Show that the binary number b=0.11…1 with 2003 1’s satisfies 0.99⋅⋅⋅9<b<0.99…9

[lfdahl]

Show that the binary number $b = 0.11 … 1$ with $2003$ $1$’s satisfies

$0.99 ··· 9 < b < 0.99…9$, where the lower bound has $602$ decimal digits $9$,

whereas the upper bound has $603$ decimal digits $9$.

 

[HallsofIvy]

Spoiler

b= 1/2+ 1/4+ ...+ 1/2^{2003}= (1/2)(1+ 1/2+ ...+ 1/2^{2002}

That's a finite geometric sequence. For the general finite geometric sequence S(N)= 1+ a+ a^2+ ...+ a^N, we can write S(N)- 1= a+ a^2+ ...+ a^{N-1}= a(1+ a+ a^2+ a^{N-1}). Add and subtract a^N inside the last parenthese: S(N)- 1= a(1+ a+ a^2+ ...+ a^{N-1}+ a^N- a^N)= a(1+ a+ ^2+ ...+ a^{N-1}+ a^N)- a^{N+1}= aS(N)- a^{N+1}.

That is, S(N)- 1= aS(N)- a^{N+1} so S(N)- aS(N)= (1- a)S(N)= 1- a^{N+1}. S(N)= (1- a^{N+1})/(1- a).

With a= 1/2 and N= 2002, That is (1- 1/2^{2003})/(1- 1/2= 2(1- 1/2^{2003})

b is 1/2 that, 1- 1/2^{2003}

 

[lfdahl]

Spoiler

b= 1/2+ 1/4+ ...+ 1/2^{2003}= (1/2)(1+ 1/2+ ...+ 1/2^{2002}

That's a finite geometric sequence. For the general finite geometric sequence S(N)= 1+ a+ a^2+ ...+ a^N, we can write S(N)- 1= a+ a^2+ ...+ a^{N-1}= a(1+ a+ a^2+ a^{N-1}). Add and subtract a^N inside the last parenthese: S(N)- 1= a(1+ a+ a^2+ ...+ a^{N-1}+ a^N- a^N)= a(1+ a+ ^2+ ...+ a^{N-1}+ a^N)- a^{N+1}= aS(N)- a^{N+1}.

That is, S(N)- 1= aS(N)- a^{N+1} so S(N)- aS(N)= (1- a)S(N)= 1- a^{N+1}. S(N)= (1- a^{N+1})/(1- a).

With a= 1/2 and N= 2002, That is (1- 1/2^{2003})/(1- 1/2= 2(1- 1/2^{2003})

b is 1/2 that, 1- 1/2^{2003}

Thankyou, HallsofIvy for your contribution,

Spoiler

- which is an important step in the suggested solution, but you still need to consider how to represent the two bounds in an adequate manner and to ensure the validity of the two inequalities. You can make it, I´m sure (Nod)

 

[johng1]

This may not be an acceptable solution since I used my calculator, but I'm quite confident that my calculator with its approximations still proves the assertion.

Spoiler

0.999.. < b < 0.999.. if and only if
1 - 0.999.. > 1 - b > 1 - 0.999.. if and only if
$(1/10)^{602}>(1/2)^{2003}>(1/10)^{603}$ if and only if
$602\ln(10)<2003\ln(2)<603\ln(10)$ if and only if
$602/2003<\ln(2)/\ln(10)<603/2003$. According to my calculator, this is saying:
$0.300549<0.301030<0.301048$

 

[lfdahl]

This may not be an acceptable solution since I used my calculator, but I'm quite confident that my calculator with its approximations still proves the assertion.

Spoiler

0.999.. < b < 0.999.. if and only if
1 - 0.999.. > 1 - b > 1 - 0.999.. if and only if
$(1/10)^{602}>(1/2)^{2003}>(1/10)^{603}$ if and only if
$602\ln(10)<2003\ln(2)<603\ln(10)$ if and only if
$602/2003<\ln(2)/\ln(10)<603/2003$. According to my calculator, this is saying:
$0.300549<0.301030<0.301048$

Thankyou, johng, for your solution, which, I think, is fully acceptable, since your deductive steps
all correlate with the suggested solution:

Spoiler

\[0.301 < \log_{10}2 < 0.30103 \\\\ 602 < 2003 \log_{10}2 < 603 \\\\ 10^{602} < 2^{2003} < 10^{603} \\\\ 1-10^{-602} < 1-2^{-2003} < 1 - 10^{-603}\\\\ 0.\underbrace{999..9}_{602} < (0.\underbrace{111..1}_{2003})_2 < 0.\underbrace{999..9}_{603}\]