a -ve number greater than infinity?


by AlbertEinstein
Tags: greater, infinity, number
AlbertEinstein
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#1
Jun24-06, 12:32 AM
P: 113
Please follow the following arguments.
5/3=1.66
5/2=2.5
5/1=5
5/0.5=10
.....
....
5/0=infinity
and then 5/(-1)= -5
What you see? As the denominator is decreased the right hand side answer increases. The denominator becomes 3 then 2,then 1, then 0,then -1 ; and the answer increases, therefore -5 must be grater than infinity.
Where's the flaw. Please illustrate.
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neutrino
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#2
Jun24-06, 01:25 AM
P: 2,048
The curve 5/x is an example of a rectangular hyperbola. It's an asymptotic curve.

http://en.wikipedia.org/wiki/Asymptote

http://mathworld.wolfram.com/Asymptote.html
matt grime
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#3
Jun24-06, 02:33 AM
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Quote Quote by AlbertEinstein
and the answer increases, therefore -5 must be grater than infinity.
Where's the flaw. Please illustrate.
In assuming that because for some set of values f(x) is increasing it is always increasing.

AKG
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#4
Jun24-06, 02:48 AM
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a -ve number greater than infinity?


Mainly, your answer doesn't work because 5 is a prime number.
Crusty
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#5
Jun24-06, 07:03 AM
P: 24
As one divides by smaller and smaller numbers, the output figure approaches but does not reach infinity, no?
arildno
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#6
Jun24-06, 07:07 AM
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Depends on your notion of smallness.
gnomedt
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#7
Jun25-06, 12:07 AM
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Please follow the following arguments.

5=5
4=4
3=3
2=2
1=1
0=0
-1=-1

What do you see? As the left hand side decreases, the right hand side is greater than or equal to zero. Therefore, -1 must be greater than or equal to zero.

Ja?

Just because a property holds true for a certain range of numbers, it doesn't mean the pattern will follow for numbers outside that range. That's the flaw.
Gokul43201
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#8
Jun28-06, 10:03 PM
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Quote Quote by AKG
Mainly, your answer doesn't work because 5 is a prime number.
That's funny.
Rach3
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#9
Jun28-06, 11:40 PM
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Quote Quote by Gokul43201
That's funny.
It is?
TriTertButoxy
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#10
Jun28-06, 11:53 PM
P: 194
Ya, its funny because AKG's answer isn't right.
ja ja
daveb
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#11
Jun29-06, 07:21 AM
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5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.
TD
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#12
Jun29-06, 07:51 AM
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Quote Quote by daveb
5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.
In absolute value, yes. But else you have to mind the sign, depending on whether you're approaching 0 from the left or right, you get -inf resp. +inf.
Jahnabi
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#13
Jul3-06, 08:59 PM
P: 5
I don't see how you can say that -1 is greater than or equal to 0. I might be horribly wrong but what I think you're doing is just equating numbers & two equal numbers are always equal .. under no circumstances can be greater than or equal to.

Look at this order:

1=1
9=9
8=8
-1=-1

Don't you see?
DaveC426913
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#14
Jul4-06, 09:57 AM
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Albert:

Please follow the following arguments.
(x^2=y)
3^2=9
2^2=4
1^2=1
0^2=0
-1^2=1

What you see? As the x value is decreased, the y value (right hand side answer) decreases. The x value becomes 3 then 2,then 1, then 0,then -1 ; and the answer (y) decreases, therefore 1 must be less than zero.

Now where's the flaw?


(Hint: the only flaw is the conclusion that a given function must result in a straight and continuous line.)
Robokapp
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#15
Jul6-06, 08:32 PM
P: 218
Okay...let me try to hit this topic.
Graphing 5/x you will have a line that going from negative infinity to positive infinity it ALWAYS DECREASES. it never increases. However, it starts out negative and ends up positive.

Lesson learned today: Don't mess with the division by zero.
d_leet
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#16
Jul6-06, 10:31 PM
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Quote Quote by Robokapp
Okay...let me try to hit this topic.
Graphing 5/x you will have a line that going from negative infinity to positive infinity it ALWAYS DECREASES. it never increases. However, it starts out negative and ends up positive.

Lesson learned today: Don't mess with the division by zero.
No you do not have a line. You have a hyperbola in the first and third quandrants of the cartesian plane.
Tyris
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#17
Jul8-06, 12:12 PM
P: 26
Here am a graph of 1/x.
clicky
5/x follows the same pattern.


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