
• Example 1: Convert $650_{10}$.
  • $650 / 16 = 40$ Remainder $10 (A)$
  • $40 / 16 = 2$ Remainder $8$
  • $2 / 16 = 0$ Remainder $2$
  • Result: $28A_{16}$.

Hexadecimal and Binary Inter-conversion

• Hexadecimal to Binary: One hexadecimal digit represents four binary digits.
  • $0: 0000, 1: 0001, 2: 0010, 3: 0011, 4: 0100, 5: 0101, 6: 0110, 7: 0111, 8: 1000, 9: 1001, A: 1010, B: 1011, C: 1100, D: 1101, E: 1110, F: 1111$.
  • Example 1: Convert $38_{16}$ to binary. $3 \rightarrow 0011$, $8 \rightarrow 1000$. Total: $00111000_2$.
  • Example 2: Convert $FB17_{16}$ to binary. $F(1111), B(1011), 1(0001), 7(0111)$. Total: $1111101100010111_2$.
• Binary to Hexadecimal: Group binary digits into sets of four starting from the LSB. Add zeros to the MSB side to complete the last grouping if needed.