Pondering $f(x) = \alpha x$ for an irrational $\alpha$.

Question

So I was helping my little brother with linear equations, specifically the slope of a line, and I stumbled on something that is a little hard for me to wrap my head around.

Obviously, we think of slope as rise over run. A slope of $\frac23$ as a rise of $2$ and a run of $3$, etc. But then I started thinking of functions like $f(x) = \pi x$. Strictly speaking, one could think of this as a rise of $\pi$ units and a run of $1$. But, since I was helping with algebra homework, I had graph paper in front of me, and this is where my question comes in play.

Looking at functions of the sort $f(x) = \alpha x$ for an irrational $\alpha$, the line would only hit a 'four corners' of the grid of the graph paper exactly once, right? Namely at the origin. Now intuitively, it would have to be true. If it hit the grid at a 'four corners' again, we would necessarily conclude that things like $\pi$ could be expressed as a ratio of integers, which we all know is not true.

But this just seems crazy! Imagine standing at the origin of the $xy$-plane with infinite sight, and imagine you were infinitely 'skinny'. If you were to personify the graph of something like $y=\pi x$ you would have to somehow be able to identify a straight path that goes on indefinitely without ever hitting a 'four corners' again. If you were to walk only on that path, you'd never hit the grid again. That's crazy. Now, clearly this is just an abstract way to really feel the idea of the denseness of $\mathbb{Q}$ in $\mathbb{R}$.

Is this intuition valid? Has anybody ever thought of this idea this way? I'd love to hear people's thoughts.

Answer 1

For a line, you can unambiguously define the slope as $\frac{y_2-y_1}{x_2-x_1}$ for any pair of non-equal points $(x_1,y_1),(x_2,y_2)$. This is nice because you no longer have to worry about gridlines.

Answer 2

So your function is $y = ax$, and I assume when you say the line hits four corners, you mean that if the grid is spaced not with an irrational scale, but with a scale of rational numbers, probably integers.

Four corners means that the line passes through some point $(x,y)$ such that $x$ and $y$ are rational numbers. The lines are assumed to be evenly spaced by some rational number, so for $x$ to be on any vertical grid line, $x$ must equal some multiple of that rational number. Similarly, for $y$ to be on a horizontal gridline, $y$ must be a multiple of that rational number that determines grid spacing.

For there to be four corners, both $x$ and $y$ must lie on those gridlines, requiring them to be rational.

Since $y = ax$, $a = y/x$ However, if a could be expressed as the ratio of two rational numbers, it itself would be a rational number! This is not allowed!

The one exception is at $0$, where $0 = 0$ and you cannot divide. I hope that makes sense, have a great day!

Answer 3

If you think of an irrational as a Dedekind cut then you are already given the recipe how to "avoid" the rationals. Say $p=(A,B)$ where $A$ and $B$ are disjoint, their union is the set of all rational numbers, and if $a\in A$ and $b\in B$ then $a<b.$ So $p$ "knows" to always be "below" $b$ for every $b\in B$, and always be "above" (i.e. bigger than) $a$, for every $a\in A.$ So when you shoot $p$ through the plane, it is already coded in its bones how exactly (from which side) it would avoid each rational. I admit I do not know what you are asking, apart from "hear people's thoughts", so I tried. There are enough many rationals, if you would rather avoid working with irrationals...there are probably even more rationals than one would need for any practical "engineering" problem. What seems crazy to me is the asymmetry, in that we start with rationals, use them to define rationals and then we could think of the irrationals as filling the gaps between the rationals, and also think of the rationals as filling the gaps between the irrationals: Why on earth there are so many more irrationals then, than rationals?