1. ## Addition of inequalities for real numbers

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

2. Originally Posted by Alfie
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
x < y
+ w < z

x + w < y + z

3. Yes, but is it possible to prove that one can add inequalities like that or is it axiomatic?

4. Originally Posted by Alfie
Yes, but is it possible to prove that one can add inequalities like that or is it axiomatic?
Yes it is possible. But we do not know the axioms and definitions.
It is sometimes done this way.
There is a set $\mathbb{P}\subseteq\mathbb{R}$ such that it is closed with respect to addition and multiplication. For any $x\in\mathbb{R}$ exactly one of these is true: $x=0,~x\in\mathbb{P} \text{ or }-x\in\mathbb{P} .$

Then we define this $x>y\text{ if and only if }(x-y)\in\mathbb{P}$.

Can you prove it from that?

5. Originally Posted by Alfie
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
We have $y = x+(y-x) = x+r_1$ and $z = w+(z-w) = w+r_2$ where $r_1,r_2>0$

Thus $y+z = x+w+r_1+r_2 > x+w$

I believe this reduces your problem to $a,b>0 \implies a+b > 0$

6. Thanks for the help, everyone.

7. 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.