Proving an Inequality Involving Real Numbers

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Discussion Overview

The discussion revolves around proving the inequality involving real numbers, specifically that if \( a, b \in \mathbb{R}^{+} \) and \( a > b \), then it follows that \( a^{-1} < b^{-1} \). The scope includes mathematical reasoning and proof techniques.

Discussion Character

  • Mathematical reasoning
  • Exploratory

Main Points Raised

  • One participant presents a proof sketch that involves showing the natural order on \( \mathbb{R} \) is preserved under multiplication by positive constants.
  • The proof sketch suggests using specific values for \( c \) to manipulate the inequality and reach the conclusion.
  • Another participant expresses interest in alternative forms of proof, indicating a desire for exploration beyond the standard approach.
  • A later reply questions the existence of other proof methods, suggesting uncertainty about the uniqueness of the presented proof.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether alternative proofs exist, with some expressing uncertainty and others suggesting the standard proof may be the only one.

Julio1
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If $a,b\in \mathbb{R}^{+}.$ Show that $a>b\implies a^{-1}<b^{-1}.$

Hello, any idea for the proof? :) Thanks
 
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A sketch of the proof : First show that the natural order on $\Bbb R$ is fixed under multiplication, i.e., if $a > b$ for $a, b \in \Bbb R$ then $ca > cb$ for any real $c > 0$ (let $a = b + k$ for some positive nonzero real $k$, where $a, b \in \Bbb R^{+}$). Conclude the proof by multiplying by $c = 1/b$ first to attain $a \cdot b^{-1} > 1$ and then multiplying by $c' = 1/a$ to get $b^{-1} > a^{-1}$.
 
Thanks :)

One question, exist other form of proof?
 
[JOKE] Sure, we know that $\log(x) = \displaystyle \int_1^x \log(t) \mathrm{d} t$ and as $1/t > 0$ for $t \in \Bbb R^+$, $\log$ is monotone increasing on $(0, \infty)$. Hence if $a > b$, $\log(a) > \log(b)$ which implies $-\log(a) < -\log(b)$ thus $\log(a^{-1}) < \log(b^{-1})$ from which we have $a^{-1} < b^{-1}$. [/JOKE]

One question, exist other form of proof?

I am not sure. The one I gave above is the standard one, I fancy.
 

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