How can I prove that the first three decimal digits of a real number between 0 and 1 can be equal to the reciprocal of 2 to the power of that number?

Question

This is the challenge problem at the end of Chapter 1 of Solow's How to Read and Do Proofs.

The problem:

Find a counter-example to the following statement: “If x is a positive real number between 0 and 1, then the first three decimal digits of x are not equal to the first three decimal digits of $2^{-x}$.

My attempt:

• The hypothesis A is that x is a real number between 0 and 1.
• The conclusion B is that the first three decimal digits of x are equal to the first three decimal digits of $2^{-x}$.
• A possible key question is, "How can I show that the first three decimal digits of a real number between 0 and 1 is identical to first three decimal digits of the reciprocal of 2 to the power of that number?

I've tried to reformulate this as a problem with graphs. I graphed the horizontally flipped version of f(x) = $2^{x}$, then looked at the graph between 0 and 1.

But I'm not sure how to continue; it's possible I need to review the necessary background knowledge.

What topic in mathematics should I study to solidify my background knowledge to learn more relevant information for answering this question? If this isn't necessary, could I receive a hint on how to proceed?

Answer 1

We are told to produce a counterexample to a somewhat arbitrary statement. The circumstances suggest that there is such a counterexample. How to find it? Why should some $x\in\ ]0,1[\ $ have a similar decimal representation as $2^{-x}$? The natural reason is that in fact $x=2^{-x}$. Therefore let's prove that there is such an $x$. Unless some trouble of the kind $0.6299997$ turns up it then should be possible to name a clear cut (i.e., having finite decimal expansion) $x$ that is a counterexample to the alleged claim.

The function $f(x):=x-2^{-x}$ has values $f(0)=-1$, $f(1)={1\over2}$, hence at least one zero $\xi\in\ ]0,1[\ $. It turns out that $\xi\doteq0.641$. Entering $x:=0.641$ into a pocket calculator gives $2^{-x}\doteq0.641268$, establishing $x$ as a counterexample to the alleged claim.

Answer 2

Hint: you can construct an easy counter example using the fact that as $x \rightarrow 0$, $2^{-x} \rightarrow 1$.

Answer 3

I started with $x=0.123$ and got $2^{-x}=0.918$

Then I considered $x=0.918$ and got $2^{-x}=0.529$

And went on applying the same function so I got the sequence

$0.123,\;0.918,\;0.529,\;0.693,\;0.618,\;0.651,\;0.636,\;0.643,\;0.640,\;0.641,\;0.641,\;0.641,\;\ldots$

Hope this is useful The fixed point $x=0.641$ is the counterexample