Prove that in PZ geometry, every PZ line has an equation of the form ax+by=c, where a, b, c are all rational numbers and a and b are not both zero. Prove that every equation of the form ax+by=c, where a, b, and c are all rational numbers and a and b are not both zero, is in fact a PZ line. Prove that if two distinct PZ lines m and n are not parallel in the usual Euclidean sense, then they have a PZ point in common.. In terms of the possible parallel axioms for geometry, which geometry does PZ geometry most resemble: projective, Euclidean, or hyperbolic geometry?
Tell me more
×
Mathematics Stack Exchange is a question and answer site for
people studying math at any level and professionals in related fields. It's 100% free, no registration required.
|
|
Hint for the first question: if $(x_1, y_1)$ and $(x_2,y_2)$ are on the same line, then either: a) $x_1=x_2$ (what is the equation for the line in this case?) or b)this line has the form $y-y_1=\frac{y_1-y_2}{x_1-x_2}(x-x_1)$ And all the numbers in sight are rational. I'll leave it to you to manipulate this equation into the appropriate form. For the second question: What does it mean, in terms of $a$, $b$, and $c$ for the lines to be parallel? When the lines are not parallel, can you find a common solution to the two equations? Keep track of your steps to show that the common solution has rational coordinates. The last question is pretty much answered by the second. |
|||
|
|