Proving an Inequality Involving Real Numbers

[Julio1]

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

Spoiler

Hello, any idea for the proof? :) Thanks

 

[mathbalarka]

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 $b^{-1} > a^{-1}$.

 

[Julio1]

Thanks :)

One question, exist other form of proof?

 

[mathbalarka]

Spoiler

[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.