A reinterpretation of the $abc$ - conjecture in terms of metric spaces?

Question

I hope it is appropriate to ask this question here:

One formulation of the abc-conjecture is

$$ c < \text{rad}(abc)^2$$

where $\gcd(a,b)=1$ and $c=a+b$. This is equivalent to ($a,b$ being arbitrary natural numbers):

$$ \frac{a+b}{\gcd(a,b)} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$

Let $d_1(a,b) = 1- \frac{\gcd(a,b)^2}{ab}$ which is a proven metric on natural numbers. Let $d_2(a,b) = 1- 2 \frac{\gcd(a,b)}{a+b}$, which I suspect to be a metric on natural numbers, but I have not proved it yet. Let $$d(a,b) = d_1(a,b)+d_2(a,b)-d_1(a,b)d_2(a,b) = 1-2\frac{\gcd(a,b)^3}{ab(a+b)}$$

Then we get the equivalent formulation of the inequality above:

$$\frac{2}{1-d_2(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$

which is equivalent to :

$$\frac{2}{1-d_2(a,b)} < \text{rad}(\frac{1}{1-d_1(a,b)}\cdot\frac{2}{1-d_2(a,b)} )^2$$

My question is if one can prove that $d_2$ and $d$ are distances on the natural numbers (without zero)?

Result: By the answer of @GregMartin, $d_2$ is a metric. By the other answer $d$ is also a metric.

Edit: By "symmetry" in $d_1$ and $d_2$, this interpretation also suggests that the following inequality is true , which might be trivial to prove or very difficult or might be wrong and may be of use or not in number theory:

$$\frac{1}{1-d_1(a,b)} < \text{rad}(\frac{2}{1-d(a,b)})^2$$

which is equivalent to

$$ \frac{ab}{\gcd(a,b)^2} < \text{rad}(\frac{ab(a+b)}{\gcd(a,b)^3})^2$$

(This is not easy to prove, as the $abc$ conjecture $c=a+b < ab < \text{rad}(abc)^2$ would follow for all $a,b$ such that $a+b < ab$.)

Second edit: Maybe the proof that $d_2,d$ are distances can be done with some sort of metric transformation, for example maybe with a Schoenberg transform (See 3.1, page 8 in https://arxiv.org/pdf/1004.0089.pdf) The idea, that this might be proved with a Schoenberg transform comes from the fact that:

$$d_1(a,b) = 1-\exp(-\hat{d}(a,b))$$ so $d_1$ is a Schoenberg transform of $\hat{d}(a,b) = \log( \frac{ab}{\gcd(a,b)^2}) = \log( \frac{\text{lcm}(a,b)}{\gcd(a,b)})$ which is proved to be a metric (see Encyclopedia of Distances, page 198, 10.3 )

Third edit: Here is some Sage Code to test the triangle inequality for triples (a,b,c) up to 100:

def d1(a,b):
    return 1-gcd(a,b)**2/(a*b)

def d2(a,b):
    return 1-2*gcd(a,b)/(a+b)

def d(a,b):
    return d1(a,b)+d2(a,b)-d1(a,b)*d2(a,b)

X = range(1,101)
for a in X:
    for b in X:
        for c in X:
            if d2(a,c) > d2(a,b)+d2(b,c):
                print "d2",a,b,c
            if d(a,c) > d(a,b)+d(b,c):
                print "d",a,b,c

so far with no counterexample.

Related: An inequality inspired by the abc-conjecture and two questions

Answer 1

$d_2$ is indeed a metric. Abbreviating $\gcd(m,n)$ to $(m,n)$, we need to show that \begin{align*} 1-\frac{2(a,c)}{a+c} &\le 1-\frac{2(a,b)}{a+b} + 1-\frac{2(b,c)}{b+c} \end{align*} or equivalently \begin{align*} \frac{2(a,b)}{a+b} + \frac{2(b,c)}{b+c} &\le 1 + \frac{2(a,c)}{a+c}. \end{align*} Furthermore, we may assume that $\gcd(a,b,c)=1$, since we can divide everything in sight by that factor.

