Prove uniqueness of $pi$'s decimal representation [closed]

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How do I prove that the decimal representation of the digits of $pi$ is unique?

For example, the number $1$ is not unique in its decimal representations since

$1.0000... = 0.999...$

irrational-numbers pi decimal-expansion

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edited Jul 19 at 22:05

RayDansh

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asked Jul 19 at 21:00

Ole Petersen

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closed as off-topic by amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel Jul 20 at 3:00

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

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  @amWhy The answer to the question you refer to does not prove uniqueness of decimal expansion for irrational numbers, it just claims so.
  – egreg
  Jul 19 at 21:29
  

  
  
  
  


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  @egreg I told the asker that the answer to the link above is far more thorough than the ones here. Regardless of the answers, the question is what's called a duplicate on MSE. Note, I had already voted to close for a second valid reason.
  – amWhy
  Jul 19 at 21:33
  

  
  
  
  


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  I disagree about the duplicate: the other question is about a quite different (albeit related) problem.
  – egreg
  Jul 19 at 21:52
  

  
  
  
  


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  Also note that $pi$ is not only irrational; it is also transcendental.
  – amWhy
  Jul 19 at 22:13
  

  
  
  
  

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How do I prove that the decimal representation of the digits of $pi$ is unique?

For example, the number $1$ is not unique in its decimal representations since

$1.0000... = 0.999...$

irrational-numbers pi decimal-expansion

share|cite|improve this question

edited Jul 19 at 22:05

RayDansh

884214

asked Jul 19 at 21:00

Ole Petersen

1637

closed as off-topic by amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel Jul 20 at 3:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @amWhy The answer to the question you refer to does not prove uniqueness of decimal expansion for irrational numbers, it just claims so.
  – egreg
  Jul 19 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg I told the asker that the answer to the link above is far more thorough than the ones here. Regardless of the answers, the question is what's called a duplicate on MSE. Note, I had already voted to close for a second valid reason.
  – amWhy
  Jul 19 at 21:33
  

  
  
  
  


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  I disagree about the duplicate: the other question is about a quite different (albeit related) problem.
  – egreg
  Jul 19 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also note that $pi$ is not only irrational; it is also transcendental.
  – amWhy
  Jul 19 at 22:13
  

  
  
  
  

add a comment | 

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How do I prove that the decimal representation of the digits of $pi$ is unique?

For example, the number $1$ is not unique in its decimal representations since

$1.0000... = 0.999...$

irrational-numbers pi decimal-expansion

share|cite|improve this question

edited Jul 19 at 22:05

RayDansh

884214

asked Jul 19 at 21:00

Ole Petersen

1637

How do I prove that the decimal representation of the digits of $pi$ is unique?

For example, the number $1$ is not unique in its decimal representations since

$1.0000... = 0.999...$

irrational-numbers pi decimal-expansion

share|cite|improve this question

edited Jul 19 at 22:05

RayDansh

884214

asked Jul 19 at 21:00

Ole Petersen

1637

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 22:05

RayDansh

884214

edited Jul 19 at 22:05

RayDansh

884214

edited Jul 19 at 22:05

RayDansh

884214

884214

asked Jul 19 at 21:00

Ole Petersen

1637

asked Jul 19 at 21:00

Ole Petersen

1637

asked Jul 19 at 21:00

Ole Petersen

1637

1637

closed as off-topic by amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel Jul 20 at 3:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel Jul 20 at 3:00

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Rob Arthan, Xander Henderson, Isaac Browne, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @amWhy The answer to the question you refer to does not prove uniqueness of decimal expansion for irrational numbers, it just claims so.
  – egreg
  Jul 19 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg I told the asker that the answer to the link above is far more thorough than the ones here. Regardless of the answers, the question is what's called a duplicate on MSE. Note, I had already voted to close for a second valid reason.
  – amWhy
  Jul 19 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I disagree about the duplicate: the other question is about a quite different (albeit related) problem.
  – egreg
  Jul 19 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also note that $pi$ is not only irrational; it is also transcendental.
  – amWhy
  Jul 19 at 22:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @amWhy The answer to the question you refer to does not prove uniqueness of decimal expansion for irrational numbers, it just claims so.
  – egreg
  Jul 19 at 21:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @egreg I told the asker that the answer to the link above is far more thorough than the ones here. Regardless of the answers, the question is what's called a duplicate on MSE. Note, I had already voted to close for a second valid reason.
  – amWhy
  Jul 19 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I disagree about the duplicate: the other question is about a quite different (albeit related) problem.
  – egreg
  Jul 19 at 21:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also note that $pi$ is not only irrational; it is also transcendental.
  – amWhy
  Jul 19 at 22:13
  

  
  
  
  

1

1

@amWhy The answer to the question you refer to does not prove uniqueness of decimal expansion for irrational numbers, it just claims so.
– egreg
Jul 19 at 21:29

@amWhy The answer to the question you refer to does not prove uniqueness of decimal expansion for irrational numbers, it just claims so.
– egreg
Jul 19 at 21:29

@egreg I told the asker that the answer to the link above is far more thorough than the ones here. Regardless of the answers, the question is what's called a duplicate on MSE. Note, I had already voted to close for a second valid reason.
– amWhy
Jul 19 at 21:33

@egreg I told the asker that the answer to the link above is far more thorough than the ones here. Regardless of the answers, the question is what's called a duplicate on MSE. Note, I had already voted to close for a second valid reason.
– amWhy
Jul 19 at 21:33

I disagree about the duplicate: the other question is about a quite different (albeit related) problem.
– egreg
Jul 19 at 21:52

I disagree about the duplicate: the other question is about a quite different (albeit related) problem.
– egreg
Jul 19 at 21:52

Also note that $pi$ is not only irrational; it is also transcendental.
– amWhy
Jul 19 at 22:13

Also note that $pi$ is not only irrational; it is also transcendental.
– amWhy
Jul 19 at 22:13

add a comment | 

4 Answers
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Take any positive number $ain [0,1]$ with decimal representation $$a =sum_n=1^infty fraca_n10^n,$$ where $a_n in 0,1,...,9$. Suppose that $$a = sum_n=1^infty fracb_n10^n$$ for some $b_nin 0,1,...,9$. Then $$sum_n=1^infty fraca_n - b_n10^n = 0.$$ Observe that $|a_n - b_n| leq 9$, so that for any $NinmathbbN$ we have $$left|sum_n=N^infty fraca_n - b_n10^nright| leq sum_n=N^infty frac910^n = frac110^N-1$$ with equality if and only if $a_n - b_n = 9$ for each $ninmathbbN$ or $a_n - b_n = -9$ for each $ninmathbbN$. Now suppose that $(a_n)neq (b_n)$. Let $N$ be the first location where the sequences differ, so that $$sum_n=1^infty fraca_n - b_n10^n = sum_n=N^infty fraca_n - b_n10^n = 0.$$ Then we have $$fracb_N - a_N10^N = sum_N+1^infty fraca_n - b_n10^n.$$ But $$left|fracb_N - a_N10^Nright| geq frac110^N geq left|sum_n=N+1^infty fraca_n - b_n10^nright|.$$ But we know that these inequalities are really equalities, so WLOG we have $a_n - b_n = 9$ for each $n > N$. Given the constraint $a_n,b_nin 0,1,...,9$, this is only possible if $a_n = 9$ and $b_n = 0$. Hence a number with a nonunique decimal representation has exactly two representations, one ending in all 0s and one ending in all 9s. To answer your question, then, the decimal representation of $pi$ does not end in all 9s or all 0s and is therefore unique.

