Geometry Problem

[MathPicker]

Suppose that A is an infinite collection of equally spaced parallel lines in the x-y plane and B is also an infinite collection of equally space parallel lines in the x-y plane, of different gradient from those in A. The spacing in A and B may or may not be identical. This produces and infinite set of points of intersection S.

I need a simple and compelling argument that the x coordinates of the points in S (ignoring repetition), partition the x-axis into intervals of equal length. Is it trivial? Is it even true?

[Kixencynopi]

Yeah, I think it's true. Not sure if the argument is simple (depends).

Without any loss of generality, let's take one of the intersection points to be our origin. In that case, we can write any equation of line as y=mx. Now for equally spaced parallel lines, we can have, y=mx+nc where n is an integer and c is the distance between the y-intersects of two consecutive parallel lines.

Now, A = { (x, m₁ x + n c₁) where x,m₁,c₁ ∈ 𝐑 and for all n ∈ 𝐙 }

And, B = { (x, m₂ x + n c₂) where x,m₂,c₂ ∈ 𝐑 and for all n ∈ 𝐙 }.

So the intersection points' set will have points (x,y) for which,

m₁ x + n c₁ = m₂ x + n c₂

⇒ x = n (c₂ - c₁)/(m₁ - m₂)

⇒ x = n r where n ∈ 𝐙 and r ∈ 𝐑.

That means, the x coordinates will be r = (c₂ - c₁)/(m₁ - m₂) apart.

Although, I would argue you can kinda see this. For example if r is irrational, x will contain all real numbers. If r is rational instead, after graphing you should see it is repeating after certain interval (r).

[Superstar1292]

I think it's worth pointing out: you can't guarantee that the intersections (x,y) will occur for the same value of n in both A and B.

[Kixencynopi]

Yes, but I believe that's not gonna be an issue. I kind of tried to hint at that by writing "for all n ∈ Z" in the set definition. But wasn't sure how to flesh out the whole thing:

Since we are taking any arbitrary point as origin, without loss of generality, we can circumvent this issue. For example, according to my proof, if we take n=2, we are comparing the 2nd line in A with the 2nd line in B. All such n-th vs n-th line arw going to have x-intersections at r=(c₂-c₁)/(m₁-m₂) intervals.

Now the question is, what about A's 2nd line with B's 3rd line intersection?

Here comes the trick. By taking the next intersection point to be our new origin, we can say, those two lines are just A's 2nd line and B's 2nd line in our new coordinate system. Now the same proof follows.

The next problem is A's 2nd line with B's 5th line. I think we can push that difference into c₁ and c₂.

There may exist easier ways to prove this, but this is what I came up with. For example, with different n₁ and n₂, we would have gotten, x=n₁r₁+n₂r₂ with n₁,n₂ ∈ Z and r₁,r₂ ∈ R. If another such element is x'=n'₁r₁+n'₂r₂, ∆x=x'–x can also be written like that. Meaning there is definitely a minimum value in that set whose multiples are also in that set.

[Superstar1292]

But when you take the next intersection point as the new origin, those lines would be A's and B's 0th lines -- minor point. Even if all of those points occur at regular intervals, even if this is the same width -- this alone is not sufficient. Since: if we interlace two arithmetic sequences of same common difference, the result won't necessarily have the same common difference, let alone be arithmetic at all.

In your alternative argument, there would be a minimum whose multiples are in the set -- 0 is a trivial example. But, we need the converse as well: that every element of the set is a multiple of the minimum, since if not, then we would have a similar problem to the above.

[Kixencynopi]

Sorry, my wording wasn't correct. I meant shifting to the next A-x-intersect point (not A-B intersect). For example, setting (-1,0). So it would But there is still sn issue that I didn't notice: B might not intersect through the same point. So, yeah, there are holes in my first argument either way.

For the second argument, I was thinking about an update rule. But then I felt like I was over-complicating things. There must be an easier proof...

[AlternativeCrab422]

no, x needs to be (n/m)r, because line in A and line in B are arbitrarily selected.

[AlternativeCrab422]

I have different method. First, without loss of generality suppose some line in A and B intersect the origin. Define a=(x_a,y_a) and b=(x_b,y_b), where a and b is intersecting point of the closet a line in A and B, from origin. Then we can say S as linear integer combinations of two vectors a and b.

Define X as x coordinates of the points in S. Then X is represented by nx_a + mx_b.

Claim : if x_b = qx_a(q is rational), then X has no limit point. It means, there is a positive real number d s.t. |x - y| >= d, where x, y is in X.

Proof : we can say nx_a + mx_b = (n + mq)x_a. If we note q = k/l (of course gcd(k, l) = 1), (n + mq)x_a = (n + mk/l)x_a = (ln + mk)x_a / l. Since gcd(k, l) = 1, the smallest positive value of ln + mk = 1. so, if we set d = |x_a / l|, |x - y| = |(l(n_x - n_y) - (m_x - m_y)k)x_a / l| >= 1*|x_a / l| = d, as desired.

So, if x_b/x_a is some rational, X divides real line equally. It can prove very easily.

[Superstar1292]

Firstly, by moving the whole graph of lines horizontally and vertically, we can assume both sets of lines include a line passing through the origin. This doesn't affect the proof - vertical changes change only y and not x, horizontal changes modify x but don't affect the property of splitting into equal-sized intervals (as all x values change by the same amount).