Note that if $a=(a,b)\alpha$ and $b=(a,b)\beta$ with $(\alpha,\beta)=1$, then $\frac{2(a,b)}{a+b} = \frac2{\alpha+\beta}$. The only unordered pairs $\{\alpha,\beta\}$ for which this is at least $\frac12$ are $\{1,1\}$, $\{1,2\}$, and $\{1,3\}$. Further, if neither $\frac{2(a,b)}{a+b}$ nor $\frac{2(b,c)}{b+c}$ is at least $\frac12$, then the inequality is automatically valid because of the $1$ on the right-hand side.

This leaves only a few cases to check. The case $\{\alpha,\beta\} = \{1,1\}$ (that is, $a=b$) is trivial. The case $\{\alpha,\beta\} = \{1,2\}$ (that is, $b=2a$) can be checked: we have $(a,c)=\gcd(a,2a,c)=1$, and so the inequality in question is \begin{align*} \frac23 + \frac{2(2,c)}{2a+c} &\le 1 + \frac2{a+c}, \end{align*} or equivalently $$ \frac{(2,c)}{2a+c} \le \frac16 + \frac1{a+c}; $$ there are only finitely many ordered pairs $(a,c)$ for which the left-hand side exceeds $\frac16$, and they can be checked by hand.

The proof for the case $\{\alpha,\beta\} = \{1,3\}$ (that is, $b=3a$) can be checked in the same way, as can the cases $a=2b$ and $a=3b$.

Answer 2

Not an answer but an observation.

Set $r_2(a,b,c)=d_2(a,c)/(d_2(a,b)+d_2(b,c))$ (when defined), and similarly for $r(a,b,c)$. Then Greg Martin's proof shows that the values of $r_2$ should be discrete, and indeed experimentally the values are in decreasing order

$(1,9/10,6/7,5/6,9/11,...)$

The same experiment done for $d$ gives

$(1,27/40,40/63,28/45,...)$

Thus, apart from trivial cases such as $a=b$ one should have the stronger triangle inequality $d(a,c)\le0.675(d(a,b)+d(b,c))$.

Answer 3

$d$ is also a metric. Proof:

First let us call a metric on natural numbers $d$ such that $d(a,b)<1$ and $d(a,b)$ is a rational number for all $a,b$ a "rational metric". Second let $d_1,d_2$ be two rational metrics such that if we set $d=d_1+d_2-d_1 d_2$ then for all $a \neq c, a \neq b$ we have $d(a,b)+d(a,c)>1$. If this is the case for $d_1,d_2$ we will call $d_1$ and $d_2$ "paired". If $d_1,d_2$ are such paired rational metrics, then $d=d_1+d_2-d_1 d_2$ is a metric. Proof:

1) $d(a,b) = 0$ iff $0 \le d_1(a,b)(1-d_2(a,b)) = -d_2(a,b) \le 0$ hence since $1-d_2(a,b)>0$ we must have $d_1(a,b) = 0$ hence $a=b$. If on the other hand $a=b$ then plugging this in $d$ and observing that $d_1(a,b)=d_2(a,b)=0$ gives us $d(a,b)=0$.

2) $d(a,b) = d(b,a)$ since $d_i(a,b) = d_i(b,a)$ for $i = 1,2$.

3) Triangle inequality: If $a=c$ or $a=b$ the triangle inequality is fullfilled and becomes an equality because of 1) : $d(b,c) \le d(a,b)+d(a,c)$ First observe that $d(x,y) < 1$ for all $x,y$. Let therefore $a\neq c, a\neq b$. Since $d_1,d_2$ are paired rational metrics we have: $d(b,c) < 1 < d(a,c)+d(a,b)$ and the triangle inequality is proved.

This proves also that $d$ is a rational metric (if $d_1,d_2$ are paired rational metrics.)

