Related Concepts: 1.01 Positional Number Systems & Base Conversions | 1.03 Signed Representation, Overflow & Two’s Complement | 2.04 Error Control, Parity Generators & Checkers

1.02 Base Complements & Subtraction Mechanics

Overview

Complement arithmetic allows computers to perform subtraction using only addition operations. This eliminates the need for physical subtraction circuitry, enabling a single binary adder circuit to perform both addition and subtraction.

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1. Formal Mathematics of Complements

For a positive number $N$ represented in base $r$ with an integer part of $n$ digits and a fractional part of $m$ digits, we define two types of complements.

1.1 The Radix Complement ($r$‘s Complement)

The radix complement ($r$‘s complement) of a number $N$ in base $r$ is defined as:

${\text{Comp}}_r(N)=\{\begin{array}{ll}{r}^n-N& \text{for }N\ne 0\\ 0& \text{for }N=0\end{array}$

Binary Implementation (2’s Complement, $r=2$)

For a binary integer ($m=0$), the 2’s complement of an $n$-bit number $N$ is $2^n-N$.

• Pencil-and-Paper Shortcut: Starting from the right (Least Significant Bit), copy all bits up to and including the first 1 exactly as they are. Then, invert all remaining bits to the left.

Decimal Implementation (10’s Complement, $r=10$)

For a decimal integer, the 10’s complement of $N$ is $10^n-N$.

• Pencil-and-Paper Shortcut: Subtract the least significant non-zero digit from $10$, and subtract all other digits to the left from $9$.

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1.2 The Diminished Radix Complement ($(r-1)$‘s Complement)

The diminished radix complement ($(r-1)$‘s complement) of a number $N$ in base $r$ is defined as:

${\text{Comp}}_{r-1}(N)=r^n-r^{-m}-N$

If $N$ has no fractional part ($m=0$), this simplifies to:

${\text{Comp}}_{r-1}(N)=r^n-1-N$

Binary Implementation (1’s Complement, $r=2$)

For a binary integer, the 1’s complement of $N$ is $(2^n-1)-N$.

• Pencil-and-Paper Shortcut: Simply invert every single bit of the binary string ($0\to 1$ and $1\to 0$).

Decimal Implementation (9’s Complement, $r=10$)

For a decimal integer, the 9’s complement of $N$ is $(10^n-1)-N$.

• Pencil-and-Paper Shortcut: Subtract each individual digit of the number from $9$.

Key Relationship:

${\text{Comp}}_r(N)={\text{Comp}}_{r-1}(N)+r^{-m}$ (For integers, $r$‘s complement is simply the $(r-1)$‘s complement plus 1 in the LSB position).

1.3 Comparison Table: Radix vs. Diminished Radix Complements

Feature	Radix ($r$‘s) Complement	Diminished Radix ($(r-1)$‘s) Complement
Mathematical Definition	$r^n-N$ (for $N\ne 0$)	$r^n-r^{-m}-N$
Binary Equivalent ($r=2$)	2’s Complement	1’s Complement
Decimal Equivalent ($r=10$)	10’s Complement	9’s Complement
Mathematical Relationship	$r$’s Complement = $(r-1)$‘s Complement $+r^{-m}$	$(r-1)$‘s Complement = $r$‘s Complement $-r^{-m}$
Ease of Hardware Generation	Slightly harder (requires an addition step $+1$ in the LSB).	Very easy (requires only inverting NOT gates in binary).
Complement Subtraction Carry Action	Discard overflow carry ($C_{out}=1$).	End-Around Carry (add $C_{out}=1$ back to LSB).
Representation of Zero in Binary	Single unique representation (0000...0000).	Two representations: $+0$ (0000...0000) and $-0$ (1111...1111).

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2. Complement Subtraction Mechanics

Subtraction of two numbers $M-N$ in base $r$ is executed by adding the complement of the subtrahend {the number being taken away, $N$} to the minuend {the number being taken from, $M$} (PYQ 2015, 2018, 2019 — 9 to 13 marks).

graph TD
    Sub["Subtraction M - N"] --> Pad["Pad M and N to Equal n-bit Length"]
    Pad --> Method{"Complement Type"}
    
    Method -->|r's Complement| Add2["Add M + r's Comp of N"]
    Add2 --> Carry2{"Carry generated?"}
    Carry2 -->|Yes: Cout = 1| Disc2["Discard Carry. Result is Positive & in True Form"]
    Carry2 -->|No: Cout = 0| Neg2["Result is Negative. Take r's Comp of Sum & add - sign"]
    
