Reply to: Question For Mathematicians

[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.

[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

[sasguru]

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

[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

[Cowboy Bob]

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

[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 .

[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

[OwlHoot]

Simple question fo mathematicians

If x > 0 and y > 0

How do you formally prove x+y > 0

Assuming x and y are real numbers, it follows from the fact that the set of real numbers is an ordered field

As someone else mentioned (I forget who), x > 0 implies x + y > y for any real y (not necessarily positive).

Also by the transitive property of a strict total order if z > y and y > t then z > t, where in particular you can take z = x + y and t = 0.

[Euro-commuter]

The exact proof depends on what axioms you assume for the "<" ordering. Say the proof will be different if you start with the axioms of an ordered group or an ordered field.

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).

[SallyAnne]

The exact proof depends on what axioms you assume for the "<" ordering. Say the proof will be different if you start with the axioms of an ordered group or an ordered field.

Ah yes, the axioms!!! I was going to say that!!

[scotspine]

The exact proof depends on what axioms you assume for the "<" ordering. Say the proof will be different if you start with the axioms of an ordered group or an ordered field.

at last. someone else who *knows*.

[bored]

The exact proof depends on what axioms you assume for the "<" ordering. Say the proof will be different if you start with the axioms of an ordered group or an ordered field.

[BlasterBates]

no no no


Given x>0 y> 0

then ....


tch

[wendigo100]

What do those little arrow things mean again?

[Euro-commuter]

Simple question fo mathematicians

If x > 0 and y > 0

How do you formally prove x+y > 0

Assume the opposite is true, i.e. there exist x and y such that
x > 0
y > 0
x+y <= 0.

then x+y-y <= 0-y
i.e. x <= -y
but y > 0 so -y < 0
therefore x < 0 which contradicts first premise.

Therefore the assumption is false, so there do not exist x any y such that
x > 0
y > 0
x+y <= 0.
Therefore x+y > 0.



PS just how many of us claimed to have a maths degree??