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#1 2013-04-10 16:30:43

Nehushtan
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Nehushtan’s challenge problems









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#2 2013-04-10 18:15:51

anonimnystefy
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Re: Nehushtan’s challenge problems

#1 is not true. Try x=y=z=-2.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#3 2013-04-10 23:21:32

Nehushtan
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Re: Nehushtan’s challenge problems



Why is it wrong?


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#4 2013-04-16 00:06:41

gAr
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Re: Nehushtan’s challenge problems



Hi,

We have



Adding those two would yield


"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense"  - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."
 

#5 2013-04-16 00:22:03

Nehushtan
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Re: Nehushtan’s challenge problems

Hi gAr.


You apparently assumed
,
,
to be positive. The result has to work for all real numbers, not just positive ones.


Last edited by Nehushtan (2013-04-16 00:25:42)


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#6 2013-04-16 06:28:33

mathdad
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Re: Nehushtan’s challenge problems

x <= y <= z implies Y >= x

thus xz <= yz

and  xz <= yz + xy

thus xz <= xy + yz

 

#7 2013-04-16 08:00:36

anonimnystefy
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Re: Nehushtan’s challenge problems

That only works when all three numbers are positive.


The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
 

#8 2013-04-16 09:54:55

mathdad
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Re: Nehushtan’s challenge problems

Case 1: (same proof as my original answer)
assume x > 0
then, since x<=y<=z, and x > 0, then y > 0 and z > 0
Y >= x, and xz <= yz
and, xz <= yz + xy
thus xz <= xy + yz

Case 2:
assume x = 0,
then, since x<=y<=z, and x = 0, then y >= 0 and z >= 0
xz = 0, xy = 0, and yz >= 0
thus xz <= xy + yz

Case 3:
Assume y = 0,
then, since x<=y<=z, and y = 0, then x <= 0 and z >= 0
thus xz <= 0, xy = 0, and yz = 0
thus, xz <= xy + yz

Case 4:
Assume z = 0,
then, since x<=y<=z, and z = 0, then x <= 0 and y <= 0
xz = 0, xy >= 0, and yz = 0
thus, xz <= xy + yz

Case 5:
Assume z < 0,
then, since x<=y<=z, and z < 0, then x < 0 and y < 0
xz > 0, xy > 0, and yz > 0, (all are positive numbers)
since |x| >= |y| >= |z|
xy >= xz >= yz

Thus, xz <= xy
and, xz <= xy + yz

 

#9 2013-04-16 22:06:50

Nehushtan
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Re: Nehushtan’s challenge problems

Splitting into cases is tedious. The short and sweet solution for #1 is



It looks like no-one is going to get #2 so I might as well post its solution as well.



New challenge problem:




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#10 2013-04-25 08:36:57

Nehushtan
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Re: Nehushtan’s challenge problems

Right, I’ll set a simpler question this time. roll







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#11 2013-05-04 09:49:24

Nehushtan
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Re: Nehushtan’s challenge problems







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