The shortest distance between the line $$x - y = 1$$ and the curve $$x^2 = 2y$$ is:

The line is $$x - y = 1$$, or equivalently $$x - y - 1 = 0$$. The parabola is $$x^2 = 2y$$, so a general point on the parabola can be written as $$\left(t, \frac{t^2}{2}\right)$$.

The distance from this point to the line $$x - y - 1 = 0$$ is $$d(t) = \frac{\left|t - \frac{t^2}{2} - 1\right|}{\sqrt{1^2 + (-1)^2}} = \frac{\left|t - \frac{t^2}{2} - 1\right|}{\sqrt{2}}$$.

Let $$g(t) = t - \frac{t^2}{2} - 1$$. To find the extremum, we set $$g'(t) = 1 - t = 0$$, giving $$t = 1$$.

At $$t = 1$$: $$g(1) = 1 - \frac{1}{2} - 1 = -\frac{1}{2}$$. Since $$g''(t) = -1 < 0$$, this is the maximum of $$g(t)$$. Since $$g(t) \leq -\frac{1}{2} < 0$$ for all $$t$$, the minimum of $$|g(t)|$$ is $$\frac{1}{2}$$, occurring at $$t = 1$$.

The shortest distance is $$\frac{1/2}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$$.