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

• MHB
• lfdahl
In summary, the binary number $b = 0.11...1$ with $2003$ $1$'s satisfies the inequality $0.99...9 < b < 0.99...9$, where the lower bound has $602$ decimal digits $9$ and the upper bound has $603$ decimal digits $9$. Both contributors, HallsofIvy and johng, have provided acceptable solutions, with johng using a calculator to prove the assertion.
lfdahl
Gold Member
MHB
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 $9$.

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}

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HallsofIvy said:
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}

- 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)

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.

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:
$0.300549<0.301030<0.301048$

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johng said:
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.

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:

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

## What is a binary number?

A binary number is a number expressed in the base-2 numeral system, which uses only two digits (0 and 1) to represent all values.