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单词 ExistenceAndUniquenessOfDecimalExpansion
释义

existence and uniqueness of decimal expansion


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

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

n<mnm-1

This fact can be proven by mathematical induction on m.

Contents:
  • 1 Proof of Existence
    • 1.1 Expansions for non-negative integers
    • 1.2 Reduction to numbers in [0,1)
    • 1.3 Expansion of numbers in [0,1)
  • 2 Proof of uniqueness
    • 2.1 Uniqueness for non-negative integers
    • 2.2 Near-uniqueness for non-negative numbers
  • 3 Every base-b expansion represents a real number

1 Proof of Existence

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

We shall assume x0, since the base-bexpansion of a negative number is by definitionthe 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

x=i=0kaibi,0ai<b,ai0

for non-negative integers x,using mathematical induction. (This proof is essentially the formalstatement 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

x-1=i=0kaibi,0ai<b,ai0,ak+1=0.

From the above equation, add 1 to both sides:

x=(a0+1)+i=1kaibi.

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

x=0+(a1+1)b+i=2kaibi.

If a1<b-1, then we can stop. Otherwise,repeat the process and continue carrying digitsuntil 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

A={n0:nx}

is bounded above by x.However, by the Archimedeanproperty, the set of natural numbers is not boundedabove, so 0A must be non-empty,and have a smallest element u(formally, by the well-ordering principle).For every nA, we havenx<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 have0x-x<1.We shall obtain the base b expansion of xas the sum of the expansion of x andx-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 firstdigit of the base-b expansion of x.Next, write

x=a1b-1+yb-1,

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

yi=aib-1+yi+1b-1,0ai=byi<b,0yi<1,y1=x.

More explicitly, we have

x-(a1b-1+a2b-2++akb-k)=b-1(y2-(a2b-1++akb-k+1))
=b-2(y3-(a3b-1++akb-k+2))
=
=b-k+1(yk-akb-1)
=b-kyk+1.

It is easy to prove that the expansion

a1b-1+a2b-2++akb-k+

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

x=i=0kaibi,0ai<b.

Now

akbki=0kaibiakbk+i=0k-1(b-1)bi=akbk+(bk-1)<(ak+1)bk,

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

Then we can consider

x-akbk=i=0k-1aibi.

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 uniquelydetermined.

2.2 Near-uniqueness for non-negative numbers

If

x=akbk++a1b+a0+a-1b-1+a-2b-2+

then a0,,ak are uniquely determined,since akbk++a1+a0is the expansion for the non-negative integer x.

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

a-1b-1a-1b-1+a-2b-2+
a-1b-1+i=2(b-1)b-i(geometric series)
=a-1b-1+(b-1)b21-b-1
=(a-1+1)b-1,

where equality on the second line occurs if and only ifa-i=b-1 for every i2.If we insist that a-i is never eventually the samedigit b-1,then this shows that the digit a-1 is uniquely determined bywhere 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 uniquelydetermined, under the assumptionPlanetmathPlanetmath that the expansiondoes not end in all digits being b-1.

If the assumption is not made, thennumbers which have an expansion ending in all digits 0have 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 xwith the base-b expansion

x=i=0kaibi+i=0a-ib-i.

This is the where we use the least upper boundsMathworldPlanetmath property of thereal 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

sn=i=0kaibi+i=0na-ib-i.

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 itslimit.

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更新时间:2025/5/4 17:10:59