    Method -->|r-1's Complement| Add1["Add M + (r-1)'s Comp of N"]
    Add1 --> Carry1{"Carry generated?"}
    Carry1 -->|Yes: Cout = 1| EndAround["End-Around Carry: Add 1 to LSB. Result is Positive"]
    Carry1 -->|No: Cout = 0| Neg1["Result is Negative. Take (r-1)'s Comp of Sum & add - sign"]

2.1 The Radix ($r$‘s) Complement Subtraction Algorithm ($M-N$)

1. Pad both $M$ and $N$ with leading zeros so they have the exact same number of digits ($n$).
2. Compute the $r$‘s complement of the subtrahend $N$: ${\text{Comp}}_r(N)=r^n-N$.
3. Add the minuend $M$ to the complement: $Sum=M+(r^n-N)=M-N+r^n$
4. Check the end carry digit ($C_{out}$):
   • Case A: Carry Occurs ($C_{out}=1$) Discard the carry (which mathematically subtracts $r^n$). The remaining digits represent the correct positive result: $Result=Sum-r^n=M-N$
   • Case B: No Carry Occurs ($C_{out}=0$) The result is negative. The remaining digits are in their complemented form. To find the true magnitude, take the $r$‘s complement of the sum and prefix a negative sign: $Result=-{\text{Comp}}_r(Sum)=-(r^n-(M-N+r^n))=-(N-M)$

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3. Step-by-Step Worked Subtraction Examples

There are two distinct outcomes to drill, and the examiner has asked for both. Case 1 is $M<N$ (no carry, negative answer); Case 2 is $M>N$ (carry generated, positive answer). Most students only practise one and lose marks on the other — especially the end-around carry step, which only ever appears in Case 2.

Case 1: $M<N$ — No Carry, Negative Result

Worked Exam Problem (PYQ 2018, 2019 — 9 to 10 marks)

Question (2019, verbatim): Perform the subtraction with the following binary numbers using (i) 2’s complement and (ii) 1’s complement. Check the answer by straight subtraction. $(100-110000{)}_2$.

Let $M=100_2$ (Decimal $4$) and $N=110000_2$ (Decimal $48$).

• Step 1: Pad to equal bit length (8-bit standard): $M=00000100_2$ $N=00110000_2$

Part A: Subtraction Using 2’s Complement ($r$‘s complement)

1. Find 2’s Complement of $N$ ($00110000_2$):
   • 1’s complement = $11001111_2$
   • Add 1 to LSB = $11010000_2$
2. Add $M+\text{2’s Comp of }N$:

      00000100   (M)
    + 11010000   (2's Comp of N)
    ----------
    0 11010100   (Sum)
    ^
    Cout = 0 (No Carry)

3. Evaluate Result: Since $C_{out}=0$, the result is negative and in 2’s complement form.
   • Take 2’s complement of the sum ($11010100_2$):
     • 1’s complement = $00101011_2$
     • Add 1 to LSB = $00101100_2$ (Decimal $44$)
   • Prefix a negative sign.
   $\text{Final Result}=-00101100_2(-44_{10})$

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Part B: Subtraction Using 1’s Complement ($(r-1)$‘s complement)

1. Find 1’s Complement of $N$ ($00110000_2$):
   • 1’s complement = $11001111_2$
2. Add $M+\text{1’s Comp of }N$:

      00000100   (M)
    + 11001111   (1's Comp of N)
    ----------
    0 11010011   (Sum)
    ^
    Cout = 0 (No Carry)

3. Evaluate Result: Since $C_{out}=0$, the result is negative.
   • Take 1’s complement of the sum ($11010011_2$):
     • 1’s complement = $00101100_2$ (Decimal $44$)
   • Prefix a negative sign.
   $\text{Final Result}=-00101100_2(-44_{10})$

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Part C: Verification by Straight Subtraction

Since $M<N$, perform $-(N-M)$ using direct binary borrowing:

   110000   (N = 48)
 - 000100   (M = 4)
 --------
   101100   (Result = 44)

Prefixing the negative sign and padding to 8-bit gives $-00101100_2$. All three methods yield the same result. $■$

Notice what did not happen here

Because no carry was generated, the 1’s complement answer needed no end-around carry. Students who have only ever seen this case walk into the exam not knowing the end-around step exists. Case 2 below is where it fires.

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Case 2: $M>N$ — Carry Generated, Positive Result

Worked Exam Problem (PYQ 2015 — 13 marks)

Question (verbatim): Perform the subtraction with the following binary numbers using (i) 2’s complement and (ii) 1’s complement. Check the answer by straight subtraction: $(1110110-111{)}_2$.

