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Let $\frac{m_1}{n_1}=\frac{m_2}{n_2}$ and $\frac{p_1}{q_1}=\frac{p_2}{q_2}$.

Let $\frac{a}{b}\,{\le}\,\frac{c}{d}{\iff}(bd<0\,{\land}\,cb\,{\le}\,ad)\,{\lor}\,(0<bd\,{\land}\,ad\,{\le}\,cb)$.

I have already proved that for the latter case, the ordering is well-defined. Now I'm trying to prove the former and the result I'm achieving is that $\frac{m_1}{n_1}\,{\le}\,\frac{p_1}{q_1}{\land}\frac{m_2}{n_2}\,{\le}\,\frac{p_2}{q_2}$ is never true unless all the fractions are equal. The proof is divided into two cases: $1) \,0<n_2q_2$ and $2)\,n_2q_2<0$. Since integers are totally ordered and $n_2q_2$ can't equal $0$, one of these must be true.

By assumption we have $n_1p_1\,{\le}\,m_1q_1$.

$1)\,0<n_2q_2$

$n_1p_1(n_2q_2)\,{\le}\,m_1q_1(n_2q_2)$

$n_1n_2(p_1q_2)\,{\le}\,q_1q_2(m_1n_2)$

By equality of $\frac{m_1}{n_1}$ and $\frac{m_2}{n_2}$, $m_1n_2=m_2n_1$. In a similar manner, $p_1q_2=p_2q_1$.

$n_1n_2(p_2q_1)\,{\le}\,q_1q_2(m_2n_1)$

$(n_1q_1)n_2p_2\,{\le}\,(n_1q_1)m_2q_2$

By definition of denominators, $n_1q_1$ is cancellable under multiplication.

$p_2n_2\,{\le}q_2m_2$

By definition of total ordering under a $bd<0$ case, if $\frac{m_2}{n_2}\,{\le}\,\frac{p_2}{q_2}$ then $m_2q_2\,{\le}\,p_2n_2$. Under the above proof, the only case when this could be true is $p_2n_2=m_2q_2$.

In the other possible case, we repeat the proof above but with $n_2q_2$ replaced by $(-1)n_2q_2$. The result is therefore $q_2m_2\,{\le}\,p_2n_2$. In this case, however, if $\frac{m_2}{n_2}\,{\le}\,\frac{p_2}{q_2}$ was true we would expect the opposite - once again the only way this could happen is if $p_2n_2=m_2q_2$

My result can't possible be correct - $\frac{1}{-2}\,{\le}\,\frac{-1}{4}$ falls under the $bd<0$ case of ${\le}$, since $(-1)*(-2)\,{\le}\,1*4$. By the above proof, if we take any rational numbers equivalent to these two - including $\frac{1}{-2}$ and $\frac{-1}{4}$ themselves, we should arrive at the fact that their ${\le}$ relation implies that $(-1)*2=1*(-4)$ which is false.

The definition of the relation can't be at fault here - we can replace ${\le}$ with ${\oplus}$ and by its definition $\frac{1}{-2}\,{\oplus}\,\frac{-1}{4}$ would still be true and so would be my proof (assuming it's consistent) - leading to $-2=-4$.

So what exactly went wrong here?

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  • $\begingroup$ In case 1) you have written : $0 < n_2 q_2$ but in the previous line it is the other way round... $\endgroup$ – Mauro ALLEGRANZA Jan 6 '17 at 17:19
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    $\begingroup$ In case 1) what about $n_1 q_1$ ? In line 7 you divide both terms by it, but what if it is negative ? $\endgroup$ – Mauro ALLEGRANZA Jan 6 '17 at 17:20

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