Triangle Inequality for d(m,n) Metric Proof

[Demon117]

1. For integers m and n, let d(m,n)=0 if m=n and d(m,n) = 1/5^k otherwise, where k is the highest power of 5 that divides m-n. Show that d is indeed a metric.

2. The attempt at a solution

Here is what I have come up with:

PROOF: Clearly by definition d(m,n) = 0 iff m=n and d(m,n)>0 for all k in Z. Next, the same highest power k of 5 that divides m-n will also divide n-m. Therefore we have d(n,m) = 1/5^k, thus d(m,n) = d(m,n) when n is not equal to m. So d is symmetric and positive definite. I must finally show that the triangle inequality holds.

I am unsure how to proceed in showing that the triangle inequality holds. Can anyone help me?

 

[snipez90]

All right so take k to be the highest power as in the problem description so that d(m,n) = 1/5^k. Let i and j be the associated powers of d(m,r) and d(r,n), respectively. We need to show that
$$\frac{1}{5^k} \leq \frac{1}{5^i} + \frac{1}{5^j}.$$
Note that it suffices to prove that k is greater than or equal to one of i or j. Without loss of generality you may assume $j \geq i \geq 0.$ Now the key of course is that k is the highest power of 5 that divides m-n, so use the metric relations given to see if another power of 5 also divides m-n and conclude the desired inequality.

 

[Demon117]

All right so take k to be the highest power as in the problem description so that d(m,n) = 1/5^k. Let i and j be the associated powers of d(m,r) and d(r,n), respectively. We need to show that
$$\frac{1}{5^k} \leq \frac{1}{5^i} + \frac{1}{5^j}.$$
Note that it suffices to prove that k is greater than or equal to one of i or j. Without loss of generality you may assume $j \geq i \geq 0.$ Now the key of course is that k is the highest power of 5 that divides m-n, so use the metric relations given to see if another power of 5 also divides m-n and conclude the desired inequality.

That is more or less what I thought of. So the way you've explained implies that I am trying to show uniqueness? I maybe I misunderstood completely.

If d(m,r) = 1/5^j for j the highest power that divides m-r and I know that d(r,n) = 1/5^i for i the highest power that divides r-n, then I have j*p = m-r and i*q = r-n, for p,q in Z. Solving for r and substituting I get iq = m-n - jp. But m-n is equivalent to k*s for some s in Z.

In all I have iq = ks - jp, so that ks = iq + jp. Where do I go from here?