I need a logical proof of the elementary statement about real numbers using order and field axioms
$((c>0)\wedge(|a|<c))\Rightarrow((-c<a)\wedge(a<c))$
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$\begingroup$ $|a|<c\implies a<c\land -a<c$ $\endgroup$– J. W. TannerCommented Jun 7, 2019 at 17:03
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3$\begingroup$ There is 'completely logical' and 'completely logical' .... How formal does this need to be? What axioms do you have? What inference rules do you have? Most importantly: What have you tried so far yourself? Where are you getting stuck? $\endgroup$– Bram28Commented Jun 7, 2019 at 17:03
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$\begingroup$ @Bram28 a>=0 implies a<c ; a<0 implies -c<a ; then stuck $\endgroup$– bumbaCommented Jun 7, 2019 at 17:16
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$\begingroup$ @bumba Whatever exactly your instructor is looking for, I am pretty sure your instructor is looking for something that explains, in more detail, why a>=0 implies a<c. In fact, it should be clear that a>=0 by itself does not imply a<c. So, what other information are you using here? That's what you need to indicate more precisely. In particular, you'll need to break down this down this reasoning all the way to the basic axioms. $\endgroup$– Bram28Commented Jun 7, 2019 at 17:25
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3$\begingroup$ The interaction between the arithmetic and the linear order on an ordered field is defined by (1) if $x>y$ then $x+z>y+z$ and (2) if $x>y$ and $z>0$ then $xz>yz. $ These are axioms. From them we prove that $-1<0<1$ and if $u<v$ then $-u>-v$.... We may define $|x|=\max (-x,x)$ and therefore $x\le |x|$ and $-x\le |-x|=|x|$.... So if $|x|<c$ then $x\le |x|<c,$ implying $x<c,$ and also $-x\le |-x|=|x|<c $, implying $-x<c$, implying $x=-(-x)>-c$. $\endgroup$– DanielWainfleetCommented Jun 7, 2019 at 17:33
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