Radix Complement

The radix complement of an $n$-digit number $N$ in base $r$ is

$r^n-N$

The number you’d add to $N$ to get the smallest power of $r$ greater than $N$. The diminished radix complement is one less:

$(r^n-1)-N$

These two complement operations are the foundation of subtraction-by-addition in digital hardware. Radix complement in base 2 is 2’s complement; in base 10 it’s 10’s complement. The diminished forms are 1’s complement and 9’s complement respectively.

Why two flavors

The diminished radix complement is easier to compute by hand. Since $r^n-1$ is the digit $r-1$ repeated $n$ times (in base 10, that’s $\underset{n}{\underbrace{99\dots 9}}$; in base 2, $\underset{n}{\underbrace{11\dots 1}}$), the diminished complement is just “complement each digit against $r-1$” — subtract each digit from $r-1$ independently, no borrows.

In binary that means flip every bit. In decimal that means $9-d$ for each digit. Trivial.

The full radix complement is then “diminished radix complement plus one”:

$r^n-N=((r^n-1)-N)+1$

Easier than computing $r^n-N$ directly.

Examples

Finding 10’s complement of 873.

Diminished (9’s) complement: $999-873=126$.

Add 1: $126+1=127$.

Check: $1000-873=127$. ✓

Finding 2’s complement of 01110011.

Diminished (1’s) complement: invert every bit → $10001100$.

Add 1: $10001100+1=10001101$.

That’s the 2’s complement representation of $-115$ in 8 bits (since $01110011=115$).

Subtraction by addition

The killer application. The standard procedure (the one this note uses throughout): to compute $M-S$ for minuend $M$ and subtrahend $S$,

1. Take the diminished radix complement of $S$ (the subtrahend).
2. Add it to $M$.
3. If a carry-out leaves the high digit, the result is positive — apply end-around carry (add the carry-out back into the low digit) to get the answer.
4. If there is no carry-out, the result is negative — take the diminished radix complement of the sum and prefix a minus sign.

This is the form that maps directly onto digital adder hardware: feed $S$ through inverters, set the carry-in to $0$ for diminished or $1$ for full radix complement, and read out the result. See 2’s Complement Arithmetic for the binary version using XOR gates.

Worked subtraction examples

Base 10, positive result: $873-218$.

9’s complement of subtrahend $218$ is $999-218=781$. Add to minuend $873$: $873+781=1654$. The leading 1 is the carry-out — apply end-around carry: $654+1=655$. Result: $873-218=655$. ✓

Base 10, negative result: $89-92$.

9’s complement of subtrahend $92$ is $99-92=7$ (working in 2 digits to match the minuend’s width). Add to minuend $89$: $89+7=96$. No carry-out, so the result is negative: take 9’s complement of $96$, which is $99-96=3$, and prefix a minus sign: $-3$. ✓

Base 2, positive result: $00110000-00010100$.

Standard binary procedure: take the 1’s complement of the subtrahend and add it to the minuend, then apply end-around carry.

1’s complement of subtrahend $00010100$ is $11101011$.

Add to minuend: $00110000+11101011=100011011$ (a 9-bit sum with a carry-out of 1).

End-around carry — add the carry-out back into the low end: $00011011+1=00011100$.

Result: $00110000-00010100=00011100$. Decimal check: $48-20=28=00011100_2$. ✓

Base 2, negative result: $1000011-1010100$.

1’s complement of the subtrahend $1010100$ is $0101011$.

Add to minuend: $1000011+0101011=1101110$ (no carry-out, so no end-around carry).

The absence of end-around carry signals a negative result. Take the 1’s complement of the sum and prefix a minus: $-(0010001)=-10001_2=-17$. Decimal check: $67-84=-17$. ✓

Why bother

Doing subtraction directly in hardware would require a separate “subtractor” circuit. Doing it by complement-and-add reuses an adder you already have. That’s a huge silicon savings — the entire CPU integer ALU uses the same adder for both ops, with one extra XOR gate row to optionally invert. For decimal it’s the same idea (using 9’s/10’s complement) — used historically in mechanical and tube calculators where a ten-input adder was much simpler than a ten-input subtractor.