Let $M=1110110_2$ (Decimal $118$) and $N=111_2$ (Decimal $7$). Pad both to 7 bits: $M=1110110_{2}N=0000111_2$

Part A: Using 2’s Complement — discard the carry

1. 2’s complement of $N$: 1’s complement of $0000111$ is $1111000$; add 1 $⟹1111001_2$.
2. Add:

      1110110   (M)
    + 1111001   (2's Comp of N)
    ----------
    1 1101111   (Sum)
    ^
    Cout = 1 (Carry generated)

3. Evaluate: A carry occurred, so discard it. The remaining 7 bits are the answer in true positive form.

$\text{Result}=1101111_2=+111_{10}$

Part B: Using 1’s Complement — add the carry back (End-Around Carry)

1. 1’s complement of $N$: invert every bit of $0000111⟹1111000_2$.
2. Add:

      1110110   (M)
    + 1111000   (1's Comp of N)
    ----------
    1 1101110   (Sum)
    ^
    Cout = 1

3. End-Around Carry: in 1’s complement you must not throw the carry away — add it back into the LSB:

      1101110
    +       1   (End-around carry)
    ---------
      1101111

$\text{Result}=1101111_2=+111_{10}$

Part C: Verification by Straight Subtraction

   1110110   (M = 118)
 - 0000111   (N = 7)
 ---------
   1101111   (Result = 111)

All three methods agree. $■$

The one rule that decides everything

Look only at the carry out of the MSB:

	Carry = 1	Carry = 0
2’s complement	Discard carry → answer is positive, read directly	Answer is negative → re-complement the sum, add minus sign
1’s complement	Add carry back to LSB → answer is positive	Answer is negative → re-complement the sum, add minus sign

Memory hook: 2’s complement throws the carry away, 1’s complement recycles it.

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4. Non-Decimal Base Multiplication

Performing arithmetic directly in bases other than decimal is a common exam requirement.

Worked Exam Problem (PYQ 2024 — 08 marks)

Question (verbatim): Multiply the following numbers in the given base without converting to decimal. (i) $(135.4{)}_6$ and $(43.2{)}_6$, (ii) $(367{)}_8$ and $(715{)}_8$.

Part (ii) is solved first because it has no radix point; part (i) adds the fraction handling.

4.1 Part (ii): $(367{)}_8\times (715{)}_8$

Direct Octal Arithmetic Rule: Perform standard digit-by-digit multiplication. When a product exceeds the base ($r=8$), divide the product by $8$. Record the remainder in the current column and carry the quotient to the next column.

Step-by-Step Multiplication:

1. Multiply $(367{)}_8$ by $5_8$ (first digit of multiplier):

   • $5\times 7=35_{10}⟹35÷8=4$ with remainder $3$ (write 3, carry 4)
   • $5\times 6=30_{10}+4=34_{10}⟹34÷8=4$ with remainder $2$ (write 2, carry 4)
   • $5\times 3=15_{10}+4=19_{10}⟹19÷8=2$ with remainder $3$ (write 3, carry 2)
   • First partial product: $2323_8$

2. Multiply $(367{)}_8$ by $1_8$ (second digit of multiplier):

   • Second partial product: $3670_8$ (shifted left one space)

3. Multiply $(367{)}_8$ by $7_8$ (third digit of multiplier):

   • $7\times 7=49_{10}⟹49÷8=6$ with remainder $1$ (write 1, carry 6)
   • $7\times 6=42_{10}+6=48_{10}⟹48÷8=6$ with remainder $0$ (write 0, carry 6)
   • $7\times 3=21_{10}+6=27_{10}⟹27÷8=3$ with remainder $3$ (write 3, carry 3)
   • Third partial product: $330100_8$ (shifted left two spaces)

4. Add the partial products in Base-8:

      002323
      003670
    + 330100
    --------
      336313

   • Column 0: $3+0+0=3_8$
   • Column 1: $2+7+0=9_{10}⟹9÷8=1$ rem $1$ (write 1, carry 1)
   • Column 2: $3+6+1+1\text{ (carry)}=11_{10}⟹11÷8=1$ rem $3$ (write 3, carry 1)
   • Column 3: $2+3+0+1\text{ (carry)}=6_8$ (write 6)
   • Column 4: $3_8$ (write 3)
   • Column 5: $3_8$ (write 3)

$\text{Final Product: }(336313{)}_8■$

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4.2 Part (i): $(135.4{)}_6\times (43.2{)}_6$ — Handling the Radix Point

The fraction trick: do not try to multiply fractional digits in place. Instead:

1. Ignore both radix points and multiply the digit strings as whole numbers.
2. Count the total number of fractional digits in the two operands ($1+1=2$ here).
3. Place the radix point that many digits from the right of the product.

