Prove that if 
and 
, then 
.
Proof.
By the transitivity of
(Theorem I.17), we have that if 
and 
, then 
.Then, if
and we have by substitution.If
and , then by substitution.If
and , then 
by transitivity of the 
relation. Hence, by definition of 
.Thus, in all cases
and implies 
Suppose a\leq b and b\leq c, yet c<a.
c0\Leftrightarrow a+(b-b)-c>0
a+(b-b)-c>0\Leftrightarrow-[(b-a)+(c-b)]>0
(Exercises 7 and 6, 1.3.3)
If a=b and b=c then b-a=0 and c-b=0 which implies
-[(b-a)+(c-b)]=-[0+0]=-0=0>0
(using axiom 4, and exercise 2, 1.3.3)
but 0>0 is a contradiction
If either a<b or b<c then -[(b-a)+(c-b)]<0
this is a contradiction of axiom 8.
If both a<b and b<c then by transitivity a<c.
Contradiction of supposition c<a
Hence, when a\leq b and b\leq c it must be that a\leq c
QED.
Hey Rori, sorry i am not very familiar with latex code
i typed up an alternate soln in Lyx and cut and pasted its contents here.
Any ideas how to integrate the symbols?