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edited Jul 19 at 22:55

answered Jul 19 at 22:30

mheldman

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=== answer 0====

Consider $3 < pi < 4$. So let $pi_0= 3$.

Consider $3.1 < pi < 3.2$ So let $b_1 = 1$ and $pi_1 = 3.1$. Note $pi - pi_1 < frac 110$

Consider $3.14 < pi < 3.15$ so let $b_2 = 4$ and $pi_2 = 3.14$.Note $pi - pi_1 < frac 1100$

Do this forever. Each step being:

Consider $pi_k-1 + frac d10^k < pi < pi_k-1 + frac d+110^k$ so let $b_k = d$ and $pi_k = pi_k-1 + frac d10^k$. Note $pi - pi_k < frac 110^k$.

As $pi$ is irrational we will never have a case where $pi_k-1 + frac d10^k = pi$ or $pi = pi_k-1 + frac d+110^k$ so we will never have a choice for any other choice of $b_k$. And so the result will be the only possible decimal expansion.

.......

The only numbers that are not unique are those that terminate with an infinite tail of $0$s and which can be rewritten by replacing the last non-zero term with one less, and then finishing with an infinite tail of $9$s.

For example: a number such as $37.345 = 37.34500000000...... = 37.344999999999......$

Let's think why.

===== answer 1 ======

$1.000..... = 0.99999.....$ is an aberation rather than the norm.

To create (rather than interpret) a decimal for a number $n= d.b_1b_2b_3.....$ you start by noting $d+ frac b_110 le n < d + frac b_1 + 110^k$ and letting $n_1 = cd+ frac b_110$.

Then you do an infinite number of comparisons. You note $n_k + fracb_k10^k le n < fracb_k + 110^k$.

And you do that for ever and you get all the $b_i$.

Note: There is only one possible value (because there is only one possible $b_k$ so that $n_k + fracb_k10^k le n < fracb_k + 110^k$) you can create and you can not create anything that that ends with an infinite number of $9$s. (For example if $n= .1$ you stare with $frac 110 le n < frac 210$ so you must say $b_1 = 1$. You can't say. Well,... $frac 010 < n le frac 110$ so I'll take $b_1=0$ and borrow forever...)

On the other hand we can interpret a decimal with a trailing number of $9$s as $0.999999..... = lim_nto infty frac 910 + frac 910^2 + .... + frac 910^n = 1$.

Now if we make a slight change to how we interpreted how to create decimals and did:

If $n_k + frac b_k10^k < n < n_k + frac b_k+110^k$ we must choose $b_k$.

But if $n_k + frac c-110^k < n = n_k + frac c10^k < n_k + frac c+110^k$ we can give ourselves a choice to let $b_k = c-1$ if we want.

Then we'd find that $n_k+1 = n_k + frac c-110^k = n - frac 110^k$ and we find $n_k+1 + frac 910^k+1 < n = n_k+1 + frac 1010^k+1$. Since we can't take $b_k+1 = 10$ we must take $b_k+1 = 9$ and we will get $9$s forever.

If that is how we make decimals then the only way we can every have two decimals for one number is if one representation terminates and the other ends with an infinite number of $9$s.

===== answer 2 =======

A decimal number, for now let's assume the number is between $0.000000.....$ and $0.9999999...... $ inclusively, is a sum of $frac b_110 + fracb_2100 + ...... + fracb_k10^k + .......$ where each $b_k$ is a digit between $0$ and $9$.

Consider the $k$th term, $b_k$ which represents $..... + fracb_k10^k+....$. And consider this is the first term we want to modify.

If we modify by increasing it by $pm d$ then we are changing the value or $n$ by $pm d*frac 110^k$.

But If you consider all the terms to the right of $b_k$ you get:

$N= (frac b_k+110^k+1 + ....... )$

And: $0 = (frac 010^k+1+ ....) le N le (frac 910^k+1+ ....) = frac 110^k*(0.9999.....) = frac 110^k$.

So if we modify $b_m$ by more than $1$, then whatever changes we make to the terms to the right, they will not be able to compensate for it.

So the absolute most we can modify $b_m$ by is $pm 1$ and we can only do that if either $N=0$ and we decrease $b_m$ by $1$ and increase $N$ to $frac 110^k$. Or if $N=frac 110^k$ and we increase $b_m$ by $1$ and decrease $N$ to $0$.

So the only numbers with multiple decimal places are those of the form:

$0.b_1b_2b_3..... b_k-1c999999999..... = 0.b_1b_2b_3....... b_k-1(c+1)000000000......$.

==== answer 3======

First: $0.999999..... = 1$

Second: If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 9$ then $k < 1$.

That should be obvious. $sum_i = 1^j-1b_i*10^-i le 1 - 10^-j+1$ And $sum_i=j+1^-infty b_i10^-i le 10^-j$ and $b_j < 9$ so $k = sum_i=1^-inftyb_i *10^-i < 1-10^-j+1 + 10^-j +9*10^-j = 1-10^-j+1 + 10*10^-j = 1$

Third if If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 0$ then $k > 0$.

And now:

Suppose a number has two valid decimal representations.

That is $n = sum_i=k^infty b_i*10^-i = sum_i=l^infty c_i*10^-i$ but the sequence of $b_i$ are not the same as the sequence of $c_i$.

Let $j$ be the index of the terms where they first differ. so $b_j ne c_j$ but $b_i = c_i$ for all $i < j$.

Then $sum_i = j^inftyb_i*10^-i = sum_i=j^infty c_i*10^-i$ or if we multiply both side by $10^j$.

$b_j + sum_k=1 b_j+k*10^-k = c_j + sum_k=1c_j+k*10^-k$.

Wolog lets assume $c_j < b_j$ so

$b_j - c_j = sum_k=1c_j+k*10^-k - sum_k=1 b_j+k*10^-k$.

