Prove the ultrametric triangle inequality |x + y₂ ≤ max{x\, \y\,}. Remark. Since max{\x|₁, \y\,} < \x\, + \y\„, the ultrametric triangle inequality implies the usual one.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Problem 6 A place p on Q is either a prime number p or the symbol ∞.
• For the symbol ∞, we will use | to denote the usual absolute value |·|. Recall
that satisfies the following properties:
1. x = 0 if and only if x = 0.
2. |xy| = |x|∞|y for all x, y ≤ Q.
3. |x+y|∞ ≤ |x|∞ + \y\∞ for all x, y ≤ Q.
• For each prime number p, we have the p-adic norm |x|p defined as follows. For n a
nonzero integer, recall that vp(n) is the exponent of p appearing in the prime factorization
of n. Namely, pº(n) | n, while pº(n)+¹ {n. Extend this definition to nonzero fractions as
follows:
(a)
N
m
Show that, if the two fractions and represent the same rational
number, then vp() = vp (m/²).
=
vp (n) — vp(m).
m
Page 10
n'
m'
(d)
V
Hence, we obtain a function vp: Q× → Z. (Recall that Qˇ consists of nonzero rational
numbers). The p-adic norm of a rational number x is defined to be
(b)
Screen Shot 202...
For example,
(c)
25
=
|x|p
1
8'
=
-Up (x)
0
24
25
13
a norm, namely:
0.
1. |x|p = 0 if and only if x =
2. xy = |x|py|p for all x, y ≤ Q.
3. |x + y₂ ≤ |x|p+ lyp for all x, y € Q.
1
p is a place
if x = 0;
if x = 0.
Hint. You need the prime factorization.
2
5
Verify that, the p-adic norm satisfies the three defining properties of
Prove the ultrametric triangle inequality
|x + y|₁ ≤ max{\x\,, \y\,}.
Remark. Since max{\x|₁, \y\,} < |x|, + \y\,, the ultrametric triangle inequality implies
the usual one.
24
=
25.
Prove that for any nonzero rational number x, we have
II |x|p =
= 1.
Transcribed Image Text:Screen Shot 2023-... Problem 6 A place p on Q is either a prime number p or the symbol ∞. • For the symbol ∞, we will use | to denote the usual absolute value |·|. Recall that satisfies the following properties: 1. x = 0 if and only if x = 0. 2. |xy| = |x|∞|y for all x, y ≤ Q. 3. |x+y|∞ ≤ |x|∞ + \y\∞ for all x, y ≤ Q. • For each prime number p, we have the p-adic norm |x|p defined as follows. For n a nonzero integer, recall that vp(n) is the exponent of p appearing in the prime factorization of n. Namely, pº(n) | n, while pº(n)+¹ {n. Extend this definition to nonzero fractions as follows: (a) N m Show that, if the two fractions and represent the same rational number, then vp() = vp (m/²). = vp (n) — vp(m). m Page 10 n' m' (d) V Hence, we obtain a function vp: Q× → Z. (Recall that Qˇ consists of nonzero rational numbers). The p-adic norm of a rational number x is defined to be (b) Screen Shot 202... For example, (c) 25 = |x|p 1 8' = -Up (x) 0 24 25 13 a norm, namely: 0. 1. |x|p = 0 if and only if x = 2. xy = |x|py|p for all x, y ≤ Q. 3. |x + y₂ ≤ |x|p+ lyp for all x, y € Q. 1 p is a place if x = 0; if x = 0. Hint. You need the prime factorization. 2 5 Verify that, the p-adic norm satisfies the three defining properties of Prove the ultrametric triangle inequality |x + y|₁ ≤ max{\x\,, \y\,}. Remark. Since max{\x|₁, \y\,} < |x|, + \y\,, the ultrametric triangle inequality implies the usual one. 24 = 25. Prove that for any nonzero rational number x, we have II |x|p = = 1.
Prove the ultrametric triangle inequality
|x+y|₁ ≤ max{ |x|p, \y\p
Remark. Since max{|x|„, \y\,} < |x|, + \y\„, the ultrametric triangle inequality implies
the usual one.
©
Transcribed Image Text:Prove the ultrametric triangle inequality |x+y|₁ ≤ max{ |x|p, \y\p Remark. Since max{|x|„, \y\,} < |x|, + \y\„, the ultrametric triangle inequality implies the usual one. ©
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