This works for any base, because shifting a radix point $k$ places is just multiplying by $r^k$, and the two shifts cancel at the end.

Step 1 — strip the points: $1354_6\times 432_6$.

Step 2 — partial products (base-6 rule: on each digit product, divide by 6, write the remainder, carry the quotient):

Multiply $1354_6$ by $2$:

• $4\times 2=8_{10}=1\times 6+2⟹$ write $2$, carry $1$
• $5\times 2=10+1=11_{10}=1\times 6+5⟹$ write $5$, carry $1$
• $3\times 2=6+1=7_{10}=1\times 6+1⟹$ write $1$, carry $1$
• $1\times 2=2+1=3⟹$ write $3$
• Partial product 1: $3152_6$

Multiply $1354_6$ by $3$ (shift left one place):

• $4\times 3=12_{10}=2\times 6+0⟹$ write $0$, carry $2$
• $5\times 3=15+2=17_{10}=2\times 6+5⟹$ write $5$, carry $2$
• $3\times 3=9+2=11_{10}=1\times 6+5⟹$ write $5$, carry $1$
• $1\times 3=3+1=4⟹$ write $4$
• Partial product 2: $4550_6$ (shifted one place)

Multiply $1354_6$ by $4$ (shift left two places):

• $4\times 4=16_{10}=2\times 6+4⟹$ write $4$, carry $2$
• $5\times 4=20+2=22_{10}=3\times 6+4⟹$ write $4$, carry $3$
• $3\times 4=12+3=15_{10}=2\times 6+3⟹$ write $3$, carry $2$
• $1\times 4=4+2=6_{10}=1\times 6+0⟹$ write $0$, carry $1⟹$ write $1$
• Partial product 3: $10344_6$ (shifted two places)

Step 3 — add the partial products in base-6:

     0003152
     0045500
  +  1034400
  ----------
     1131452

• Col 0: $2$
• Col 1: $5$
• Col 2: $1+5+4=10_{10}=1\times 6+4⟹$ write $4$, carry $1$
• Col 3: $3+5+4+1=13_{10}=2\times 6+1⟹$ write $1$, carry $2$
• Col 4: $0+4+3+2=9_{10}=1\times 6+3⟹$ write $3$, carry $1$
• Col 5: $0+0+0+1=1$
• Col 6: $1$

Step 4 — restore the radix point (2 fractional digits total, counted from the right):

$(135.4{)}_6\times (43.2{)}_6=(11314.52{)}_6■$

Sanity check (for your own confidence only — never write this in the answer script, the question forbids converting): $(135.4{)}_6=59.667$, $(43.2{)}_6=27.333$, product $=1630.889$; and $(11314.52{)}_6=1296+216+108+6+4+\frac{5}{6}+\frac{2}{36}=1630.889$. ✓

Exam Traps in Non-Decimal Multiplication

• The question says “without converting to decimal.” Converting to decimal, multiplying, and converting back will lose most of the marks even if the final number is right. Show the digit-by-digit carries.
• Every digit you write must be $<r$. A "$6$" or "$7$" in a base-6 answer is an instant error signal.
• Carry the quotient, write the remainder — students routinely do it the other way round under time pressure.
• Fractional digits are counted, not computed. One fractional digit in each operand always gives two in the product.

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Past Year Questions (PYQs)

Question (as asked)	Years	Marks	Solved in
Subtract $(1110110-111{)}_2$ using 2’s and 1’s complement, check by straight subtraction	2015	13	§3 Case 2
Subtract $(100-11000{)}_2$ using 2’s and 1’s complement, check by straight subtraction	2018	9	§3 Case 1 (same method)
Subtract $(100-110000{)}_2$ using 2’s and 1’s complement, check by straight subtraction	2019	10	§3 Case 1
Multiply $(135.4{)}_6\times (43.2{)}_6$ and $(367{)}_8\times (715{)}_8$ without converting to decimal	2024	8	§4.1, §4.2

Pattern to notice: the subtraction question appears in three of the last ten papers and always demands all three methods (2’s complement, 1’s complement, straight subtraction). Answering with only one method caps you at roughly a third of the marks.