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A -ve number greater than infinity?

  1. Jun 24, 2006 #1
    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.
     
  2. jcsd
  3. Jun 24, 2006 #2
  4. Jun 24, 2006 #3

    matt grime

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    In assuming that because for some set of values f(x) is increasing it is always increasing.
     
  5. Jun 24, 2006 #4

    AKG

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    Mainly, your answer doesn't work because 5 is a prime number.
     
  6. Jun 24, 2006 #5
    As one divides by smaller and smaller numbers, the output figure approaches but does not reach infinity, no?
     
  7. Jun 24, 2006 #6

    arildno

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    Depends on your notion of smallness.
     
  8. Jun 25, 2006 #7
    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.
     
  9. Jun 28, 2006 #8

    Gokul43201

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    :rofl: :rofl: That's funny.
     
  10. Jun 28, 2006 #9
    It is? :confused:
     
  11. Jun 28, 2006 #10
    Ya, its funny because AKG's answer isn't right.
    ja ja
     
  12. Jun 29, 2006 #11
    5/0 does not equal infinity. The limit of 5/x as x approaches 0 equals inifinty.
     
  13. Jun 29, 2006 #12

    TD

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    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.
     
  14. Jul 3, 2006 #13
    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?
     
  15. Jul 4, 2006 #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: Jul 4, 2006
  16. Jul 6, 2006 #15
    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.
     
  17. Jul 6, 2006 #16
    No you do not have a line. You have a hyperbola in the first and third quandrants of the cartesian plane.
     
  18. Jul 8, 2006 #17
    Here am a graph of 1/x.
    clicky
    5/x follows the same pattern.
     
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