Two's complement signed numbers: Visual walkthrough

Before we start, two words we'll use constantly:

A bit is one switch: it is either 0 (off) or 1 (on).
A number system is a rule that turns a row of switches into one number. We're going to invent such a rule for negatives.

Step 1 — The circle of all switch patterns

WHAT: We arrange every 4-bit pattern 0000 … 1111 around a ring, 0000 at the top.

WHY a circle and not a line? Because a machine with 4 switches cannot hold a 17th pattern. When you count past 1111 you don't get 10000 — that fifth switch doesn't exist, so you drop back to 0000. A line has ends; this thing has no end, it loops. A circle is the honest picture.

PICTURE: Look at the wheel below. Walking clockwise = adding 1. Walking counter-clockwise = subtracting 1. Notice the join between 1111 and 0000 — that seam is the whole secret.

Figure — Two's complement signed numbers

Recall Why must it wrap?

A machine with $n$ switches has how many total patterns, and what happens when you add 1 to the last one? ::: Exactly $2^{n}$ patterns; adding 1 to 111…1 overflows past the last switch and lands back on 000…0 — the wheel closes.

Step 2 — Going backwards from zero

WHAT: We physically walk one seat backward from zero.

WHY this pins down $- 1$ : $- 1$ is defined as "the thing you reach by subtracting 1 from 0". There is no arguing with that. Whatever seat we land on is $- 1$ for this machine — and that seat is 1111.

PICTURE: The red arrow steps 0000 → 1111. So 1111 is $- 1$ . Keep going: 1110 is $- 2$ , 1101 is $- 3$ , and so on down the top-left of the wheel.

Step 3 — Split the wheel: who is positive, who is negative?

WHAT: We colour the wheel into two halves by the leftmost switch.

WHY split exactly there? Because in binary the leftmost switch is 0 for the lower half (0000–0111) and 1 for the upper half (1000–1111). That's not a coincidence we imposed — it's just how counting works. So the MSB already separates "small counts" from "far-around-the-back counts". We simply reinterpret "far around the back" as negative.

PICTURE: Right half glows green (0…7), left half glows red (the negatives). The dividing line runs through 0000 at the top and between 0111 and 1000 at the bottom.

Two questions this raises, answered in the next steps:

Exactly what number is each negative seat? (Step 4)
Why is one half slightly bigger than the other? (Step 5)

Step 4 — Turning " $U - 2^{n}$ " into the negative weight

WHAT: For any red (top-bit-1) pattern, we compute its negative name by subtracting one full lap.

WHY subtract $2^{n}$ ? One full lap around the wheel is $2^{n}$ seats. Going forward a full lap brings you back to the same seat, so the value " $U$ " and the value " $U - 2^{n}$ " point at the same place. Since the red seats are past the halfway mark, their $U$ is large and positive; subtracting a lap makes it the small negative we want.

V = clockwise count U - one full lap 2^{n}

Now watch the algebra turn into the parent's "negative weight". Write $U$ by splitting off the top switch:

U = value of the top switch b_{n - 1} 2^{n - 1} + lower switches rest

For a red seat $b_{n - 1} = 1$ , so $b_{n - 1} 2^{n - 1} = 2^{n - 1}$ . Substitute:

V = (2^{n - 1} + rest) - 2^{n} = rest + = - 2^{n - 1} (2^{n - 1} - 2^{n}) = top switch’s TRUE weight - 2^{n - 1} + rest

Term by term: the top switch, which looked like it was worth $+ 2^{n - 1}$ , is really worth $- 2^{n - 1}$ once we subtract the lap. That single flip in sign is the parent note's headline definition.

PICTURE: For 1111, the blue bar $+ 8$ (the raw top-switch value) gets a full lap $16$ subtracted, sliding it to $- 8$ ; add the lower 111 $= 7$ and you land on $- 8 + 7 = - 1$ . ✓ — matching Step 2.

Recall The one substitution that does everything

Why is the top bit's weight $- 2^{n - 1}$ and not $+ 2^{n - 1}$ ? ::: Because for a negative seat we subtract one lap $2^{n}$ ; $+ 2^{n - 1} - 2^{n} = - 2^{n - 1}$ , so the top switch flips to a negative weight.

