Question For Mathematicians

[bored]

I assumed a strict total order, i.e. asymmetric and transitive. But it does indeed have to be at least a group. Surely though the proof for a strictly stronger space like a ring or a field would be just the same as for a group? Only the basic group properties are being used here. (note to the less technical: what I mean by that is that a proof relating to addition and subtraction is just the same whether you are able to do multiplication or not).

Your proof indeed almost derives from the axioms - you haven't proved that if x>0, then -x<0.

A simple proof that starts from the axioms would be like this:

a) x > 0 => x + y > 0 + y = y
b) x + y > y, y > 0 => x + y > 0

[Euro-commuter]

Your proof indeed almost derives from the axioms - you haven't proved that if x>0, then -x<0.

Implicit in the definitions of unity and inverse, no?

What you prove on the way, and what is taken as obvious, has always intrigued me. Technically, all of mathematics is tautologically implicit in the axioms. A "proof" is an exposition of the steps in detail small enough for mathematicians to follow, and preferably too great for others to follow .

[Cowboy Bob]

What if > is overloaded C++ style...

[bored]

Implicit in the definitions of unity and inverse, no?

What you prove on the way, and what is taken as obvious, has always intrigued me. Technically, all of mathematics is tautologically implicit in the axioms. A "proof" is an exposition of the steps in detail small enough for mathematicians to follow, and preferably too great for others to follow .

The statement is not part of the axioms, so it must be proved. The proof is trivial but so is the proof to the original question.

x > 0 => x + (-x) > 0 + (-x) => 0 > -x QED

[sasguru]

My A-level maths is not good enough to follow some of these arguments.

[BlasterBates]

ok ok I'll rephrase it

R is the set of real numbers

so what we are proving is the following

The set X = {x:R | x > 0}

for all x1, x2 in X the following is true x1 + x2 > 0

[richard-af]

ok ok I'll rephrase it

R is the set of real numbers

so what we are proving is the following

The set X = {x:R | x > 0}

for all x1, x2 in X the following is true x1 + x2 > 0

Just when I thought I'd learnt the ruddiments of awk, this pops up.