A -ve number greater than infinity?

  • #1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #3
matt grime
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AlbertEinstein said:
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.
 
  • #4
AKG
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Mainly, your answer doesn't work because 5 is a prime number.
 
  • #5
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As one divides by smaller and smaller numbers, the output figure approaches but does not reach infinity, no?
 
  • #6
arildno
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Depends on your notion of smallness.
 
  • #7
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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.
 
  • #8
Gokul43201
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AKG said:
Mainly, your answer doesn't work because 5 is a prime number.
:rofl: :rofl: That's funny.
 
  • #9
Rach3
Gokul43201 said:
:rofl: :rofl: That's funny.
It is? :confused:
 
  • #10
Ya, its funny because AKG's answer isn't right.
ja ja
 
  • #11
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5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.
 
  • #12
TD
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daveb said:
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.
 
  • #13
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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?
 
  • #14
DaveC426913
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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.)
 
Last edited:
  • #15
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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.
 
  • #16
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Robokapp said:
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.
 
  • #17
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Here am a graph of 1/x.
http://www.mathsrevision.net/gcse/1overx.gif [Broken]
5/x follows the same pattern.
 
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