Proving the Inequality: 0<A<B Implies 1/B<1/A

[solakis1]

Is the following proof ,proving ; \(\displaystyle \forall A\forall B[0<A<B\Longrightarrow \frac{1}{B}<\frac{1}{A}]\) correct??

PROOF:

Let, 0<a<b................1

Let \(\displaystyle \frac{1}{a}<\frac{1}{b}\)

Let,\(\displaystyle \frac{1}{a}=\frac{1}{b}\)

But for \(\displaystyle \frac{1}{a}=\frac{1}{b}\Longrightarrow\neg(\frac{1}{a}<\frac{1}{b})\)

Hence we have: \(\displaystyle \frac{1}{a}<\frac{1}{b}\wedge\neg(\frac{1}{a}<\frac{1}{b})\). a contradiction

Thus \(\displaystyle \neg(\frac{1}{a}=\frac{1}{b})\)..................2

For \(\displaystyle \frac{1}{a}<\frac{1}{b}\Longrightarrow (ab)\frac{1}{a}<(ab)\frac{1}{b}\Longrightarrow b<a(\),since a,b are different from zero and 0<ab

And by using (1) b<a& a<b => b<b = b<b & ~(b<b) ,a contradiction

Thus \(\displaystyle \neg(\frac{1}{a}<\frac{1}{b})\)..................3

Therefor ,from (2) and (3) we have: \(\displaystyle \neg(\frac{1}{a}=\frac{1}{b})\).and \(\displaystyle \neg(\frac{1}{a}<\frac{1}{b})\) =>\(\displaystyle \neg((\frac{1}{a}=\frac{1}{b})\vee(\frac{1}{a}<\frac{1}{b}))\Longrightarrow\frac{1}{b}<\frac{1}{a}\),by using the trichotomy law

 

[Valkarie]

Therefore, for all A,B\in \mathbb{R}[0<A<B\Longrightarrow \frac{1}{B}<\frac{1}{A}]No, this proof is not correct. It does not make logical sense and the chain of reasoning is very hard to follow. Additionally, you never actually prove that "for all A,B\in \mathbb{R}[0<A<B\Longrightarrow \frac{1}{B}<\frac{1}{A}]".