existence and uniqueness of decimal expansion

The existence and uniqueness of decimal expansions (or more generally, base-b expansions) is taken for granted by most grade school students, but they are facts that need to be rigorously proven if one wants to understand the real numbers thoroughly.

We mention the following fact about natural numbersMathworldPlanetmath n,m0, which we will use many times implicitly:


This fact can be proven by mathematical induction on m.

1 Proof of Existence

Let x be a number for which we want to write a base-b expansion for any natural number b greater than one.

We shall assume x0, since the base-b expansion of a negative number is by definition the negative of the expansion for its absolute valueMathworldPlanetmathPlanetmathPlanetmath.

1.1 Expansions for non-negative integers

First we prove the existence of expansions of the form


for non-negative integers x, using mathematical induction. (This proof is essentially the formal statement of how to do additionPlanetmathPlanetmath by base-b digits.)

The number x=0 obviously has the expansion 0.

Suppose that we know the existence of expansions for a number x-1. We prove the existence of an expansion for x.

Let x-1 be expanded as


From the above equation, add 1 to both sides:


If a0<b-1, then we are done. Otherwise, (a0+1)=b, and therefore we may write instead


If a1<b-1, then we can stop. Otherwise, repeat the process and continue carrying digits until we reach some i for which ai<b-1. Since ak+1=0, this process is guaranteed to stop. At the end we will have expressed x in base b.

1.2 Reduction to numbers in [0,1)

Let x be the greatest integer less than or equal to x, otherwise known as the floor of x. We prove that the floor of x exists.

The set


is bounded above by x. However, by the Archimedean property, the set of natural numbers is not bounded above, so 0A must be non-empty, and have a smallest element u (formally, by the well-ordering principle). For every nA, we have nx<u. The latter condition is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to nu-1<x, so u-1 is the maximum element of A. In other words, x=u-1.

Since u-1x<u, we have 0x-x<1. We shall obtain the base b expansion of x as the sum of the expansion of x and x-x.

1.3 Expansion of numbers in [0,1)

Given x[0,1), let a1=bx. Then 0a1bx<b, so we can take a1 as the first digit of the base-b expansion of x. Next, write


and observe that 0x-a1b-1=b-1(bx-bx)<b-1, so it is possible to get the next digit of the expansion by expanding y. We do this recursively, leading to these recursive relations:


More explicitly, we have

x-(a1b-1+a2b-2++akb-k) =b-1(y2-(a2b-1++akb-k+1))

It is easy to prove that the expansion


converges to x:

0x-i=1kaib-i=b-kyk+1<1bk0,as k.

(Formally, the “0” part appeals to the Archimedean property.)

2 Proof of uniqueness

2.1 Uniqueness for non-negative integers

Suppose that




and the intervals [akbk,(ak+1)bk) are disjoint for each value of ak, so ak is uniquely determined by where x lies in amongst these intervals.

Then we can consider


Repeating the previous argumentMathworldPlanetmathPlanetmath with k replaced by k-1, we see that ak-1 is uniquely determined. Then we can consider x-akbk-ak-1bk-1 and so on. Continuing this way, we see that all the digits ai are uniquely determined.

2.2 Near-uniqueness for non-negative numbers



then a0,,ak are uniquely determined, since akbk++a1+a0 is the expansion for the non-negative integer x.

The argument to prove that a-i are uniquely determined proceeds similarly as before. We have

a-1b-1 a-1b-1+a-2b-2+
a-1b-1+i=2(b-1)b-i (geometric series)

where equality on the second line occurs if and only if a-i=b-1 for every i2. If we insist that a-i is never eventually the same digit b-1, then this shows that the digit a-1 is uniquely determined by where the original number x in the disjoint intervals [a-1b-1,(a-1+1)b-1).

This argument may be repeated, to show that a-i are uniquely determined, under the assumptionPlanetmathPlanetmath that the expansion does not end in all digits being b-1.

If the assumption is not made, then numbers which have an expansion ending in all digits 0 have an alternate expansion ending in all digits b-1, but other numbers still have unique base-b expansions.

3 Every base-b expansion represents a real number

We also want to prove that for every sequenceMathworldPlanetmath of digits ak,ak-1,,a1,a0,a-1,a-2, there exists a real number x with the base-b expansion


This is the where we use the least upper boundsMathworldPlanetmath property of the real numbers. (So far we have only used the Archimedean property, so what we have done so far is also valid for .)

Consider the sequence {sn} with the


This sequence, considered as a set, is bounded above, for snakbk++a0+1. So it has a least upper bound x. Since the sequence {sn} is also increasing, its least upper bound is the same as its limit.

Title existence and uniqueness of decimal expansion
Canonical name ExistenceAndUniquenessOfDecimalExpansion
Date of creation 2013-03-22 15:42:12
Last modified on 2013-03-22 15:42:12
Owner stevecheng (10074)
Last modified by stevecheng (10074)
Numerical id 8
Author stevecheng (10074)
Entry type TheoremMathworldPlanetmath
Classification msc 11A99
Synonym every decimal expansion represents a real number
Related topic CantorsDiagonalArgument
Related topic DecimalExpansion