No $b_j - c_j$ is an positive integer and $b_j-c_j ge 1$.

Note $0 le b_j+k;c_j+kle 9$ so $0 = .00000000 le sum_k=1 b_j+k*10^-k le .999999...... = 1$ and likewise $0 le sum_k=1 c_j+k*10^-k le 1$

So $b_j - c_j = sum_k=1 c_j+k*10^-k - sum_k=1 b_j+k*10^-k le 1 - 0$.

So $b_j - c_j = 1$. and that means $sum_k=1 c_j+k*10^-k = 1$ and $sum_k=1 b_j+k*10^-k = 0$.

So $c_j+k$ are all $9$s and $b_j+k$ are all $0$ and $b_j = c_j + 1$ and for all $i < j$ $b_j = c_j$.

Those are the only numbers with two decimal representations. In other words. Only terminating decimals, that can be represented with infinite trailing $0$ or infinite trailing $9$s have ore than one representation.

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edited Jul 20 at 1:22

answered Jul 19 at 22:39

fleablood

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Because $pi$ is irrational and the only real numbers greater than $0$ which have $2$ distinct decimal representations are those that can be written as $frac a10^b$, with $ainmathbb N$ and $binmathbbZ^+$. In particular, they're all rational.

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answered Jul 19 at 21:02

José Carlos Santos

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  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  

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Double decimal representation happens only with rational numbers ending with $xxx9999999.....$ such as $$ 25=24.9999999...$$

As you know $pi$ is irrational and the decimal representation of irrational numbers are unique.

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answered Jul 19 at 21:09

Mohammad Riazi-Kermani

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  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  


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  No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
  – fleablood
  Jul 20 at 1:09
  

  
  
  
  

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4 Answers
4

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4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

Take any positive number $ain [0,1]$ with decimal representation $$a =sum_n=1^infty fraca_n10^n,$$ where $a_n in 0,1,...,9$. Suppose that $$a = sum_n=1^infty fracb_n10^n$$ for some $b_nin 0,1,...,9$. Then $$sum_n=1^infty fraca_n - b_n10^n = 0.$$ Observe that $|a_n - b_n| leq 9$, so that for any $NinmathbbN$ we have $$left|sum_n=N^infty fraca_n - b_n10^nright| leq sum_n=N^infty frac910^n = frac110^N-1$$ with equality if and only if $a_n - b_n = 9$ for each $ninmathbbN$ or $a_n - b_n = -9$ for each $ninmathbbN$. Now suppose that $(a_n)neq (b_n)$. Let $N$ be the first location where the sequences differ, so that $$sum_n=1^infty fraca_n - b_n10^n = sum_n=N^infty fraca_n - b_n10^n = 0.$$ Then we have $$fracb_N - a_N10^N = sum_N+1^infty fraca_n - b_n10^n.$$ But $$left|fracb_N - a_N10^Nright| geq frac110^N geq left|sum_n=N+1^infty fraca_n - b_n10^nright|.$$ But we know that these inequalities are really equalities, so WLOG we have $a_n - b_n = 9$ for each $n > N$. Given the constraint $a_n,b_nin 0,1,...,9$, this is only possible if $a_n = 9$ and $b_n = 0$. Hence a number with a nonunique decimal representation has exactly two representations, one ending in all 0s and one ending in all 9s. To answer your question, then, the decimal representation of $pi$ does not end in all 9s or all 0s and is therefore unique.

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edited Jul 19 at 22:55

answered Jul 19 at 22:30

mheldman

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Take any positive number $ain [0,1]$ with decimal representation $$a =sum_n=1^infty fraca_n10^n,$$ where $a_n in 0,1,...,9$. Suppose that $$a = sum_n=1^infty fracb_n10^n$$ for some $b_nin 0,1,...,9$. Then $$sum_n=1^infty fraca_n - b_n10^n = 0.$$ Observe that $|a_n - b_n| leq 9$, so that for any $NinmathbbN$ we have $$left|sum_n=N^infty fraca_n - b_n10^nright| leq sum_n=N^infty frac910^n = frac110^N-1$$ with equality if and only if $a_n - b_n = 9$ for each $ninmathbbN$ or $a_n - b_n = -9$ for each $ninmathbbN$. Now suppose that $(a_n)neq (b_n)$. Let $N$ be the first location where the sequences differ, so that $$sum_n=1^infty fraca_n - b_n10^n = sum_n=N^infty fraca_n - b_n10^n = 0.$$ Then we have $$fracb_N - a_N10^N = sum_N+1^infty fraca_n - b_n10^n.$$ But $$left|fracb_N - a_N10^Nright| geq frac110^N geq left|sum_n=N+1^infty fraca_n - b_n10^nright|.$$ But we know that these inequalities are really equalities, so WLOG we have $a_n - b_n = 9$ for each $n > N$. Given the constraint $a_n,b_nin 0,1,...,9$, this is only possible if $a_n = 9$ and $b_n = 0$. Hence a number with a nonunique decimal representation has exactly two representations, one ending in all 0s and one ending in all 9s. To answer your question, then, the decimal representation of $pi$ does not end in all 9s or all 0s and is therefore unique.

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edited Jul 19 at 22:55

answered Jul 19 at 22:30

mheldman

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up vote
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accepted

up vote
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Take any positive number $ain [0,1]$ with decimal representation $$a =sum_n=1^infty fraca_n10^n,$$ where $a_n in 0,1,...,9$. Suppose that $$a = sum_n=1^infty fracb_n10^n$$ for some $b_nin 0,1,...,9$. Then $$sum_n=1^infty fraca_n - b_n10^n = 0.$$ Observe that $|a_n - b_n| leq 9$, so that for any $NinmathbbN$ we have $$left|sum_n=N^infty fraca_n - b_n10^nright| leq sum_n=N^infty frac910^n = frac110^N-1$$ with equality if and only if $a_n - b_n = 9$ for each $ninmathbbN$ or $a_n - b_n = -9$ for each $ninmathbbN$. Now suppose that $(a_n)neq (b_n)$. Let $N$ be the first location where the sequences differ, so that $$sum_n=1^infty fraca_n - b_n10^n = sum_n=N^infty fraca_n - b_n10^n = 0.$$ Then we have $$fracb_N - a_N10^N = sum_N+1^infty fraca_n - b_n10^n.$$ But $$left|fracb_N - a_N10^Nright| geq frac110^N geq left|sum_n=N+1^infty fraca_n - b_n10^nright|.$$ But we know that these inequalities are really equalities, so WLOG we have $a_n - b_n = 9$ for each $n > N$. Given the constraint $a_n,b_nin 0,1,...,9$, this is only possible if $a_n = 9$ and $b_n = 0$. Hence a number with a nonunique decimal representation has exactly two representations, one ending in all 0s and one ending in all 9s. To answer your question, then, the decimal representation of $pi$ does not end in all 9s or all 0s and is therefore unique.

