Question For Mathematicians

[Bagpuss]

Isn't that one of the fundamental assumptions of maths?

[Ollie]

If x>0 and y>0

then 0<x and 0<y

Therefore x+y>0 and 0<x+y


Also 0+0<x and 0+0<y

Therefore x>0+0 and y>0+0

So x+y is also > 0+0

Got it?

[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??

[wendigo100]

What do those little arrow things mean again?

[BlasterBates]

no no no


Given x>0 y> 0

then ....


tch

[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.

[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*.

[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!!

[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).

[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.