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What is the largest number less than 1?

  1. Nov 26, 2003 #1
    How can you represent the largest # that is less than 1?
     
  2. jcsd
  3. Nov 26, 2003 #2
    lim x
    x-->1
     
  4. Nov 26, 2003 #3
    Can you represent that in decimal form?
     
  5. Nov 26, 2003 #4
    Wouldn't .9 (with a line over the 9, sorry, I haven't read the "How to post math functions" thread) be sufficient?

    __
    .9
     
  6. Nov 26, 2003 #5
    The line over it means infinity.

    So couldn't you also say .999... ?
     
  7. Nov 26, 2003 #6

    Hurkyl

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    Among the real numbers (or the rational numbers, or the irrational numbers) there is no largest number less than 1.

    Among the integers, 0 is the largest number less than 1.
     
  8. Nov 26, 2003 #7
    How can there be no largest # less than 1? that doesn't make any sense.
     
  9. Nov 26, 2003 #8

    Hurkyl

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    Why would you think there is a largest?

    Here's a short proof there isn't a largest number less than 1:

    Assume that there is a largest number less than 1. Let's represent this number by [tex]\inline{x}[/tex].

    Now, consider the number [tex]\inline{y=(1+x)/2}[/tex]

    First, notice that [tex]\inline{y>x}[/tex]:

    [tex]
    y = \frac{1+x}{2} > \frac{x+x}{2} = x
    [/tex]

    Now, notice that [tex]\inline{y<1}[/tex]:

    [tex]
    y = \frac{1+x}{2} < \frac{1+1}{2} = 1
    [/tex]

    So consider carefully what we have proven:

    If we assume there is a largest number less than 1, we can find a number less than 1, yet larger than the largest number less than 1.

    That is a contradiction; our assumption that there is a largest number less than 1 must be false.
     
  10. Nov 26, 2003 #9
    Hurkyl,
    I understand your reasoning, and it does make complete sense...

    I do have a question, however..
    Is there a number greater than .999... and less than 1?
    How would you represent that number?
     
  11. Nov 26, 2003 #10

    chroot

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    The number [tex]\inline{0.99\overline{9}}[/tex] is equal to one.

    The limit [tex]\inline{\lim_{x \rightarrow 1} x}[/tex] also equals one.

    Hurkyl is correct; there is no largest real number less than one. No matter how many nines you put in a row, I can make a number with more. In the limit as the number of nines reaches infinity, the resulting quantity is one.

    - Warren
     
  12. Nov 27, 2003 #11
    I can't believe that there is no largest # less than 1. There has to be, in theory.

    If you can say that .999... = 1 then how can you not in mathamatics represent the largest # that is less than 1?

    the .999... = 1 rule only works because you NEVER reach the end of infinity.

    I like sound laws and such, and I love math and physics... and at the same time am a hard core programmer at heart. Logic is key to me.

    I could never write .999... = 1, I could say 1 = 1 and .999... = .999... but not 1 = .999...

    I would say that .999... is the largest # less than 1 because as you go out there in decimal places whatever place your at you can just add a .----1 to that to achieve your value of 1... but yea, you'll never even approach infinity cause it extends forever. Damn you infinity!
     
  13. Nov 27, 2003 #12

    krab

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    You may love math, but you don't know what it is if you don't realize that math is based on rigor. If you want to hold on to your opinion about there being a largest number less than 1, then you must find a fault with Hurkyl's proof.

    If you don't understand Hurkyl's proof, here's another thing to think about. Consider the mapping

    [tex]y={1\over 1-x}[/tex]

    and for x, use the real number interval [0->1) (This means, all the real numbers from 0 to 1, but excluding 1.) This interval gets mapped to [1->infinity). You will see that there being no largest real number x less than 1 is analogous to saying there is no largest real number y. (I am assuming you agree that there is no largest real number.)
     
  14. Nov 27, 2003 #13

    jcsd

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    All the proofs so far have been perfectly adequate, but here's another one:

    for a nunber x where:

    [tex]0 < x < 1[/tex]

    we know that:

    [tex]0 < \sqrt{x} < 1[/tex]

    and

    [tex] x < \sqrt{x}[/tex]

    Lets say that there is a largest number between 0 and 1, what is it's sqaure root? if it greater than it's square root, it is greater than 1, if it is equal to it's square root it is equal to 1 and if it's square root is greater than itself then we have generated a number that is larger than the largest number less than 1 so it can't be the largest number less than 1!
     
  15. Nov 27, 2003 #14

    HallsofIvy

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    Logic is key? Your last sentence has no logic in it at all. You seem to be confusing logic with handwaving. In particular what is your DEFINITION of .999...?
     
    Last edited: Apr 21, 2011
  16. Nov 27, 2003 #15
    Sorry, the largest WHOLE number less than 1 is, by definition, 0.
    .9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 etc. infinitum is not a whole number
     
  17. Nov 27, 2003 #16

    jcsd

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    0.9... recurring IS a whole number it is equal to 1.
     
  18. Nov 27, 2003 #17
    um.sure it is......how then can you explan that .999 infinitum will NEVER reach 1?
    Must be that new math
     
  19. Nov 27, 2003 #18

    jcsd

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    No it's very old maths, most peope are taught the following sometime during their secondary eductaion:

    x = 0.99999.....

    =>

    10x = 9.9999.... =>

    10x -x = 9x = 9 =>

    x = 1
     
  20. Nov 27, 2003 #19

    Hurkyl

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    Why?

    I don't see the connection.

    No, the rule works because it is a logical consequence of the definitions.

    Can you write [tex]\inline{1/2=2/4}[/tex]? [tex]\inline{1}[/tex] and [tex]\inline{0.\bar{9}}[/tex] are two different representations of the same number, just like [tex]\inline{1/2}[/tex] and [tex]\inline{2/4}[/tex].

    This is where your problem lies; you are imagining [tex]\inline{0.\bar{9}}[/tex] as some sort of process instead of as a number.

    While it is certainly true you can get the value [tex]\inline{0.\bar{9}}[/tex] through a process (such as taking the limit of [tex]\inline{0.9, 0.99, 0.999, \ldots}[/tex]), [tex]\inline{0.\bar{9}}[/tex] is a number. It does not change, it does not approach anything; it is simply a number.
     
  21. Nov 27, 2003 #20

    HallsofIvy

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    No one needs to explain it- it's not true.

    .999 "infinitum", by which I presume you mean the infinite sequence of 9s, is, by definition, the infinite series .9+ .09+ ...+ 9(.1)n+... which can be proven to be exactly equal to 1 (it's a very easy geometric series- you should have learned how to sum those in secondary school).
     
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