Question

Metric Space

Enrichment Problem E2.1. In this problem, we consider a new metric not considered in the lectures. We set X = Q the set of rational numbers, and fix a prime number p E N. For 91.92 € Q, we define 8(91,92) (the p-adic distance) as follows. If qı = 42, we define 8p(91,92) = 0. Otherwise, write 91 - 92 = a/b+ 0 in lowest terms. At most one of a and b can contain a power of p in its prime factorization. Extracting this power of p, we can write 91 – 92 = p. where neither of c and d is divisible by p (and either c = a or b = d). The distance 0,91,92) is then defined to equal p". For example, 8(20/3,0) = 1/4, and 82(3/20,0) = 4. Note also that, by definition, 0,91,92) equals (91-92,0). (a) Using the fact that 8(0,2 + y) <max(8,(0,2), 8,(0,y)), prove that is a distance on Q. (b) Complete your proof by proving the inequality 80, x+y)<max( 0,), (0, y)).

Answer 1

ie. where ne z for a fix pecime PE IN we have to show so(h, y) is a distance on a ¥*, Y EQ 1. Sp(x,y) 7,0 and Sp (x,y)=0 iff x=y + x, y EQ 2. Sp(x,y) = Sp (y,x) &?, Y EQ 3. Sp (0,7) Sp(x,y) + Sp (Y,Z) + x,y,z €9 proof: 1 x-y= pra where (a, p) = (6, 7) = 1 m. → S(x,y) = 1 1 1 (by det) -> $,7) 7,0 ( nyo for any nez) when aty if x=y then S (x, y) =0 (by def") 2. Sp (x, y) = 1 6 where x-y = pn as with Sp(y,x) = 1 km (y-x = - pre = pn (-a) = 3 2 2 2 Sp (z,y)= Sp (y,x) and (6, p=1 part ca) 3. We have to une given in (-a, p)=1 (9,p)=1) - Question. ie. Sy(x,y) = Sp (oxty) = max Sp Co, x), sp (Oy)} we know that max{A, B} < A+B &A, BEZ » Spla,y)= Sp Coa+y) & Sp(0,x) + Sp Co,y) ( Triangle mequality Now to show part (6) ie. To show S₂(x+y) = max sp (o, x), sp (OY)g. Compute S (0, x) = pm where x-o= x= pm & with (PC) =(p, d) = 1 and sp (o, y) = ps where yo=y=p & with CP,e) = (ef) = 1 and sea MEZL where C=-a Now WWLOG assume mis max sp (0x), splo, y)} = sploy) Sp (0, x+y) = 1 1 85 2x+y = p me toppe = ps/proses te mes Splo,x+y) = Sploy) max?scow).sey Hence the claim ole F P P+) 778 splaty, o) = p 2 = 1/35 = 2(04) -