Solution

The common topology determines which elements have absolute value below one, because

∣ x ∣_{i} < 1 ⟺ x^{n} ⟶ 0

in that topology. It therefore also determines all comparisons:

∣ x ∣_{i} < ∣ y ∣_{i}

exactly when

∣ x / y ∣_{i} < 1

Choose

a \in K^{\times}

with

∣ a ∣_{1} > 1

; then also

∣ a ∣_{2} > 1

. For every

x \in K^{\times}

m, n

, the preceding observation gives

∣ x ∣_{1}^{n} < ∣ a ∣_{1}^{m} ⟺ ∣ x ∣_{2}^{n} < ∣ a ∣_{2}^{m} .

Thus the two real numbers

\frac{l o g ∣ x ∣ _{1}}{l o g ∣ a ∣ _{1}} and \frac{l o g ∣ x ∣ _{2}}{l o g ∣ a ∣ _{2}}

have the same rational upper cuts and are equal. Setting

c = \frac{l o g ∣ a ∣ _{1}}{l o g ∣ a ∣ _{2}} > 0

gives

∣ x ∣_{1} = ∣ x ∣_{2}^{c}

for every

x

Solved by gpt-5.6-sol high.

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