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edited Jul 19 at 22:55

answered Jul 19 at 22:30

mheldman

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Take any positive number $ain [0,1]$ with decimal representation $$a =sum_n=1^infty fraca_n10^n,$$ where $a_n in 0,1,...,9$. Suppose that $$a = sum_n=1^infty fracb_n10^n$$ for some $b_nin 0,1,...,9$. Then $$sum_n=1^infty fraca_n - b_n10^n = 0.$$ Observe that $|a_n - b_n| leq 9$, so that for any $NinmathbbN$ we have $$left|sum_n=N^infty fraca_n - b_n10^nright| leq sum_n=N^infty frac910^n = frac110^N-1$$ with equality if and only if $a_n - b_n = 9$ for each $ninmathbbN$ or $a_n - b_n = -9$ for each $ninmathbbN$. Now suppose that $(a_n)neq (b_n)$. Let $N$ be the first location where the sequences differ, so that $$sum_n=1^infty fraca_n - b_n10^n = sum_n=N^infty fraca_n - b_n10^n = 0.$$ Then we have $$fracb_N - a_N10^N = sum_N+1^infty fraca_n - b_n10^n.$$ But $$left|fracb_N - a_N10^Nright| geq frac110^N geq left|sum_n=N+1^infty fraca_n - b_n10^nright|.$$ But we know that these inequalities are really equalities, so WLOG we have $a_n - b_n = 9$ for each $n > N$. Given the constraint $a_n,b_nin 0,1,...,9$, this is only possible if $a_n = 9$ and $b_n = 0$. Hence a number with a nonunique decimal representation has exactly two representations, one ending in all 0s and one ending in all 9s. To answer your question, then, the decimal representation of $pi$ does not end in all 9s or all 0s and is therefore unique.

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edited Jul 19 at 22:55

answered Jul 19 at 22:30

mheldman

54616

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 22:55

edited Jul 19 at 22:55

edited Jul 19 at 22:55

answered Jul 19 at 22:30

mheldman

54616

answered Jul 19 at 22:30

mheldman

54616

answered Jul 19 at 22:30

mheldman

54616

54616

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=== answer 0====

Consider $3 < pi < 4$. So let $pi_0= 3$.

Consider $3.1 < pi < 3.2$ So let $b_1 = 1$ and $pi_1 = 3.1$. Note $pi - pi_1 < frac 110$

Consider $3.14 < pi < 3.15$ so let $b_2 = 4$ and $pi_2 = 3.14$.Note $pi - pi_1 < frac 1100$

Do this forever. Each step being:

Consider $pi_k-1 + frac d10^k < pi < pi_k-1 + frac d+110^k$ so let $b_k = d$ and $pi_k = pi_k-1 + frac d10^k$. Note $pi - pi_k < frac 110^k$.

As $pi$ is irrational we will never have a case where $pi_k-1 + frac d10^k = pi$ or $pi = pi_k-1 + frac d+110^k$ so we will never have a choice for any other choice of $b_k$. And so the result will be the only possible decimal expansion.

.......

The only numbers that are not unique are those that terminate with an infinite tail of $0$s and which can be rewritten by replacing the last non-zero term with one less, and then finishing with an infinite tail of $9$s.

For example: a number such as $37.345 = 37.34500000000...... = 37.344999999999......$

Let's think why.

===== answer 1 ======

$1.000..... = 0.99999.....$ is an aberation rather than the norm.

To create (rather than interpret) a decimal for a number $n= d.b_1b_2b_3.....$ you start by noting $d+ frac b_110 le n < d + frac b_1 + 110^k$ and letting $n_1 = cd+ frac b_110$.

Then you do an infinite number of comparisons. You note $n_k + fracb_k10^k le n < fracb_k + 110^k$.

And you do that for ever and you get all the $b_i$.

Note: There is only one possible value (because there is only one possible $b_k$ so that $n_k + fracb_k10^k le n < fracb_k + 110^k$) you can create and you can not create anything that that ends with an infinite number of $9$s. (For example if $n= .1$ you stare with $frac 110 le n < frac 210$ so you must say $b_1 = 1$. You can't say. Well,... $frac 010 < n le frac 110$ so I'll take $b_1=0$ and borrow forever...)

On the other hand we can interpret a decimal with a trailing number of $9$s as $0.999999..... = lim_nto infty frac 910 + frac 910^2 + .... + frac 910^n = 1$.

Now if we make a slight change to how we interpreted how to create decimals and did:

If $n_k + frac b_k10^k < n < n_k + frac b_k+110^k$ we must choose $b_k$.

But if $n_k + frac c-110^k < n = n_k + frac c10^k < n_k + frac c+110^k$ we can give ourselves a choice to let $b_k = c-1$ if we want.

Then we'd find that $n_k+1 = n_k + frac c-110^k = n - frac 110^k$ and we find $n_k+1 + frac 910^k+1 < n = n_k+1 + frac 1010^k+1$. Since we can't take $b_k+1 = 10$ we must take $b_k+1 = 9$ and we will get $9$s forever.

If that is how we make decimals then the only way we can every have two decimals for one number is if one representation terminates and the other ends with an infinite number of $9$s.

===== answer 2 =======

A decimal number, for now let's assume the number is between $0.000000.....$ and $0.9999999...... $ inclusively, is a sum of $frac b_110 + fracb_2100 + ...... + fracb_k10^k + .......$ where each $b_k$ is a digit between $0$ and $9$.

Consider the $k$th term, $b_k$ which represents $..... + fracb_k10^k+....$. And consider this is the first term we want to modify.

If we modify by increasing it by $pm d$ then we are changing the value or $n$ by $pm d*frac 110^k$.

But If you consider all the terms to the right of $b_k$ you get:

$N= (frac b_k+110^k+1 + ....... )$

And: $0 = (frac 010^k+1+ ....) le N le (frac 910^k+1+ ....) = frac 110^k*(0.9999.....) = frac 110^k$.

So if we modify $b_m$ by more than $1$, then whatever changes we make to the terms to the right, they will not be able to compensate for it.

So the absolute most we can modify $b_m$ by is $pm 1$ and we can only do that if either $N=0$ and we decrease $b_m$ by $1$ and increase $N$ to $frac 110^k$. Or if $N=frac 110^k$ and we increase $b_m$ by $1$ and decrease $N$ to $0$.

