1
$\begingroup$

I wish to prove the following inequality: $a\leq b\wedge b\leq c\Longrightarrow a\leq c$.

I tried the following proof.

Proof:

From the trichotomy law we have,that $a\leq b\Longrightarrow\neg(b<a)$.

Also $b\leq c\Longrightarrow\neg(c<b)$.

But : $\neg(c<b)\Longrightarrow\neg(c<b)\vee\neg(b<a)\Longrightarrow\neg(c<b\wedge b<a)$, by using De Morgan.

And, since $c<b\wedge b<a\Longrightarrow c<a$, we have $\neg(c<a)\Longrightarrow a\leq c$ ,by using the trichotomy law.

$\endgroup$
1
  • $\begingroup$ It seems like something that one would prove from axioms. What are the axioms you are using? $\endgroup$
    – Wojowu
    Oct 26, 2015 at 18:24

1 Answer 1

Reset to default
1
$\begingroup$

From $\neg(c<b\wedge b<a),$ we cannot directly conclude that $\neg(c<a).$ Rather, the reverse implication is equivalent to transitivity.

A better way to proceed, I'd say, is to use/prove that $$x\le y\iff(x=y\vee x<y)$$ for real numbers $x$ and $y$. This will allow you to proceed directly by cases, each of which is almost trivial.

$\endgroup$
2
  • $\begingroup$ Why can,t we directly conclude that $\neg(c<a)$ $\endgroup$
    – chris
    Oct 27, 2015 at 9:57
  • 1
    $\begingroup$ Well, for example, $\neg(2<1\wedge1<3)$ is true, but $\neg(2<3)$ does not follow. $\endgroup$
    – Cameron Buie
    Oct 27, 2015 at 12:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .