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I can't seem to get the negation, $\lnot (a < b < 0)$, right.

I thought I could break it into 3 parts: a < b, a < 0, b < 0, but that leaves me with a > b or a > 0 or b > 0 (greater or equal)

The exercise question was: For a and b in $\mathbb{R}$, prove:

$(a < b < 0) \implies a^2 > b^2$

I wanted to see what the contrapositive was, and this came up. It would make sense if (b < a < 0), or if (0 < a < b), but I'm not quite sure how to get to this point when just negating (a < b < 0).

What confuses me further is that if I said b is between a and c, it would still be written as $(a < b < c)$; its negation would be b is not between a and c, which could be written as $(b < a < c) \land (b > c > a)$.

So my question is, how is $\lnot (a < b < 0)$ normally negated? Is it just context dependent? In this case, I would just like to know how to express the contrapositive, though I realize you don't have to do so for the proof.

Thanks in advance. Apologies if this is a dumb question.

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    $\begingroup$ Technically, $a<b<0$ is an abuse of notation, meaning $(a<b)\land (b<0)$. So the negation is $(a\geq b)\lor (b\geq 0)$ $\endgroup$ – Thomas Andrews Jun 26 '14 at 8:09
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Let's assume the following axiom:

  • Negatives Reverse Inequalities: If $x < y$ and $z < 0$, then $xz > yz$.

Now suppose that $a < b < 0$. Then since $a < b$ and $a < 0$, it follows from the above axiom that $a^2 > ab$. Likewise, since $a < b$ and $b < 0$, it follows from the above axiom that $ab > b^2$. Combining these two results, it follows by the transitivity of the $>$ relation that $a^2 > b^2$, as desired. $~~\blacksquare$


Although unnecessary, here's the contrapositive (just apply DeMorgan's Law): \begin{align*} (a < b < 0) \implies (a^2 > b^2) &\quad\equiv\quad ((a < b) \land (b < 0)) \implies (a^2 > b^2) \\ &\quad\equiv\quad \neg(a^2 > b^2) \implies \neg((a < b) \land (b < 0)) \\ &\quad\equiv\quad \neg(a^2 > b^2) \implies (\neg(a < b) \lor \neg(b < 0)) \\ &\quad\equiv\quad (a^2 \leq b^2) \implies ((a \geq b) \lor (b \geq 0)) \\ \end{align*}

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We have $$\begin{align} \neg(a < b < 0) &\iff \neg(\,(a < b) \,\&\, (b < 0)\,) \\&\iff (\,\neg(a < b)\,) \vee (\,\neg(b < 0)\,) \\&\iff (a \geq b) \vee (b \geq 0) \,\,. \end{align}$$

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  • $\begingroup$ Wouldn't (a≥b)∨(b≥0) make the statement mean something different in this context? Thanks! $\endgroup$ – user157000 Jun 26 '14 at 8:33
  • $\begingroup$ No it wouldn't...that's what the proof above shows. Use De Morgans Law to obtain the second equivalence @user157000 $\endgroup$ – afedder Jun 28 '14 at 2:56

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