So the only numbers with multiple decimal places are those of the form:

$0.b_1b_2b_3..... b_k-1c999999999..... = 0.b_1b_2b_3....... b_k-1(c+1)000000000......$.

==== answer 3======

First: $0.999999..... = 1$

Second: If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 9$ then $k < 1$.

That should be obvious. $sum_i = 1^j-1b_i*10^-i le 1 - 10^-j+1$ And $sum_i=j+1^-infty b_i10^-i le 10^-j$ and $b_j < 9$ so $k = sum_i=1^-inftyb_i *10^-i < 1-10^-j+1 + 10^-j +9*10^-j = 1-10^-j+1 + 10*10^-j = 1$

Third if If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 0$ then $k > 0$.

And now:

Suppose a number has two valid decimal representations.

That is $n = sum_i=k^infty b_i*10^-i = sum_i=l^infty c_i*10^-i$ but the sequence of $b_i$ are not the same as the sequence of $c_i$.

Let $j$ be the index of the terms where they first differ. so $b_j ne c_j$ but $b_i = c_i$ for all $i < j$.

Then $sum_i = j^inftyb_i*10^-i = sum_i=j^infty c_i*10^-i$ or if we multiply both side by $10^j$.

$b_j + sum_k=1 b_j+k*10^-k = c_j + sum_k=1c_j+k*10^-k$.

Wolog lets assume $c_j < b_j$ so

$b_j - c_j = sum_k=1c_j+k*10^-k - sum_k=1 b_j+k*10^-k$.

No $b_j - c_j$ is an positive integer and $b_j-c_j ge 1$.

Note $0 le b_j+k;c_j+kle 9$ so $0 = .00000000 le sum_k=1 b_j+k*10^-k le .999999...... = 1$ and likewise $0 le sum_k=1 c_j+k*10^-k le 1$

So $b_j - c_j = sum_k=1 c_j+k*10^-k - sum_k=1 b_j+k*10^-k le 1 - 0$.

So $b_j - c_j = 1$. and that means $sum_k=1 c_j+k*10^-k = 1$ and $sum_k=1 b_j+k*10^-k = 0$.

So $c_j+k$ are all $9$s and $b_j+k$ are all $0$ and $b_j = c_j + 1$ and for all $i < j$ $b_j = c_j$.

Those are the only numbers with two decimal representations. In other words. Only terminating decimals, that can be represented with infinite trailing $0$ or infinite trailing $9$s have ore than one representation.

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edited Jul 20 at 1:22

answered Jul 19 at 22:39

fleablood

60.4k22575

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=== answer 0====

Consider $3 < pi < 4$. So let $pi_0= 3$.

Consider $3.1 < pi < 3.2$ So let $b_1 = 1$ and $pi_1 = 3.1$. Note $pi - pi_1 < frac 110$

Consider $3.14 < pi < 3.15$ so let $b_2 = 4$ and $pi_2 = 3.14$.Note $pi - pi_1 < frac 1100$

Do this forever. Each step being:

Consider $pi_k-1 + frac d10^k < pi < pi_k-1 + frac d+110^k$ so let $b_k = d$ and $pi_k = pi_k-1 + frac d10^k$. Note $pi - pi_k < frac 110^k$.

As $pi$ is irrational we will never have a case where $pi_k-1 + frac d10^k = pi$ or $pi = pi_k-1 + frac d+110^k$ so we will never have a choice for any other choice of $b_k$. And so the result will be the only possible decimal expansion.

.......

The only numbers that are not unique are those that terminate with an infinite tail of $0$s and which can be rewritten by replacing the last non-zero term with one less, and then finishing with an infinite tail of $9$s.

For example: a number such as $37.345 = 37.34500000000...... = 37.344999999999......$

Let's think why.

===== answer 1 ======

$1.000..... = 0.99999.....$ is an aberation rather than the norm.

To create (rather than interpret) a decimal for a number $n= d.b_1b_2b_3.....$ you start by noting $d+ frac b_110 le n < d + frac b_1 + 110^k$ and letting $n_1 = cd+ frac b_110$.

Then you do an infinite number of comparisons. You note $n_k + fracb_k10^k le n < fracb_k + 110^k$.

And you do that for ever and you get all the $b_i$.

Note: There is only one possible value (because there is only one possible $b_k$ so that $n_k + fracb_k10^k le n < fracb_k + 110^k$) you can create and you can not create anything that that ends with an infinite number of $9$s. (For example if $n= .1$ you stare with $frac 110 le n < frac 210$ so you must say $b_1 = 1$. You can't say. Well,... $frac 010 < n le frac 110$ so I'll take $b_1=0$ and borrow forever...)

On the other hand we can interpret a decimal with a trailing number of $9$s as $0.999999..... = lim_nto infty frac 910 + frac 910^2 + .... + frac 910^n = 1$.

Now if we make a slight change to how we interpreted how to create decimals and did:

If $n_k + frac b_k10^k < n < n_k + frac b_k+110^k$ we must choose $b_k$.

But if $n_k + frac c-110^k < n = n_k + frac c10^k < n_k + frac c+110^k$ we can give ourselves a choice to let $b_k = c-1$ if we want.

Then we'd find that $n_k+1 = n_k + frac c-110^k = n - frac 110^k$ and we find $n_k+1 + frac 910^k+1 < n = n_k+1 + frac 1010^k+1$. Since we can't take $b_k+1 = 10$ we must take $b_k+1 = 9$ and we will get $9$s forever.

If that is how we make decimals then the only way we can every have two decimals for one number is if one representation terminates and the other ends with an infinite number of $9$s.

===== answer 2 =======

A decimal number, for now let's assume the number is between $0.000000.....$ and $0.9999999...... $ inclusively, is a sum of $frac b_110 + fracb_2100 + ...... + fracb_k10^k + .......$ where each $b_k$ is a digit between $0$ and $9$.

Consider the $k$th term, $b_k$ which represents $..... + fracb_k10^k+....$. And consider this is the first term we want to modify.

If we modify by increasing it by $pm d$ then we are changing the value or $n$ by $pm d*frac 110^k$.

But If you consider all the terms to the right of $b_k$ you get:

$N= (frac b_k+110^k+1 + ....... )$

And: $0 = (frac 010^k+1+ ....) le N le (frac 910^k+1+ ....) = frac 110^k*(0.9999.....) = frac 110^k$.

So if we modify $b_m$ by more than $1$, then whatever changes we make to the terms to the right, they will not be able to compensate for it.

So the absolute most we can modify $b_m$ by is $pm 1$ and we can only do that if either $N=0$ and we decrease $b_m$ by $1$ and increase $N$ to $frac 110^k$. Or if $N=frac 110^k$ and we increase $b_m$ by $1$ and decrease $N$ to $0$.

