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.
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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. $\displaystyle 0 < v-u$ $\displaystyle 0 < y-x$ $\displaystyle 0< v-u+y-x = (v+y) - (u+x)$ $\displaystyle u+x < v+y$ $\displaystyle 0<y-x$ $\displaystyle 0(z)>(y-x)z$ $\displaystyle 0>yz-xz$ $\displaystyle xz>yz$
Last edited by Gamma; Apr 28th 2009 at 08:37 PM. Reason: Quote
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