# Intro to Analysis

• Sep 6th 2009, 10:53 PM
haikuambulance
Intro to Analysis
I just started this class and for some reason the material isn't clicking at all. I was wondering if anyone could give me some sort of hint on how to approach this problem. I would GREATLY appreciate it.

http://img.photobucket.com/albums/v1...e/homework.jpg
• Sep 7th 2009, 04:06 AM
mr fantastic
Quote:

Originally Posted by haikuambulance
I just started this class and for some reason the material isn't clicking at all. I was wondering if anyone could give me some sort of hint on how to approach this problem. I would GREATLY appreciate it.

http://img.photobucket.com/albums/v1...e/homework.jpg

I will do the first bit:

\$\displaystyle x = qx + (1 - q) x < qx + (1 - q)y\$ for \$\displaystyle x < y\$.

\$\displaystyle y = qy + (1 - q) y > qx + (1 - q)y\$ for \$\displaystyle x < y\$.

Therefore \$\displaystyle x < qx + (1 - q)y < y\$ for \$\displaystyle x < y\$.
• Sep 7th 2009, 04:54 AM
HallsofIvy
Here's how I would do it. Since x< y, y- x> 0. Since q< 1, subtracting q from each part, 0< 1-q. Since 0< q, -q< 0 and 1- q< 1. Setting p= 1- q, we also have 0< p< 1. multiplying each part by the positive number y- x, 0(y-x)< p(y-x)< 1(y-x) or 0< py- px< y- x. Adding x to each part, x< py- px+ x< y. -px+ x= (1- p)x= qx and py= (1-q)y so that is x< (1- q)y+ qx< y.

(I first did that using "q" rather than "p" and arrived at x< qy+ (1-q)x< y which is true but not what is wanted. That when I switched to p= 1- q and did the same proof.)