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#1 2006-03-05 10:54:30

razor
Member
Registered: 2006-02-24
Posts: 6

Help

We have the real  positive(only) numbers space of dimension n.We have also a relation of two members(sorry),continuous,absoletuly monotonic(only >),cursive,in this space.
Also x,y are arrays and x>y in this space and 0<a<1.
How can we proove that  a*x+(1-a)*y>y ?.   (*:  multiplication)

(G. Aliprantis, D. Brown and O. Burhinshaw, Existence and Optimality of Competitive Equilibrium, Springer-Verlag, 1990.)

sorry for my english.

Last edited by razor (2006-03-05 10:59:29)

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#2 2006-03-05 12:55:23

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Help

ax + (1-a)y > ay + (1-a)y = y(a + 1 - a) = y
ax + (1-a)y > y

All that needed was x > y.

Last edited by Ricky (2006-03-05 12:56:13)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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