U only have to solve part a I included all parts just in case U needed more information 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 |·|. Recallthat 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|, defined as follows. For n anonzero integer, recall that vp(n) is the exponent of p appearing in the prime factorizationof n. Namely, pup(n) | n, while pº(n)+¹ {n. Extend this definition to nonzero fractions asfollows:(a)Up (™):= vp(n) — vp(m).Show that, if the two fractions andnumber, then vp() = vp (m ² ).nmn'represent the same rational 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 |·|. Recallthat 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 anonzero integer, recall that vp(n) is the exponent of p appearing in the prime factorizationof n. Namely, pº(n) | n, while pº(n)+¹ {n. Extend this definition to nonzero fractions asfollows:(a)NmShow that, if the two fractions and represent the same rationalnumber, then vp() = vp (m/²).=vp (n) — vp(m).mPage 10n'm'(d)VHence, we obtain a function vp: Q× → Z. (Recall that Qˇ consists of nonzero rationalnumbers). The p-adic norm of a rational number x is defined to be(b)Screen Shot 202...For example,(c)25=|x|p18'=-Up (x)0242513a 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.1p is a placeif x = 0;if x = 0.Hint. You need the prime factorization.25Verify that, the p-adic norm satisfies the three defining properties ofProve the ultrametric triangle inequality|x + y|₁ ≤ max{\x\,, \y\,}.Remark. Since max{\x|₁, \y\,}

Question

U only have to solve part a I included all parts just in case U needed more information 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 |·|. Recallthat 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|, defined as follows. For n anonzero integer, recall that vp(n) is the exponent of p appearing in the prime factorizationof n. Namely, pup(n) | n, while pº(n)+¹ {n. Extend this definition to nonzero fractions asfollows:(a)Up (™):= vp(n) — vp(m).Show that, if the two fractions andnumber, then vp() = vp (m ² ).nmn'represent the same rational 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 |·|. Recallthat 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 anonzero integer, recall that vp(n) is the exponent of p appearing in the prime factorizationof n. Namely, pº(n) | n, while pº(n)+¹ {n. Extend this definition to nonzero fractions asfollows:(a)NmShow that, if the two fractions and represent the same rationalnumber, then vp() = vp (m/²).=vp (n) — vp(m).mPage 10n'm'(d)VHence, we obtain a function vp: Q× → Z. (Recall that Qˇ consists of nonzero rationalnumbers). The p-adic norm of a rational number x is defined to be(b)Screen Shot 202...For example,(c)25=|x|p18'=-Up (x)0242513a 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.1p is a placeif x = 0;if x = 0.Hint. You need the prime factorization.25Verify that, the p-adic norm satisfies the three defining properties ofProve the ultrametric triangle inequality|x + y|₁ ≤ max{\x\,, \y\,}.Remark. Since max{\x|₁, \y\,} < |x|, + \y\,, the ultrametric triangle inequality impliesthe usual one.24=25.Prove that for any nonzero rational number x, we haveII |x|p == 1.

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Answered: Image /qna-images/answer/0c3daa32-8dab-43bd-a390-2cb14001266a.jpg