So the only numbers with multiple decimal places are those of the form:

$0.b_1b_2b_3..... b_k-1c999999999..... = 0.b_1b_2b_3....... b_k-1(c+1)000000000......$.

==== answer 3======

First: $0.999999..... = 1$

Second: If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 9$ then $k < 1$.

That should be obvious. $sum_i = 1^j-1b_i*10^-i le 1 - 10^-j+1$ And $sum_i=j+1^-infty b_i10^-i le 10^-j$ and $b_j < 9$ so $k = sum_i=1^-inftyb_i *10^-i < 1-10^-j+1 + 10^-j +9*10^-j = 1-10^-j+1 + 10*10^-j = 1$

Third if If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 0$ then $k > 0$.

And now:

Suppose a number has two valid decimal representations.

That is $n = sum_i=k^infty b_i*10^-i = sum_i=l^infty c_i*10^-i$ but the sequence of $b_i$ are not the same as the sequence of $c_i$.

Let $j$ be the index of the terms where they first differ. so $b_j ne c_j$ but $b_i = c_i$ for all $i < j$.

Then $sum_i = j^inftyb_i*10^-i = sum_i=j^infty c_i*10^-i$ or if we multiply both side by $10^j$.

$b_j + sum_k=1 b_j+k*10^-k = c_j + sum_k=1c_j+k*10^-k$.

Wolog lets assume $c_j < b_j$ so

$b_j - c_j = sum_k=1c_j+k*10^-k - sum_k=1 b_j+k*10^-k$.

No $b_j - c_j$ is an positive integer and $b_j-c_j ge 1$.

Note $0 le b_j+k;c_j+kle 9$ so $0 = .00000000 le sum_k=1 b_j+k*10^-k le .999999...... = 1$ and likewise $0 le sum_k=1 c_j+k*10^-k le 1$

So $b_j - c_j = sum_k=1 c_j+k*10^-k - sum_k=1 b_j+k*10^-k le 1 - 0$.

So $b_j - c_j = 1$. and that means $sum_k=1 c_j+k*10^-k = 1$ and $sum_k=1 b_j+k*10^-k = 0$.

So $c_j+k$ are all $9$s and $b_j+k$ are all $0$ and $b_j = c_j + 1$ and for all $i < j$ $b_j = c_j$.

Those are the only numbers with two decimal representations. In other words. Only terminating decimals, that can be represented with infinite trailing $0$ or infinite trailing $9$s have ore than one representation.

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edited Jul 20 at 1:22

answered Jul 19 at 22:39

fleablood

60.4k22575

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up vote
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=== answer 0====

Consider $3 < pi < 4$. So let $pi_0= 3$.

Consider $3.1 < pi < 3.2$ So let $b_1 = 1$ and $pi_1 = 3.1$. Note $pi - pi_1 < frac 110$

Consider $3.14 < pi < 3.15$ so let $b_2 = 4$ and $pi_2 = 3.14$.Note $pi - pi_1 < frac 1100$

Do this forever. Each step being:

Consider $pi_k-1 + frac d10^k < pi < pi_k-1 + frac d+110^k$ so let $b_k = d$ and $pi_k = pi_k-1 + frac d10^k$. Note $pi - pi_k < frac 110^k$.

As $pi$ is irrational we will never have a case where $pi_k-1 + frac d10^k = pi$ or $pi = pi_k-1 + frac d+110^k$ so we will never have a choice for any other choice of $b_k$. And so the result will be the only possible decimal expansion.

.......

The only numbers that are not unique are those that terminate with an infinite tail of $0$s and which can be rewritten by replacing the last non-zero term with one less, and then finishing with an infinite tail of $9$s.

For example: a number such as $37.345 = 37.34500000000...... = 37.344999999999......$

Let's think why.

===== answer 1 ======

$1.000..... = 0.99999.....$ is an aberation rather than the norm.

To create (rather than interpret) a decimal for a number $n= d.b_1b_2b_3.....$ you start by noting $d+ frac b_110 le n < d + frac b_1 + 110^k$ and letting $n_1 = cd+ frac b_110$.

Then you do an infinite number of comparisons. You note $n_k + fracb_k10^k le n < fracb_k + 110^k$.

And you do that for ever and you get all the $b_i$.

Note: There is only one possible value (because there is only one possible $b_k$ so that $n_k + fracb_k10^k le n < fracb_k + 110^k$) you can create and you can not create anything that that ends with an infinite number of $9$s. (For example if $n= .1$ you stare with $frac 110 le n < frac 210$ so you must say $b_1 = 1$. You can't say. Well,... $frac 010 < n le frac 110$ so I'll take $b_1=0$ and borrow forever...)

On the other hand we can interpret a decimal with a trailing number of $9$s as $0.999999..... = lim_nto infty frac 910 + frac 910^2 + .... + frac 910^n = 1$.

Now if we make a slight change to how we interpreted how to create decimals and did:

If $n_k + frac b_k10^k < n < n_k + frac b_k+110^k$ we must choose $b_k$.

But if $n_k + frac c-110^k < n = n_k + frac c10^k < n_k + frac c+110^k$ we can give ourselves a choice to let $b_k = c-1$ if we want.

Then we'd find that $n_k+1 = n_k + frac c-110^k = n - frac 110^k$ and we find $n_k+1 + frac 910^k+1 < n = n_k+1 + frac 1010^k+1$. Since we can't take $b_k+1 = 10$ we must take $b_k+1 = 9$ and we will get $9$s forever.

If that is how we make decimals then the only way we can every have two decimals for one number is if one representation terminates and the other ends with an infinite number of $9$s.

===== answer 2 =======

A decimal number, for now let's assume the number is between $0.000000.....$ and $0.9999999...... $ inclusively, is a sum of $frac b_110 + fracb_2100 + ...... + fracb_k10^k + .......$ where each $b_k$ is a digit between $0$ and $9$.

Consider the $k$th term, $b_k$ which represents $..... + fracb_k10^k+....$. And consider this is the first term we want to modify.

If we modify by increasing it by $pm d$ then we are changing the value or $n$ by $pm d*frac 110^k$.

But If you consider all the terms to the right of $b_k$ you get:

$N= (frac b_k+110^k+1 + ....... )$

And: $0 = (frac 010^k+1+ ....) le N le (frac 910^k+1+ ....) = frac 110^k*(0.9999.....) = frac 110^k$.

So if we modify $b_m$ by more than $1$, then whatever changes we make to the terms to the right, they will not be able to compensate for it.

So the absolute most we can modify $b_m$ by is $pm 1$ and we can only do that if either $N=0$ and we decrease $b_m$ by $1$ and increase $N$ to $frac 110^k$. Or if $N=frac 110^k$ and we increase $b_m$ by $1$ and decrease $N$ to $0$.