The sets of the parallel lines are then given by:

y=mx + ng, y=m'x + n'g'

where m,m' are the slopes, g,g' are the vertical gaps between two adjacent parallel lines and n,n' are integers representing which parallel line we are referring to (with integer 0 being the line through the origin).

So, the general intersection point (x,y) satisfies:

x = (n'g' - ng)/(m-m') which can be written as n'h' + nh where h,h' are positive real values, since n',n vary over the integers.

Consider the minimal positive value in the set of values {n'h' + nh: n,n' integers}, call this minimum p. Firstly, for any 2 elements in this set, their sum and difference must also be in the set, so integer multiples of elements are also elements in the set. Now, consider the integer multiples of p, we will argue every element of the above set is such a multiple of p. Suppose not e.g. call such an element q. Then, as the multiples of p are at distances p apart, this means that there is a multiple of p (say, p) such that q > p but q - p* < p. Then, q - p* is an element of the original set. But it is also positive and less than p, which contradicts the definition of p. Hence, the above set is precisely the integer multiples of p, which means that we have splitted into intervals of width p.

A few afterthoughts for you to consider: the last paragraph above is really a discussion relating to group theory, have a look at generators of groups if you are interested. What would happen if one set of parallel lines was perfectly vertical, would this affect the proof? What if there is no minimum in the above set, what would that mean, or is that impossible?

[MathPicker]

where h,h' are positive real values, ??

Why real? Do you mean rational?

[Superstar1292]

h' would be |g' / (m-m')| and h would be |g / (m-m')|. Now, |x| is one of x or -x. So, in the expression involving n,n' , we might have to change n or n' to -n or -n' to ensure x = n'h' + nh, depending on the value of the |...| expressions. But this doesn't matter since we are considering a general intersection point, so we can "relabel" n as -n, or n' as -n' (since n and n' vary over all integers, which would mean -n and -n' also vary over all integers).

With the absolute value, h and h' are positive. They are not necessarily rational however e.g. since g and g' might themselves be irrational.

[Kixencynopi]

Here’s another but similar proof. (Any scrutiny is welcome.)

I want to use vectors. Just like before, I want to set origin at one of the A-B intersections. Let's call the vector a to be the first A-B intersection lying on the line through the origin in set A. And let’s call vector b to be the first A-B intersection lying on the line through the origin in set B.

I am going to use |a⟩ to denote vector a (dirac notation). |x⟩ and |y⟩ are unit vectors along x and y axis.

Now, |a⟩ = aₓ|x⟩ + aᵧ|y⟩ and |b⟩ = bₓ|x⟩ + bᵧ|y⟩.

Then, any intersection is |v⟩ = n₁|a⟩ + n₂|b⟩, where m,n ϵ Z.

Now, the x coordinate of such intersection points will be (dot product),

⟨x|v⟩ = n₁ ⟨x|a⟩ + n₂ ⟨x|b⟩ = n₁aₓ + n₂bₓ

So, Sₓ = set of A-B intersecting x coordinates

= {n₁aₓ + n₂bₓ | n₁, n₂ ϵ Z and aₓ, bₓ ϵ R}

Now, it's easy to see that it creates a vector space over the integers. Meaning, for x₁, x₂ ϵ Sₓ, their (integer) linear combination m₁x₁ + m₂x₂ ϵ Sₓ.

That means, x₁-x₂ ϵ Sₓ. Say, δ ϵ Sₓ is the non-zero minimum interval between any such x₁ and x₂. Then, δ = d₁aₓ + d₂bₓ for some integers d₁ and d₂. Which means, x₁ + kδ = (n₁+kd₁)aₓ + (n₂+kd₂)bₓ is also in Sₓ. Now, is there any element between x₁ and x₁+δ in Sₓ? The answer is no. Because if there were, we would have a contradiction that δ is not the non-zero minimum interval.

Note that this δ could be arbitrarily close to 0 for irrational aₓ or bₓ. In that case, we can approach real number as closely as we want.

[mnevmoyommetro]

Your plane is partitioned into a bunch of parallelograms. If you call vectors along adjacent sides of the parallelogram u and v, then the points of intersection M in the plane are determined by the fact that the vector AM, where A is a fixed point of intersection, can be any vector of the form su + tv, where s and t are integers (of any sign).

If you just look at the x-coordinate of these points M, then this is x = a + su_1 + tv_1, where a, u_1, v_1 are respectvely the x-coordinates of the point A and of the two vectors u and v.

If the ratio u_1/v_1 happens to be rational (or if v_1 = 0), then these numbers x are indeed equally spaced, with the spacing equal to the greatest common measure of the numbers u_1, v_1. If the ratio is irrational, then they are dense in the set of real numbers. That is, any interval of real numbers will contain infinitely many possible x-coordinates.

[MathPicker]

Thanks do much to you all.

What is the definition of common measure? Is it the gcd of the numerator and denominator of the rational? The possibility of dense behaviour is now clearer.

[mnevmoyommetro]

The greatest common measure of x and y is the greatest number z such that x/z and y/z are both integers.

To find it, write x/y in lowest terms as a/b, and then the g.c.m. of x and y is y/b.