# Cartesian slope rationality

#### Loren

If a line coincides with at least two vertices on an infinite Cartesian grid, its slope is rational.

If a line coincides with only one vertex on an infinite Cartesian grid, its slope is irrational.

If a line coincides with no vertices on an infinite Cartesian grid, its slope is either irrational or rational.

#### skipjack

Forum Staff
Consider the slope in terms of the (integer) coordinates of the vertices mentioned.

#### DarnItJimImAnEngineer

If a line coincides with at least two vertices on an infinite Cartesian grid, its slope is rational.
Agreed. Wouldn't that also mean it must coincide with an infinite number of vertices?

If a line coincides with only one vertex on an infinite Cartesian grid, its slope is irrational.
If it were rational and passed through one vertex, it would pass through others, so agreed.

If a line coincides with no vertices on an infinite Cartesian grid, its slope is either irrational or rational.
I can absolutely picture the rational case (slope = 1 is an easy example).

I don't know how to prove an irrational number could pass through zero vertices.
Prove $\exists \varepsilon : \{n\pi + \varepsilon, n\in \mathbb{Z}\} \cap \mathbb{Z} = \emptyset$

• 1 person

#### DarnItJimImAnEngineer

...wait, if $n\pi+Â½$ is not an integer, then $2n\pi+1$ is not an integer, so $2n\pi$ is not an integer, which is true because $\pi$ is irrational.

Did I just do it?

#### Loren

Agreed. Wouldn't that also mean it must coincide with an infinite number of vertices?

If it were rational and passed through one vertex, it would pass through others, so agreed.

I can absolutely picture the rational case (slope = 1 is an easy example).

I don't know how to prove an irrational number could pass through zero vertices.
Prove $\exists \varepsilon : \{n\pi + \varepsilon, n\in \mathbb{Z}\} \cap \mathbb{Z} = \emptyset$
I believe your contentions above per rationals. How about a rational slope with zero coincidences? I think your next post can apply to that.

#### DarnItJimImAnEngineer

y = x + 0.5 will not pass through any integer vertices.

#### Loren

y = x + 0.5 will not pass through any integer vertices.
Your equation agrees with my mind.

#### v8archie

Math Team
On the other hand, since the OP never mentions integers, we can have the gridlines where we want, although I presume that they are evenly spaced. Thus every line passes through at least one vertex.

#### DarnItJimImAnEngineer

On the other hand, since the OP never mentions integers, we can have the gridlines where we want, although I presume that they are evenly spaced. Thus every line passes through at least one vertex. • 1 person

#### OOOVincentOOO

Does this apply to problem?

Tree Gaps and Orchard Problems - Numberphile:
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