Step 5 — The lopsided wheel (the degenerate seat)

WHAT: We look at the single seat with no partner: 1000.

WHY it's special: Its value is $- 2^{n - 1}$ (for $n = 4$ : $- 8$ ). Try to negate it — flip to 0111, add 1 → 1000 again! It maps to itself. There is no $+ 8$ seat on a 4-bit wheel to hold its opposite, because the wheel only reaches $+ 7$ .

PICTURE: The lone red seat 1000 at the bottom-left has a curved arrow looping back onto itself — negating it doesn't move it. This is the " $- 2^{n - 1}$ has no twin" fact drawn out.

Step 6 — Deriving "flip, then bump" on the wheel

WHAT: We prove the negation rule using one identity, then read it off the wheel.

WHY start from all-ones? Because all-ones is a number we know cold: $n 111 \dots 1 = 2^{n} - 1$ (it's one seat short of a full lap). So:

every column has one 1 x + \overset{x}{ˉ} = all ones = one short of a lap 2^{n} - 1

Rearrange to isolate $\overset{x}{ˉ}$ , then add 1 to both sides:

\overset{x}{ˉ} = (2^{n} - 1) - x ⟹ flip, then bump \overset{x}{ˉ} + 1 = a full lap, minus x 2^{n} - x

And " $2^{n} - x$ " is where you land if you walk $x$ steps counter-clockwise from 0000 — which is the definition of $- x$ on the wheel:

- x = \overset{x}{ˉ} + 1 (mod 2^{n})

Term by term: $\overset{x}{ˉ}$ overshoots the target by exactly 1 seat (because all-ones is one short of a lap); the " $+ 1$ " bump closes that last gap and drops you onto $- x$ .

PICTURE: Start at 0101 ( $+ 5$ ). Flip → 1010 (that's $2^{n} - 1 - x$ , one seat short). Bump +1 → 1011, which the wheel labels $- 5$ . The two short arrows show "flip lands near, bump lands exactly".

Step 7 — Why one adder does subtraction too (the carry vanishes)

WHAT: Add 0111 ( $7$ ) and 1011 ( $- 5$ ) with an ordinary adder and watch the overflow bit.

WHY the carry is safe to drop: The two seats sum to 1 0010 — a 1 popped out of the top (a 5th switch). But that leftmost 1 is worth $2^{4} = 16$ = one full lap, and one full lap brings you back to the same seat. Mathematically $2^{n} \equiv 0 (mod 2^{n})$ : it's invisible to a 4-switch machine. So dropping it is not sloppy — it's exact.

70111 + - 5 1011 = = one lap 2^{n}, drop it 1 = 2 0010 = 2 ✓

PICTURE: Two clockwise/counter-clockwise walks on the wheel meeting at 0010; the discarded carry is drawn as a full grey lap that returns to start, labelled "invisible: $2^{n} \equiv 0$ ".

The one-picture summary

Recall Feynman retelling — the whole walkthrough in plain words

Say it back to a 12-year-old, no symbols. ::: We only have switches, so there's no minus sign to store. Put all the switch patterns around a clock. Zero is at the top. Going forward one step means "add one"; going back one step means "subtract one". Step back from zero and you land on all-switches-on — so we say that pattern is minus one. We paint the whole back half of the clock "negative", and the leftmost switch tells you which half you're in. To find the negative of a number, flip every switch — that gets you almost there, one seat short — then add one to land exactly on the negative seat. And when you add two of these clock-numbers and a 1 pops off the far left, just ignore it: that 1 is one whole lap around the clock, which brings you right back to where you were, so it changes nothing. That's the whole trick — one clock, one adder, negatives for free.

Connections

Parent: Two's complement — the rules this page draws out
Binary number system — the plain clockwise counting on the wheel
Sign-magnitude representation — the naive split that gives two zeros (a symmetric, gap-having wheel)
One's complement — flip only, no bump: lands one seat short (the missing $+ 1$ of Step 6)
Full adder — the circuit doing the clockwise/counter-clockwise walk of Step 7
Modular arithmetic — the " $2^{n} \equiv 0$ " that makes the carry a full invisible lap
Overflow and carry flags — the other signal to watch when dropping the carry