Balanced Ternary Representation

Summary: Euler’s theorem (§330–§331): every integer — positive, negative, or zero — has a unique representation as ${\sum }_k{c}_k{3}^k$ with digits $c_k\in \{-1,0,+1\}$. The proof mirrors the binary case: the formal Laurent product ${\prod }_{k\ge 0}(x^{-3^k}+1+x^{3^k})$ has every coefficient equal to $1$ at every power of $x$, positive or negative. Application: weighing on a two-pan balance with weights $1,3,9,27,81,\dots$ pounds.

Sources: chapter16

Last updated: 2026-05-11

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Statement

${\prod }_{k\ge 0}(x^{-3^k}+1+x^{3^k})={\sum }_{n\in \mathbb{Z}}{x}^n.$

Equivalently, every integer $n\in \mathbb{Z}$ has a unique representation

$n={\sum }_{k\ge 0}{c}_k{3}^k,c_k\in \{-1,0,+1\}.$

(Source: chapter16, §331.)

Euler’s proof (§331)

Let

$P={\prod }_{k\ge 0}(x^{-3^k}+1+x^{3^k})=\cdots +c{x}^{-3}+b{x}^{-2}+a{x}^{-1}+1+\alpha x+\beta x^2+\gamma x^3+\delta x^4+ϵx^5+\cdots .$

The product is a formal Laurent series — both positive and negative powers of $x$ appear because of the $x^{-3^k}$ terms.

Substituting $x^3$ for $x$ drops the $k=0$ factor:

${\prod }_{k\ge 1}(x^{-3^k}+1+x^{3^k})=\frac{P}{x^{-1}+1+x}.$

The left side is

$P(x^3)=\cdots +c{x}^{-9}+b{x}^{-6}+a{x}^{-3}+1+\alpha x^3+\beta x^6+\gamma x^9+\cdots .$

Hence

$P=(x^{-1}+1+x)\cdot P(x^3)=\cdots +a{x}^{-4}+a{x}^{-3}+a{x}^{-2}+b{x}^{-1}\cdot (\dots )\cdot \dots +1+x+\alpha x^2+\alpha x^3+\alpha x^4+\beta x^5+\beta x^6+\beta x^7+\gamma x^8+\gamma x^9+\gamma x^{10}+\cdots ,$

after multiplying the three-term factor through. Comparing with the original expansion of $P$ term by term:

$\alpha =1,\beta =\alpha ,\gamma =\alpha ,\delta =\alpha ,ϵ=\beta ,\zeta =\beta ,\dots \text{and}a=1,b=a,c=a,\dots$

All coefficients (positive and negative powers) equal $1$. Therefore

$P={\sum }_{n\in \mathbb{Z}}{x}^n=\cdots +x^{-3}+x^{-2}+x^{-1}+1+x+x^2+x^3+\cdots .$

Combinatorial interpretation

Expanding $\prod (x^{-3^k}+1+x^{3^k})$ as a sum over choices of one of three terms from each factor:

$P={\sum }_{\begin{array}{c}(c_k{)}_{k\ge 0}\\ c_k\in \{-1,0,+1\}\end{array}}{x}^{{\sum }_k{c}_k{3}^k}.$

The coefficient of $x^n$ is the number of sequences $(c_k)$ with $\sum c_k{3}^k=n$. Euler’s identity says this count is always $1$ — every integer has a unique balanced-ternary representation.

Application: weighing with ternary weights on a two-pan balance (§330)

A set of weights $1,3,9,27,81,\dots$ pounds suffices to weigh any whole number of pounds on a two-pan balance: each weight goes either on the opposite pan from the goods (digit $+1$), on the same pan as the goods (digit $-1$), or off the scale (digit $0$). Euler’s examples (§330):

$1=12=3-13=34=3+15=9-3-16=9-37=9-3+18=9-19=9$

$10=9+111=9+3-112=9+3.$

With $n$ weights ($1,3,\dots ,3^{n-1}$, total $(3^n-1)/2$), any integer up to $(3^n-1)/2$ can be weighed — a much faster growth than binary’s $2^n-1$. Six ternary weights ($1,3,9,27,81,243$, total $364$) reach $364$; six binary weights reach only $63$.

Comparison to binary (§329 vs §331)

	Binary	Balanced ternary
Identity	$\prod (1+x^{2^k})=1/(1-x)$	$\prod (x^{-3^k}+1+x^{3^k})={\sum }_{n\in \mathbb{Z}}{x}^n$
Digits	$\{0,1\}$	$\{-1,0,+1\}$
Range with $n$ weights	$0$ to $2^n-1$	$-(3^n-1)/2$ to $(3^n-1)/2$
Physical	one-pan scale	two-pan scale

Modern reading

The trick — make a symmetric digit set $\{-1,0,+1\}$ so that the product covers all integers, not just non-negative ones — is exactly the modern notion of balanced ternary (also called signed-digit ternary). It is used in some digital-arithmetic algorithms (Booth multiplier, non-adjacent form for elliptic-curve scalar multiplication) precisely because the symmetric digit set eliminates the need for a separate sign bit. Euler’s two pages (§330–§331) are the earliest known appearance of the system.

Related pages

• chapter-16-on-the-partition-of-numbers
• partition-of-numbers
• partition-generating-functions
• binary-representation-theorem
• geometric-series