What remains to show is that $d_2(a,b) = 1-\frac{2 \gcd(a,b)}{a+b}$, $d_1(a,b) = 1-\frac{\gcd(a,b)^2}{ab}$ are paired (rational) metrics, hence it remains to show that $d(a,b) = 1- \frac{2 \gcd(a,b)^3}{ab(a+b)}$ satisfies:

$$d(a,c)+d(a,b)>1, \text{ whenever } a\neq c, a \neq b$$

The last inequality is equivalent to, after some algebra:

$$\frac{abc(a+b)(a+c)}{2} - \gcd(a,b)^3c(a+c) - \gcd(a,c)^3b(a+b)>0$$

Let $U=\gcd(a,b,c)$. Then there exist natural numbers $R,S,T,A,B,C$ such that:

$$RU = \gcd(a,b), SU = \gcd(a,c), TU = \gcd(b,c), a = RSUA, b = RTUB, c = STUC$$

Plugging this in the last inequality and after some algebra, we find:

$$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0 $$

We can pair each of the positive summand with a negative summand to give for example:

$$(A^3*B*C*R^2*S^2*T-2*A*C*R^2)=(A^2*B*S^2*T - 2)*A*C*R^2$$

The condition $a \neq b$ translates to $SA \neq TB$ and similarily $a \neq c$ translates to $RA \neq TC$. Suppose that $A^2*B*S^2*T - 2 \le 0$. The case $A^2*B*S^2*T=1$ contradicts $SA \neq TB$. Hence we can only have at most $A^2*B*S^2*T=2$ which leads to $A=S=1$, $BT=2$ and plugging this in the definition of $a,b$ we get $b=2a$ and $d(a,b)=\frac{2}{3}$.

Now we must show that the other pairings give the desired result:

$$( A^2*B^2*C*R^2*S*T^2-2*B**2*S*T)=(A^2*C*R^2*T - 2)*B^2*S*T$$ A similar argument to the above leads to: If $A^2*C*R^2*T = 2$, then $A=R=1$, $CT=2$ which leads to (with $S=A=1$) $a=RSUA=U,b=RTUB=2U=2a,c=STUC=2U=2a$ and it follows that $d(a,c)=\frac{2}{3}$, so $d(a,b)+d(a,c)=\frac{4}{3}>1$, and this case is done.

If $A^2*C*R^2*T > 2$ and $A^2*B*S^2*T=2$ then $$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0 $$ is true.

If $A^2*C*R^2*T > 2$ and $A^2*B*S^2*T>2$ then $$1/2*(A^3*B*C*R^2*S^2*T + A^2*B^2*C*R^2*S*T^2 + A^2*B*C^2*R*S^2*T^2 + A*B^2*C^2*R*S*T^3 - 2*A*C*R^2 - 2*A*B*S^2 - 2*C^2*R*T - 2*B^2*S*T)*R^2*S^2*T*U^5 > 0 $$ is true. This shows, that $d_1,d_2$ are paired metrics and completes the proof.

Answer 4

This question has already very good answers. I justed wanted to highlight that it is possible to shorten the proofs, using the following:

If $X_a = \{ a/k | 1 \le k \le a \}$ then $X_a \cap X_b = \gcd(a,b)$, which is straightforward to prove. Then $d_1(a,b) = 1-\gcd(a,b)^2/(ab) = 1-|X_a \cap X_b|^2 / (|X_a||X_b|)$ is the squared cosine metric (see Encyclopedia of Distances) and $d_2(a,b) = 1-2\gcd(a,b)/(a+b) = 1-2|X_a \cap X_b| / (|X_a|+|X_b|)$ is the Sorensen Metric (Encyclopedia of Distances). Hence $d_1,d_2$ are metrics of the form $d_i = 1- s_i$ where $s_i$ is a similarity. But then $s=s_1 \cdot s_2$ is also a similarity and $d=d_1 +d_2 -d_1 d_2 = 1-s=1-s_1 s_2$ is a metric.