So the only numbers with multiple decimal places are those of the form:

$0.b_1b_2b_3..... b_k-1c999999999..... = 0.b_1b_2b_3....... b_k-1(c+1)000000000......$.

==== answer 3======

First: $0.999999..... = 1$

Second: If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 9$ then $k < 1$.

That should be obvious. $sum_i = 1^j-1b_i*10^-i le 1 - 10^-j+1$ And $sum_i=j+1^-infty b_i10^-i le 10^-j$ and $b_j < 9$ so $k = sum_i=1^-inftyb_i *10^-i < 1-10^-j+1 + 10^-j +9*10^-j = 1-10^-j+1 + 10*10^-j = 1$

Third if If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 0$ then $k > 0$.

And now:

Suppose a number has two valid decimal representations.

That is $n = sum_i=k^infty b_i*10^-i = sum_i=l^infty c_i*10^-i$ but the sequence of $b_i$ are not the same as the sequence of $c_i$.

Let $j$ be the index of the terms where they first differ. so $b_j ne c_j$ but $b_i = c_i$ for all $i < j$.

Then $sum_i = j^inftyb_i*10^-i = sum_i=j^infty c_i*10^-i$ or if we multiply both side by $10^j$.

$b_j + sum_k=1 b_j+k*10^-k = c_j + sum_k=1c_j+k*10^-k$.

Wolog lets assume $c_j < b_j$ so

$b_j - c_j = sum_k=1c_j+k*10^-k - sum_k=1 b_j+k*10^-k$.

No $b_j - c_j$ is an positive integer and $b_j-c_j ge 1$.

Note $0 le b_j+k;c_j+kle 9$ so $0 = .00000000 le sum_k=1 b_j+k*10^-k le .999999...... = 1$ and likewise $0 le sum_k=1 c_j+k*10^-k le 1$

So $b_j - c_j = sum_k=1 c_j+k*10^-k - sum_k=1 b_j+k*10^-k le 1 - 0$.

So $b_j - c_j = 1$. and that means $sum_k=1 c_j+k*10^-k = 1$ and $sum_k=1 b_j+k*10^-k = 0$.

So $c_j+k$ are all $9$s and $b_j+k$ are all $0$ and $b_j = c_j + 1$ and for all $i < j$ $b_j = c_j$.

Those are the only numbers with two decimal representations. In other words. Only terminating decimals, that can be represented with infinite trailing $0$ or infinite trailing $9$s have ore than one representation.

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edited Jul 20 at 1:22

answered Jul 19 at 22:39

fleablood

60.4k22575

=== answer 0====

Consider $3 < pi < 4$. So let $pi_0= 3$.

Consider $3.1 < pi < 3.2$ So let $b_1 = 1$ and $pi_1 = 3.1$. Note $pi - pi_1 < frac 110$

Consider $3.14 < pi < 3.15$ so let $b_2 = 4$ and $pi_2 = 3.14$.Note $pi - pi_1 < frac 1100$

Do this forever. Each step being:

Consider $pi_k-1 + frac d10^k < pi < pi_k-1 + frac d+110^k$ so let $b_k = d$ and $pi_k = pi_k-1 + frac d10^k$. Note $pi - pi_k < frac 110^k$.

As $pi$ is irrational we will never have a case where $pi_k-1 + frac d10^k = pi$ or $pi = pi_k-1 + frac d+110^k$ so we will never have a choice for any other choice of $b_k$. And so the result will be the only possible decimal expansion.

.......

The only numbers that are not unique are those that terminate with an infinite tail of $0$s and which can be rewritten by replacing the last non-zero term with one less, and then finishing with an infinite tail of $9$s.

For example: a number such as $37.345 = 37.34500000000...... = 37.344999999999......$

Let's think why.

===== answer 1 ======

$1.000..... = 0.99999.....$ is an aberation rather than the norm.

To create (rather than interpret) a decimal for a number $n= d.b_1b_2b_3.....$ you start by noting $d+ frac b_110 le n < d + frac b_1 + 110^k$ and letting $n_1 = cd+ frac b_110$.

Then you do an infinite number of comparisons. You note $n_k + fracb_k10^k le n < fracb_k + 110^k$.

And you do that for ever and you get all the $b_i$.

Note: There is only one possible value (because there is only one possible $b_k$ so that $n_k + fracb_k10^k le n < fracb_k + 110^k$) you can create and you can not create anything that that ends with an infinite number of $9$s. (For example if $n= .1$ you stare with $frac 110 le n < frac 210$ so you must say $b_1 = 1$. You can't say. Well,... $frac 010 < n le frac 110$ so I'll take $b_1=0$ and borrow forever...)

On the other hand we can interpret a decimal with a trailing number of $9$s as $0.999999..... = lim_nto infty frac 910 + frac 910^2 + .... + frac 910^n = 1$.

Now if we make a slight change to how we interpreted how to create decimals and did:

If $n_k + frac b_k10^k < n < n_k + frac b_k+110^k$ we must choose $b_k$.

But if $n_k + frac c-110^k < n = n_k + frac c10^k < n_k + frac c+110^k$ we can give ourselves a choice to let $b_k = c-1$ if we want.

Then we'd find that $n_k+1 = n_k + frac c-110^k = n - frac 110^k$ and we find $n_k+1 + frac 910^k+1 < n = n_k+1 + frac 1010^k+1$. Since we can't take $b_k+1 = 10$ we must take $b_k+1 = 9$ and we will get $9$s forever.

If that is how we make decimals then the only way we can every have two decimals for one number is if one representation terminates and the other ends with an infinite number of $9$s.

===== answer 2 =======

A decimal number, for now let's assume the number is between $0.000000.....$ and $0.9999999...... $ inclusively, is a sum of $frac b_110 + fracb_2100 + ...... + fracb_k10^k + .......$ where each $b_k$ is a digit between $0$ and $9$.

Consider the $k$th term, $b_k$ which represents $..... + fracb_k10^k+....$. And consider this is the first term we want to modify.

If we modify by increasing it by $pm d$ then we are changing the value or $n$ by $pm d*frac 110^k$.

But If you consider all the terms to the right of $b_k$ you get:

$N= (frac b_k+110^k+1 + ....... )$

And: $0 = (frac 010^k+1+ ....) le N le (frac 910^k+1+ ....) = frac 110^k*(0.9999.....) = frac 110^k$.

So if we modify $b_m$ by more than $1$, then whatever changes we make to the terms to the right, they will not be able to compensate for it.

