How do you prove the addition of inequalites property for real numbers:
If x < y and w < z, then x + w < y + z, for all real numbers w,x,y,z.
Thanks
Yes it is possible. But we do not know the axioms and definitions.
It is sometimes done this way.
There is a set $\displaystyle \mathbb{P}\subseteq\mathbb{R}$ such that it is closed with respect to addition and multiplication. For any $\displaystyle x\in\mathbb{R}$ exactly one of these is true: $\displaystyle x=0,~x\in\mathbb{P} \text{ or }-x\in\mathbb{P} .$
Then we define this $\displaystyle x>y\text{ if and only if }(x-y)\in\mathbb{P} $.
Can you prove it from that?
One of the defining properties of an "ordered field" is that if a< b then, for any number c, a+ c< b+ c.
Now, if you have x< y and w< z, then you can start with x+ w< y+ w from the property I just stated. If then from w< z, w+ y< z+ y. Finally, you use the "transitivity" of inequality, "if a< b and b< c then a< c" to argue that since x+ w< y+ w and w+ y< z+ y, you get x+ w< z+ y= y+ z.