# Order of Numbers Proofs

• April 28th 2009, 08:29 PM
noles2188
Order of Numbers Proofs
Prove that if u < v and x < y, then u+x < v+y.

Also if x < y and z < 0, then y*z < x*z.
• April 28th 2009, 08:36 PM
Gamma
Quote:

Originally Posted by noles2188
Prove that if u < v and x < y, then u+x < v+y.

Also if x < y and z < 0, then y*z < x*z.

I dunno what axioms you can use.
$0 < v-u$
$0 < y-x$
$0< v-u+y-x = (v+y) - (u+x)$
$u+x < v+y$

$0
$0(z)>(y-x)z$
$0>yz-xz$
$xz>yz$