Does .999~ equal 1?

College Confidential Community
61

<p>BookAddict, you are right. The function is undefined at x = 1.</p>

<p>Sigh: it is unfortunate that people could not understand one of the most fundamental proofs given.</p>

<p>Let me regurge it, in the hope that repetition indeed does get a point across:</p>

<p>Let x = 0.9- (- to denote repeating)</p>

<p>x = 0.9- ... A
10*x* = 9.9- (I realize that many people in this thread do not understand the properties of infinity...)
10*x*-x = 9
9*x* = 9
x = 1 ... B</p>

<p>Since A = B,
0.9- = 1</p>

<p>...Or, realize that:
1/3 = 0.3-
3/3 = 0.9-
1 = 0.9-</p>

<p>Don't bother making it complicated with convergence...</p>

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<p>
[quote]
Don't bother making it complicated with convergence...

[/quote]
</p>

<p>Unfortunately, you have to. This is a case where the weakness of the decimal system shows its face: It's not such a great thing. There are numbers out there that cannot be written as decimals, and numbers out there that can be written several ways. </p>

<p>Most of the proofs here are intuitive quasi-proofs, but they illustrate the point. You have to get down to the definitions to ensure that you are doing what's right.</p>

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<p>What? Are there rules on CC that mandate me that it's insufficient to use the elementary proofs? Please -- considering I'm seeing people perplexed by the concept of limits, I wouldn't even bother with convergence. If they do not understand the most basic of proofs, they are hopeless.</p>

64

<p>
[quote]
Are there rules on CC that mandate me that it's insufficient to use the elementary proofs?

[/quote]
</p>

<p>Haha no such rules.</p>

<p>But for example,</p>

<p>
[quote]
...Or, realize that:
1/3 = 0.3-
3/3 = 0.9-
1 = 0.9-

[/quote]
</p>

<p>This is not a proof. I could say: 0/3=0, 0/2=0, 0/1=0, 0/0=0. But it wouldn't be true.</p>

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<p>0/0 is undefined. 3/3 is defined as 1. This is a fundamental flaw in your reasoning.</p>

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<p>ameivil: But the function is defined at .9~, so then doesn't that mean that .9~ and 1 are not equal?</p>

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<p>How does that follow?</p>

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<p>
[quote]
ameivil: But the function is defined at .9~, so then doesn't that mean that .9~ and 1 are not equal?

[/quote]
</p>

<p>yeah im not following that.</p>

69

<p>
[quote]
0/0 is undefined. 3/3 is defined as 1. This is a fundamental flaw in your reasoning.

[/quote]
</p>

<p>Exactly! That's why proofs by "example" don't necessarily work. Yours gives the right result, but it can be used to give the wrong result as well, like I showed with my example.</p>

70

<p>Well, you'd restrict the domain to all x except where x = 0 in x/x.</p>

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<p>multiply 2 x1... multiply 2 x .999~</p>

<p>square .99999999~</p>

<p>square 1</p>

<p>divide .999~ by 2</p>

<p>divide 1 by 2; after some math equations, you find that the distance between 0 and 1 is the same as between 0 and .99999~</p>

<p>it's zeno's dichatomy, or soemthing like that</p>

<p>good ol' Taylor Polynomials, lol</p>

72

<p>...whoops., didn't see the last 3 pages of this thread, lol. Anyhue, my argument, while not very thoroughly supported (i jsut got back from swim practice after 3 horus of sleep; yay!), still stands; :-).</p>

<p>you can argue that if you start at point A and approach point B by halves of the distance you are away (so, if you were 1 inch away, then .5 inches, then .25 inches. etc.), you will never actually reach point B.</p>

<p>However, it IS possible to go from point a to b, and as you approach A you are always approaching it a "half the distance to b" at a time, though maybe not at a certain rate, and, since you actualyl can reach b, you have jsut done an "infinite" amount of halving to finally reach point b.</p>

<p>say point a is one wall of a room and point b is an opposite wall. Now walk from one to the other; you cna do it, right? Well, you are ultiamtely doing an infinite number of halving... </p>

<p>...trippy</p>

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<p>"But the function is defined at .9~, so then doesn't that mean that .9~ and 1 are not equal?"</p>

<p>Well, the function is defined as you get infinetely close to 1 (it gets infinitely close to 3), but not at exactly 1.</p>

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<p>When you get to real analysis, at some point you might construct the real numbers as equivalence classes of Cauchy sequences of rational numbers. Under that definition, you can define .999... as the equivalence class of the following sequence:</p>

<p>{.9, .99, .999, .9999, ...}</p>

<p>This is equivalent to the sequence {1, 1, 1, 1, ...} because
the sequence {1-.9, 1-.99, 1-.999, ...} converges to zero. </p>

<p>Therefore .999... = 1.</p>

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<p>It took 5 pages to conclude this...and somewhere I read that CC is full of top 5% of students.</p>

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<p>I wonder if someone asked about Fermat's Last Theorem whether or not we could prove that too?</p>