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

correct?

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

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- Oct 11th 2009, 04:23 AMmetlxMore proofs
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

correct?

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 - Oct 11th 2009, 06:01 AMHallsofIvy
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.

Quote:

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.

Quote:

3) (a>b) ^ (c>0) => ac > bc

Quote:

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.