Proving order relation of real number

[Lily@pie]

Homework Statement

If a>0, then -a>0. Moreover, if a<b, then -a>-b

Homework Equations

axioms of real numbers can be applied

The Attempt at a Solution

I am not sure what am I suppose to prove. So what I did was trying to prove "if a<b, then -a>-b" given that "If a>0, then -a>0."

I tried the direct proof method:
Suppose a<b, since we are given -a<a. So, -a<a<b. But I don't know how to bring -b in as the question didn't specify anything about b. So do I do 2 cases as in when b>0 and b<0?

I've also tried proof by contradiction:
Assuming a<b and -a<-b
But I don't know how to start ><

Need help! Thanks sooo much

 

[HallsofIvy]

I believe those are two different problems:
1) Prove that: If a> 0 then -a< 0.

2) Prove that: if a> b then -a< -b.

Now, how are "-a" and "-b" defined?

 

[Lily@pie]

So firstly I need to prove this statement "If a>0, -a<0" ?

what do you mean by how is it defined?? Like -a is defined to be <0?

 

[Lily@pie]

Or like -a = -a + 0
and -b = -b + 0
-b = -b + (a+(-a))
-b = (-b+a)+(-a) > -a??

 

[Lily@pie]

Oh! I found out a way but not sure about it

-b = -b+0
-b = -b+(a+(-a))
-b = (-b+a)+(-a)
Since a<b, (-b+a)<0
So, -b = (-b+a)+(-a) < -a
Therefore, -b < -a when a<b...

Is this prove alright??

 

[HallsofIvy]

Or like -a = -a + 0
and -b = -b + 0
-b = -b + (a+(-a))
-b = (-b+a)+(-a) > -a??

Better. -a is defined as the "additive inverse" of a or simply a+ (-a)= 0.

Oh! I found out a way but not sure about it


-b = -b+0
-b = -b+(a+(-a))
-b = (-b+a)+(-a)
Since a<b, (-b+a)<0
So, -b = (-b+a)+(-a) < -a
Therefore, -b < -a when a<b...

Is this prove alright??

Pretty good. Slightly simpler would be
a< b
a+(-a)< b+ (-a)
0< b+ -a

Can you finish that?

Proving "if 0< a then -a< 0" should be simpler.

 

[Lily@pie]

Yup yup ^^

a< b
a+(-a)< b+ (-a)
0< b+ -a
0+(-b)<b+ -a + -b
-b < (b+(-b))+-a
-b<-a

yay! Thanks so much