So the absolute most we can modify $b_m$ by is $pm 1$ and we can only do that if either $N=0$ and we decrease $b_m$ by $1$ and increase $N$ to $frac 110^k$. Or if $N=frac 110^k$ and we increase $b_m$ by $1$ and decrease $N$ to $0$.

So the only numbers with multiple decimal places are those of the form:

$0.b_1b_2b_3..... b_k-1c999999999..... = 0.b_1b_2b_3....... b_k-1(c+1)000000000......$.

==== answer 3======

First: $0.999999..... = 1$

Second: If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 9$ then $k < 1$.

That should be obvious. $sum_i = 1^j-1b_i*10^-i le 1 - 10^-j+1$ And $sum_i=j+1^-infty b_i10^-i le 10^-j$ and $b_j < 9$ so $k = sum_i=1^-inftyb_i *10^-i < 1-10^-j+1 + 10^-j +9*10^-j = 1-10^-j+1 + 10*10^-j = 1$

Third if If $k = sum_i=1^-inftyb_i *10^-i$ and $b_i = 0.... 1$ and if any $b_j ne 0$ then $k > 0$.

And now:

Suppose a number has two valid decimal representations.

That is $n = sum_i=k^infty b_i*10^-i = sum_i=l^infty c_i*10^-i$ but the sequence of $b_i$ are not the same as the sequence of $c_i$.

Let $j$ be the index of the terms where they first differ. so $b_j ne c_j$ but $b_i = c_i$ for all $i < j$.

Then $sum_i = j^inftyb_i*10^-i = sum_i=j^infty c_i*10^-i$ or if we multiply both side by $10^j$.

$b_j + sum_k=1 b_j+k*10^-k = c_j + sum_k=1c_j+k*10^-k$.

Wolog lets assume $c_j < b_j$ so

$b_j - c_j = sum_k=1c_j+k*10^-k - sum_k=1 b_j+k*10^-k$.

No $b_j - c_j$ is an positive integer and $b_j-c_j ge 1$.

Note $0 le b_j+k;c_j+kle 9$ so $0 = .00000000 le sum_k=1 b_j+k*10^-k le .999999...... = 1$ and likewise $0 le sum_k=1 c_j+k*10^-k le 1$

So $b_j - c_j = sum_k=1 c_j+k*10^-k - sum_k=1 b_j+k*10^-k le 1 - 0$.

So $b_j - c_j = 1$. and that means $sum_k=1 c_j+k*10^-k = 1$ and $sum_k=1 b_j+k*10^-k = 0$.

So $c_j+k$ are all $9$s and $b_j+k$ are all $0$ and $b_j = c_j + 1$ and for all $i < j$ $b_j = c_j$.

Those are the only numbers with two decimal representations. In other words. Only terminating decimals, that can be represented with infinite trailing $0$ or infinite trailing $9$s have ore than one representation.

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edited Jul 20 at 1:22

answered Jul 19 at 22:39

fleablood

60.4k22575

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share|cite|improve this answer

edited Jul 20 at 1:22

edited Jul 20 at 1:22

edited Jul 20 at 1:22

answered Jul 19 at 22:39

fleablood

60.4k22575

answered Jul 19 at 22:39

fleablood

60.4k22575

answered Jul 19 at 22:39

fleablood

60.4k22575

60.4k22575

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up vote
0
down vote

Because $pi$ is irrational and the only real numbers greater than $0$ which have $2$ distinct decimal representations are those that can be written as $frac a10^b$, with $ainmathbb N$ and $binmathbbZ^+$. In particular, they're all rational.

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answered Jul 19 at 21:02

José Carlos Santos

114k1698177

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

Because $pi$ is irrational and the only real numbers greater than $0$ which have $2$ distinct decimal representations are those that can be written as $frac a10^b$, with $ainmathbb N$ and $binmathbbZ^+$. In particular, they're all rational.

share|cite|improve this answer

answered Jul 19 at 21:02

José Carlos Santos

114k1698177

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Because $pi$ is irrational and the only real numbers greater than $0$ which have $2$ distinct decimal representations are those that can be written as $frac a10^b$, with $ainmathbb N$ and $binmathbbZ^+$. In particular, they're all rational.

share|cite|improve this answer

answered Jul 19 at 21:02

José Carlos Santos

114k1698177

Because $pi$ is irrational and the only real numbers greater than $0$ which have $2$ distinct decimal representations are those that can be written as $frac a10^b$, with $ainmathbb N$ and $binmathbbZ^+$. In particular, they're all rational.

share|cite|improve this answer

answered Jul 19 at 21:02

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 21:02

José Carlos Santos

114k1698177

answered Jul 19 at 21:02

José Carlos Santos

114k1698177

answered Jul 19 at 21:02

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  

2

2

The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
– amWhy
Jul 19 at 21:13

The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so. You, in particular, should have already searched and found such a proof-request-duplicate, prior to immediately answering the question.
– amWhy
Jul 19 at 21:13

add a comment | 

up vote
0
down vote

Double decimal representation happens only with rational numbers ending with $xxx9999999.....$ such as $$ 25=24.9999999...$$

As you know $pi$ is irrational and the decimal representation of irrational numbers are unique.

share|cite|improve this answer

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
  – fleablood
  Jul 20 at 1:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

Double decimal representation happens only with rational numbers ending with $xxx9999999.....$ such as $$ 25=24.9999999...$$

As you know $pi$ is irrational and the decimal representation of irrational numbers are unique.

share|cite|improve this answer

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
  – fleablood
  Jul 20 at 1:09
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Double decimal representation happens only with rational numbers ending with $xxx9999999.....$ such as $$ 25=24.9999999...$$

As you know $pi$ is irrational and the decimal representation of irrational numbers are unique.

share|cite|improve this answer

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

Double decimal representation happens only with rational numbers ending with $xxx9999999.....$ such as $$ 25=24.9999999...$$

As you know $pi$ is irrational and the decimal representation of irrational numbers are unique.

share|cite|improve this answer

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

answered Jul 19 at 21:09

Mohammad Riazi-Kermani

27.5k41852

27.5k41852

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
  – fleablood
  Jul 20 at 1:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
  – amWhy
  Jul 19 at 21:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
  – fleablood
  Jul 20 at 1:09
  

  
  
  
  

2

2

The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
– amWhy
Jul 19 at 21:13

The problem here, and with the other answer, is that you fail to prove why an irrational number (like $pi$) has a unique decimal representation. You just claim it is so.
– amWhy
Jul 19 at 21:13

No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
– fleablood
Jul 20 at 1:09

No, but it sets up how to think about the problem and how to reframe it in a more general and, hopefully, easy and intuitive statement to prove.
– fleablood
Jul 20 at 1:09

add a comment |