using kronecker's theorem can we prove there's some power of two yielding a number whose initial digits equal my social security number?

Question

I just watched the "Great Courses" series of lectures in number theory, in which Professor Burger stated that

for any irrational number r, the sequence ({n * r}) where n >= 0 is dense in the interval [0,1)

we can prove there's some power of two yielding a number whose initial digits equal my social security number.

The sketch of the proof given in the course used the fact that log 2 is an irrational number. I tried to flesh out the whole proof to really understand it and I got stuck.

My attempt at a proof

First assume my social security number is 566...

From Kronecker:

there exists m | m * log 2 = X.566...

Since log 2^x = x * log 2, we have

m * log 2   =    X.566...
log 2 ^ m = X.566...

What to do next? Or maybe I need to start on a different tack ? I'd be very grateful for any tips or guidance.

( kroenecker proof details here )

See here, or, for an even earlier incarnation of this question here. — Jyrki Lahtonen, Jul 22 '15 at 22:21
$\lg 5.66\approx 0.752816$ and $\lg 5.67\approx0.753583$, so you want an integer $n$ such that $X.75282\le n\lg2\le X.75358$ for some integer $X$. — Jyrki Lahtonen, Jul 22 '15 at 22:36
The answer could depend on the rest of your SSN, you better post the remaining digits. — Nate, Jul 22 '15 at 23:21

Ben Grossmann · Accepted Answer · 2015-07-22T22:51:26.800

1

Let $N$ be the integer formed out of the required digits (for instance, your SSN). We want the first digits of $2^n$ to line up with those of $N$. Verify for yourself that it is equivalent to check that there exists a (non-negative) integer $k$ such that $$ 10^k \cdot N \leq 2^n \leq 10^k \cdot (N + 1) $$ Taking the log base $10$ of all of these, we see that $n$ is the exponent we want if there exists an integer $k$ such that we have the inequalities $$ k + \log_{10} N \leq n \log_{10}2 \leq k + \log_{10}(N + 1) $$ Perhaps now you can relate what you're looking for to Kronecker's theorem.

edited Jul 22 '15 at 22:51

answered Jul 22 '15 at 22:46

Ben Grossmann

225,327

i understood the two steps that you wrote down.. but still not seeing how i can relate this to Kronecker. But i understand a bit better than before.. so thanks ! I will keep thinking about it till it (hopefully) sinks in. – Chris Bedford Jul 23 '15 at 00:40
You need to show that the condition that I wrote here amounts to checking that $${\log_{10}(N)}\leq {\log_{10}(2)n} \leq {\log_{10}(N+1)}$$ – Ben Grossmann Jul 24 '15 at 11:52
Thanks. I'm getting close to wrapping my head around it. I think what you are getting at in the comment above is that the fractional parts of the two bounds "log-base-10(N)" and "log-base-10(N+1)" are two points along the interval [0,1). by kroenecker, there exists a multiple M of (irrational) log-base10(2) for which the fractional part of M falls within that interval. So let's say M is comprised of an integer part "I" and a fractional part "F". Then M = I.F (where I and F are strings of digits). Then we can recover the "k" in your first two equalities by setting k = I – Chris Bedford Jul 25 '15 at 06:18
Yep! ${}{}{}{}{}{}{}{}$ – Ben Grossmann Jul 25 '15 at 23:18
woo hoo ! this was a fun one.. thanks! – Chris Bedford Jul 26 '15 at 06:16

using kronecker's theorem can we prove there's some power of two yielding a number whose initial digits equal my social security number?

1 Answers1