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Somebody wants to write an essay on the non-uniqueness of the decimal representation of real numbers?
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<p>This is completely correct. A real number can be expressed in more than one way in decimal notation. There's no reason to bother about definitions. It's simply the way it is.</p>
<p>This is a basic definition of number theory: any number can be written in the form</p>
<p>Do you realize that all of those "proofs" deal with something which implies "getting close to" (for example, limits). All of those functions work on the assumption that something extremely close to something else IS that something else. It's pointless to prove something using something else which uses it. It's like if someone were to ask you what the word anonymity means and you answering "anonymous".</p>
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Do you realize that all of those "proofs" deal with something which implies "getting close to" (for example, limits). All of those functions work on the assumption that something extremely close to something else IS that something else. It's pointless to prove something using something else which uses it. It's like if someone were to ask you what the word anonymity means and you answering "anonymous".
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<p>That's incorrect. For example,</p>
<p>lim 1/x as x approaches inf is 0 - not close to 0, not infinitiely close to 0, but 0. The mathematics are well-defined. They work on no assumptions.</p>
<p>It's not so much that the function doesn't exist at that point but rather that we can't evaluate it. If you take the limit of 1/x as x approaches 0 you simply get infinity. Therefore, it does exist, but its not something we can get a number for, nor can we graph it. (Although in your example you could simply devide through by x)</p>
<p>[edit - moose, you should wait to get further into math before you make ignorant statements.]</p>
<p>Sorry my example is wrong (well it's not, but it's harder to explain). Lets say I said this instead: y=(2x+1)(x-1)/(x-1). In this example, you can divide by (x-1), to draw the graph (It should be a line). However, it would have a discontinuity at 1. The graph would thus exist at .999~, but not at 1.</p>
<p>Ok different from mathematical proofs, take this situation.
Your an engineer making the hubble telescope for NASA, which will be used to zoom in on other universes millions of lightyears away. The mirror is to be build perfect. The concave mirror is to be build with a perfect ONE (lets say thickness) , anything less or more, and the whole 15 Billion project is wasted.</p>
<p>i really dont know a whole lot on this subject, but it seems to me like you cant really treat 0.999~ like a regular number, just think, if you were working it out on paper, you cant truly multiply it by a number and then perform some other mathematical operation to that product because you would never compute the exact product, you'd be sitting there writing a repeating number for eternity, it just doesnt seem like it abides by the same rules as other non repeating numbers</p>
<p>ABSOLUTELY NOT. .99999repeating=.99999repeating, not 1. Calculus is stoopid. Then I guess 99.99999repeating equals 100, if that were to hold. That's dumb. I don't care what any math teacher says, .999repeating has, does and will forevermore, equal .999repeating. Bottom line.</p>
<p>the only thing "stoopid" about this argument is the people who are too naive to not realize that .999~=1. You can prove mathematically that it equals one. The basis is that .999~ is a infinite sequence, and the only way to rationally understand what that number equals is by finding out what it is arbitrarily close to... what number is .999~ arbitrarily close to...1.</p>
<p>the same reason why we say .333~=1/3 -- .333 doesn't = 1/3 nor does .3333333333. .333~ arbitrarily gets close to 1/3, just as .666~ arbitrarily gets close to 2/3.</p>
<p>1/3 + 2/3 = 1 (.333~ + .666~ = .999~ = 1) </p>
<p>the same reason why 1 + 1/infinity = 1 as 1/infinity arbitrarily goes to 0.</p>
<p>you can also prove this with calculus...but i havn't done that since first semester first year.</p>
<p>btw, to the person who made that argument about NASA...its not perfect. Its as perfect as we can get to with our own measurement tools. It could be an inch perfect to the 1/10000 (or however precise our instruments are) but its never going to actually be "perfect."</p>
<p>BookAdict, as I said in my post (whether or not your example is wrong) it does exist, we just can't evaluate it. It's true that there is no number when x = 0 in the function 1/x. However, if you take the limits as x tends towards 0 you find that 1/x = infinity. This is a real result. Imagine dividing 1 such that each bit you're dividing it into is zero. You would need infinite many pieces to give one. </p>
<p>To all those who disagree with the idea that .999~=1 are wrong. To the person who discussed the telescope, your argument is horrible. What you are essentially saying is that in order for this telescope to work it must be INFINITELY precise. That is, down below the smallest partition of space it must be correct (which is impossible because space is quantized). This is in no way the same as saying .999~ recurring forever is equal to one. </p>
<p>Everyone who thinks that this is rounding, it is not. Everyone who thinks that this is imprecise is wrong. It is perfectly precise, just hard to grasp.</p>
<p>schoens: you are right about the limit as x approaches 0 of 1/x. But that's different than my second example. In that example, you would say that the limit as x approaches 1 of y=(2x+1)(x-1)/(x-1) is 1. That is true, but that doesn't mean the graph equals one at that point. Compare the graph of 1/x to (2x+1)(x-1)/(x-1), and you'll see the difference. Unless, of course, 0/0=infinity. (You get 0/0 by plugging one into the above equation.) If I remember correctly, it doesn't, but I could be wrong.</p>
<p>Vyse: You're argument isn't really valid, because you could say that 1-.9~ is .000~1. But proving whether .000~1 =0 or not is the same as proving if .99~=1, which we're trying to do now.</p>
<p>BookAddict, you're quite right. Your example is different form mine. However, the limit of the function still exists. In order to find Lim x->1 (2x+1)(x-1)/(x-1) you must use something called L'Hopital's rule. It states that for any two functions f(x), and g(x) that both tend towards zero at some value c Lim x->c f(x)/g(x) = Lim x->c f'(x)/g'(x). Therefore, in your situation Lim x->1 (2x+1)(x-1)/(x-1) = Lim x->1 d[(2x+1)(x-1)]/dx / d[(x-1)]/dx = 2(x-1) + 2x + 1 = 4x-1 = 3. Therefore, the limit of your function at x = 1 is 3. So, it does exist but cant compute it directly.</p>
<p>oops...I said the limit was 1, not 3. I wasn't thinking...</p>
<p>But still, the rest of what I said is ok, I think. I agree that there is a limit at x=1. But just because there is a limit at x=1 doesn't mean that the line exists at x=1, does it? I don't think so, because you can't plug in 1. Once again, 1/x is different, because you get 1/0, which equals infinity. In my equation, you get 0/0. </p>
<p>I'm sorry, if I don't make sense or I'm completely wrong. I did make it up to Calc 2, but I haven't done any Calc in like 2 months. (That may not seem long, but I forget things quickly. :) )</p>