MHB Understanding Part (c) of the Protractor Postulate: Explaining r=30

[pholee95]

I'm stuck on this problem. Can anyone please help me understand?

Consider the model of geometry where point means rational point in the Euclidean plane and all of our other terms have their normal interpretation. This model doesn't satisfy the Ruler Postulate because there isn't a one-to-one correspondence with R. It also doesn't satisfy part (c) of the Protractor Postulate. Explain why it doesn't satisfy this part of the postulate by considering the line through (0,0) and (1,0), the upper half-plane, and the number r = 30. (Hint: Use a little piece of trig and think about the point E in this case.)

*I know that part (c) of the protractor postulate states this: For each real number r, 0 < r < 180, and for each half-plane H bounded by AB there exists a unique ray AE such that E is in H and μ(angleBAE) = r◦.

 

[Evgeny.Makarov]

Consider the model of geometry where point means rational point in the Euclidean plane

I assume this means point with rational coordinates.

part (c) of the protractor postulate states this: For each real number r, 0 < r < 180, and for each half-plane H bounded by AB there exists a unique ray AE such that E is in H and μ(angleBAE) = r◦.

A ray is, first of all, a set of points. Are there any points with rational coordinates on the ray that forms $30^\circ$ with the $Ox$ axis?

 

[pholee95]

I assume this means point with rational coordinates.

A ray is, first of all, a set of points. Are there any points with rational coordinates on the ray that forms $30^\circ$ with the $Ox$ axis?

There are none right?

 

[Evgeny.Makarov]

There are none right?

Right, except $(0,0)$. The axiom requires that at least one point $E$ lies on the ray, which is not the case here. But you have to prove that there are no rational points on the ray.

 

[pholee95]

Right, except $(0,0)$. The axiom requires that at least one point $E$ lies on the ray, which is not the case here. But you have to prove that there are no rational points on the ray.

Ah. I understand it now. Thank you so much for your help!