Proving that $\mathbb{P}(\{a$ and $b$ are co-prime}$)$ $=0$ for $a,b$ following the Uniform distribution over [$n, 2n] $ as $n \rightarrow \infty$

Question

I was investigating the following problem and came across an interest result that I would be interested in proving rigorously.

Consider $a,b \in \mathbb{N}$, and plot the point $(a,b)$ on the cartesian plane. If we join a line from $(0,0)$ to $(a,b)$, what is the percentage chance that no other integer co-ordinates lie on that line?

While investigating the above, this led me to consider what happens as the point $(a,b)$ get larger and so I decided to consider what happens in the limit as $a \rightarrow \infty$ and $b \rightarrow \infty$. Intuitively, it feels as though the chance that a line from $(0,0)$ should become less and less likely that it hits $(a,b)$ without having first passed through at least one integer co-ordinate.

This led to me to take the conjecture that, in the limit, the percentage chance of this point satisfying this requirement is $0$%.

In trying to prove this, I came across the following result (which I have not seen a proof for, and so could turn out to be wrong):

Claim: if you draw a line from $(x_1,y_1)$ to $(x_2,y_2)$, then the number of integer co-ordinates on this line will be $ \gcd(|x_2 - x_1|, |y_2 - y_1|)+1$

Therefore, since our starting point is $0$, this tells us that the number of integer co-ordinates on a line from $(0,0)$ to $(a,b)$ must be $\gcd(a,b)+1$. Since the start and end points are both integers, this tells us that a point has no integer co-ordinates along it if and only if $\gcd(a,b)+1 = 2$ (ie. the only integers on the line are the start and end points). Therefore, by rearranging this, we are essentially trying to show that the probability of $\gcd(a,b) = 1$ approaches $0$ as $a,b \rightarrow \infty$.

So if we can show that the probability of $a$ and $b$ being coprime approaches $0$ in the limit (which seems reasonable), then we are done.

Is there a way for me to formalise an argument of this nature (and is what I have done up until this point reasonable)?

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Note$_1$: the source for the claim is Quora and it was given without a proof, this does make me somewhat sceptical so as an additional question, I would be interested in knowing whether or not this result is actually try and if there is a proof reference that I could be pointed towards.

Note$_2$: I have tested the above claim for some small numbers and it seems to hold up, but I haven't been able to find anything completely rigorous to show that it holds more generally.

Note$_3$: As pointed out by lulu in the comments, this post will likely be useful as we are specifically considering the edge case of what is computed here.

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Edit: as mentioned by @Slugger, we can't really talk about the notion of probability without first considering the distribution of our choices of $a,b$. Really, we only want $a,b$ to be large without any bias beyond this. Therefore, it makes sense to consider a discrete uniform distribution from $n$ to $2n$. We can then take $n \rightarrow \infty$ and this should be equivalent to the problem that I am trying to resolve (note that my choice of the upper bound being $2n$ was arbitrary as in the limit this would be equivalent to a discrete uniform distribution from, for example, $n$ to $3n$).

Answer 1

As stated I think the claim is simply wrong. For any $a,b$ we have that $$ \mathbb{P} (a,b,\text{ co-prime}) = \begin{cases} 1 &\text{if $a,b$ are co-prime} \\ 0& \text{otherwise} \end{cases}. $$ Now, $\lim_{a,b\to \infty} f(a,b) = 0$ if and only if for all sequences $(a_n,b_n)$ with $a_n,b_n \to \infty$ we have $\lim_{n\to \infty} f(a_n,b_n) = 0$. However, we may pick $a_n = p_n$, the $n$-th prime, and $b_n = p_n + 1$. Then $$ \lim_{n\to \infty} \mathbb{P}(a_n,b_n \text{ co-prime}) = 1, $$ so that $\lim_{a,b\to\infty} \mathbb{P}(a,b\text{ co-prime})$ cannot be $0$.

Additionally, the sequence $a_n =p_n$ and $b_n = 2p_n$ shows that in fact the limit does not even exist.

I think the intuitive thing you are trying to prove is not captured by the limit you actually give.

Answer 2

Let's consider the suggestion in the question that we might just as well consider $[n,3n]$ rather than $[n,2n].$

Now suppose that in the limit as $n \to\infty,$ the probability that $a$ and $b$ are coprime when $(a,b)$ is uniformly selected from $[n,3n]\times[n,3n]$ goes to $p.$

In particular, choose $N$ large enough so that for all $n>N,$ the probability that $a$ and $b$ are coprime is between $p - 0.01$ and $p + 0.01$ and also let $N$ be large enough so that $(2n + 1)/(3n) < \frac12.$ Then for any $n>N,$ the probability that $a$ and $b$ are coprime if they are selected instead from $[1,3n]\times[1,3n]$ is no greater than the probability of landing on a coprime pair in the subset $[n,3n]\times[n,3n]$ (which is less than $\frac12 (p + 0.01)$ plus the probability of landing outside $[n,3n]\times[n,3n],$ which is less than $\frac59.$ Hence the probability that $a$ and $b$ are coprime when selected from $[1,3n]\times[1,3n]$ is less than $$ \frac59 + \frac12 (p + 0.01) = \frac59 + 0.005 + \frac12 p < 0.57 + \frac12 p. $$

But we know that in the limit, the probability of picking a coprime pair from $[1,3n]\times[1,3n]$ is $6/\pi^2 > 0.6.$ So $p$ cannot be $0.$

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Obviously you can just keep trying new ways to define the limit -- and by the way, any distribution that has $a$ and $b$ independent and is not the uniform distribution over $[1,n]\times[1,n]$ will end up covering only a kind of "cone," not a whole "ball" around the origin of $\mathbb N\times\mathbb N$ -- but at some point I think we have to ask what is the point that we're trying to make.