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Is x<=ky equivalent to x<y? 
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#1
Jan1413, 02:03 PM

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. [itex]\exists k<1 \,\, s.t. \,\, x \leq k\cdot y \, \Leftrightarrow \, x<y[/itex] So far, I've always seen statements like this to be false, but I'm not sure why. Thanks, Jesse 


#2
Jan1413, 02:27 PM

P: 2,812

is k also a positive real?



#3
Jan1413, 02:27 PM

P: 5

Sorry, yes. 0<k<1.



#4
Jan1413, 02:38 PM

P: 2,812

Is x<=ky equivalent to x<y?
I cant say if its true or not but try to visualize it on a xy plane where the area under the line x=y is x<y versus the line x=ky.
Perhaps that will help you decide. 


#5
Jan1413, 10:18 PM

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[/itex] where k is arbitrarily close to 1 (but not equal) be the same as saying [itex]x<1 \cdot y[/itex]?
It seems to me that changing the [itex]\leq[/itex] to [itex]<[/itex] 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. 


#6
Jan1413, 10:33 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,353

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. 


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