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

*Related*

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?