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In lecture I learned the following definition of valuation: Let $(K,+,\cdot)$ be a field and let $(G,+)$ be a totally ordered group. A map $v: K \longrightarrow G\cup\{\infty\}$ is a valuation if the following properties hold:

  • $v(ab) = v(a)+v(b)$
  • $v(a+b) \ge \min\{v(a),v(b)\}$
  • $v(a) = \infty \iff a = 0$

Then we proved that:

  1. $v(1) = 0$
  2. $v(a^{-1}) = -v(a)$
  3. $v(-a) = v(a)$
  4. $v(a - b) \ge \min\{v(a), v(b)\}$

Later in the lecture we used the "easy fact" that

$$ \text{If }v(a) \ne v(b), \text{ then } v(a+b) = \min\{v(a), v(b)\}$$ I do not understand why this is true. I looked on Wikipedia and there they define a valuation to so that it has this property. But is it possible to derive this from my definition?

user26857
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3nondatur
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1 Answers1

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If $v(a) >v(b)$ and $v(a+b) > v(b)$ then $$v(b)= v(a+b-a)\ge \min(v(-a),v(a+b))> v(b)$$

reuns
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  • You can use Approach0 which allows the search of mathematical expressions, and typing $\nu(a+b) = \min(\nu(a),\nu(b))$ would give you a good look at candidate duplicates for this question (since there's no attempt, this can be called a duplicate if it's been answered before). You can then mark as duplicate as move on. The duplicate found above is on the first page of the Approach0 search list. Next time, if you think a question is a duplicate (note : I think abstract-algebra and elementary-number-theory must always be checked IMO) then you can use this website. – Sarvesh Ravichandran Iyer Apr 30 '21 at 18:59