### Be Reasonable

Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.

### Good Approximations

Solve quadratic equations and use continued fractions to find rational approximations to irrational numbers.

### Rational Roots

Given that a, b and c are natural numbers show that if sqrt a+sqrt b is rational then it is a natural number. Extend this to 3 variables.

Suppose we have constructed a Munchkin road. The first tile is required to have a corner at the point $(0,0)$. It is clear when we think about it that the destination of the road (i.e. the top/bottom vertex of the last triangle) can be reached by tracing a path from $(0,0)$ along edges of the shapes that make up the road. We set up an induction argument. $(0,0)$ is obviously of the form $(\frac{a+b\sqrt{3}}{2},\frac{c+d\sqrt{3}}{2})$. Now suppose we choose a path from $(0,0)$ to the destination of the road. Pick a vertex on this path, which has coordinates $(x,y)$ say. All edges in the road have length $1$.Bearing in mind that the triangles are equilateral, it is clear that the angle to the horizontal of any edge must be a multiple of $30^{\circ}$. Therefore the next vertex in the path must be one of $(x\pm 1,y)$, $(x,y\pm 1)$, $(x\pm\frac{1}{2},y\pm\frac{\sqrt{3}}{2})$ or $(x\pm\frac{\sqrt{3}}{2},y\pm\frac{1}{2})$. By induction, then, any vertex on the path, and in particular the destination of the path, is of the form$(\frac{a+b\sqrt{3}}{2},\frac{c+d\sqrt{3}}{2})$.