how to prove that:
1) (a<b) ^ (b<c) => a<c
I tried this:
b = a +x , x € Q+
b = a + x < c => a<c
2) a>b => a+c > b+c
3) (a>b) ^ (c>0) => ac > bc
4) (a>b) ^ (c<0) => ac < bc
Any help is much appreciated
There are many different ways to define inequality. I presume the definition you are using is "x< y if and only if there exist a positive number z such that x+ z= y".
Okay, if a< b then there exist positive x such that a+x= b. Since b< c, there exist positive y such that b+ y= c. No, you cannot simply say "a+ x< c therefore a< c" unless you quote a theorem to that effect. And it looks to me like that is the theorem you are trying to prove here!
What you can do is replace b in "b+ y= c" by a+ x. Then a+x+ y= a+ (x+y)= c. Since the sum of two positive numbers is positive, that says that a< c.
You've switched from "<" to ">". That confused me!2) a>b => a+c > b+c
So there exist positive x such that b+ x= a. Now add c to both sides of that equation.
So there exist positive x such that b+ x= a. Multiply both sides by c to get c(b+ x)= bc+ cx= ac. And note that the product of positive numbers is positive.3) (a>b) ^ (c>0) => ac > bc
If c< 0 then there exist positive x such that c+ x= 0. But -c is the unique number such that c+(-c)= 0 so x= -c and we have proved that -c is positive.4) (a>b) ^ (c<0) => ac < bc
Therefore, by (3), a(-c)> b(-c). That is, there exist positive x such that -bc+ x= ac. Now add ac and bc to both sides.