# Is x<=ky equivalent to x<y?

by tragicmuffin
Tags: analysis, contraction mapping, lipschitz
 Share this thread:
 P: 5 I have a question on something that has come up many times for me in other contexts, such as Lipschitz transforms, and recently contraction mappings and the fixed point theorem. Is the following statement always false, and why? Let x and y be two positive real numbers. $\exists k<1 \,\, s.t. \,\, x \leq k\cdot y \, \Leftrightarrow \, x  P: 3,002 is k also a positive real?  P: 5 Sorry, yes. 0  P: 3,002 Is x<=ky equivalent to x  P: 5 I guess the thing that I don't understand is that if k is strictly less than 1, can't we still make k as close to 1 as we want? Then wouldn't saying [itex]x \leq k \cdot y$ where k is arbitrarily close to 1 (but not equal) be the same as saying $x<1 \cdot y$? It seems to me that changing the $\leq$ to $<$ would account for k not being equal to 1, only close. In the example with the lines, the same idea would hold. We can make k arbitrarily close to 1, and thus make the line x=ky arbitrarily close to the line x=y.
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,569
 Quote by tragicmuffin I have a question on something that has come up many times for me in other contexts, such as Lipschitz transforms, and recently contraction mappings and the fixed point theorem. Is the following statement always false, and why? Let x and y be two positive real numbers. [itex]\exists k<1 \,\, s.t. \,\, x \leq k\cdot y \, \Leftrightarrow \, x
If k< 1 then, mutiplying both sides by the positive number y, ky< y. So if x< ky, it is certainly true that x< y.

Conversely, if x< y, then x/y< 1 and so there exist k such that x/y< k< 1. That is, we have k< 1 and, multiplying both sides of x/y< k, x< ky.

 Related Discussions Introductory Physics Homework 1 Calculus & Beyond Homework 3 Engineering, Comp Sci, & Technology Homework 1 Math & Science Software 6 General Math 8