Solution Dec. 8, 2006

Let $f(x)\,$ be a function of a real variable with real values.
If $x\,$ is irrational then $f(x)=0\,$
If $x\,$ is rational $m/n\,$ in lowest terms, $f(x)=1/n\,$
$f(0)=1\,$
Show that $f(x)\,$ is nowhere differentiable.

Proof:

If $f(x)\,$ is differentiable then its derivative at $x\,$ is given by this limit:

$f'(x) = \lim_{h\to 0}\,\frac {f(x+h)-f(x)}{h}\qquad\qquad (1)$

Suppose $x\,$ is rational. Take a sequence of irrational numbers that converges to 0, e.g., $h_k = \frac{\sqrt{2}}{2^k}$

$f(x)=1/n\,$ for some integer $n\,$ since $x\,$ is rational.
$f(x+h_k)=0,\,$ since $x+h_k\,$ is irrational.

Using these values in (1) we get: $f'(x)=\lim_{h_k\to 0} \frac{-1/n}{h_k}$ Which does not exist.

Now consider $x\,$ is irrational.
$f(x)=0\,$
As we take the limit in (1), anytime h is rational the quotient is 0. So if the derivative exists it is 0.

$x\,$ will have an infinite sequence of rational decimals which converges to it, e.g., for $x=\pi\,$ we have:
$x_k={3,31/10,314/100,3141/1000,\cdots}\,$

Define: $h_k = x_k - x\,$ and apply this to (1)
$f'(x) = \lim_{h_k\to 0}\, \frac{f(x_k)}{h_k}$
$f(x_k)\ge 1/10^k,\,$ cancellation to lowest terms can only make the denominator smaller.
$h_k< 1/10^k,\,$ so we have:
$\frac{f(x_k)}{h_k}>1\,$
All these large values for the quotient are in between values of 0 mentioned above. Therefore the limit does not exist.

$\square$

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