Does .999...=1?

Colleges & Universities Massachusetts Institute of Technology
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<p>but look,
1/3=0.333…is just approximation
you cannot define 1/3 using decimals, you can say that 1/3=.3333…with an absolute error (i think that’s how you call it) of ±1/oo (one over infinity), no? which is 0, or i’d better say, ALMOST 0</p>

<p>that’s why we can say that 1=.999… ± 1/oo , which is ALMOST equal</p>

<p>ok, arguing with Ben makes no sence b/c he definately knows more than i do, AND it could probably cost me admission to Caltech ;), but i still got some questions about this whole thing</p>

<p>i was tought that between any two real numbers there are infinite number of values, but the distance between those two numbers would be only one</p>

<p>by that i mean, that the distance between .9 and 1 is .1, but there are infinite number of values between them</p>

<p>the distance between .9999 and 1.0000 is .0001, but between them there are still infinite number of values.
that;s why i conlude that, although between .999…and 1.000…the distance is 0, there are still infinite number of values between them, right?
now, lets calculate the distance betwen .999…and one of those infinite number of values in between .999…and 1
then let’s calculate the distance between that point and another one, and then lets calculate the distance between the second point and 1
so, as i know, the distnce between .999…and first point (n1) would be
sqrt((n1-.999…)^2), the distnce between n1 and n2 would be sqrt((n2-n1)^2), and the distnce between n2 and 1 would be sqrt((1-n2)^2), so the total disatance between .999…and 1 would be
sqrt((n1-.999…)^2) + sqrt((n2-n1)^2) + sqrt((1-n2)^2) which can be equal to 0 only if 1=.999…=n1=n2, which is impossible because n1 and n2 were parts of the infinite number of values other than 1 or .999…</p>

<p>now…im not saying that .999…does not equal to 1, but rather im asking you to explain ^this^</p>

<p>As for Ben:
“At this point I have to run… I’ll try to think of the shortest proof I can that no real number has more than two decimial expansions. On the other hand, Hriundeli, if you haven’t finished calculus, I doubt the technical arguments about infinite series that would be involved would make much sense… perhaps it would be better for you to learn a little more… otherwise it would just be wasted time for you and me.”…</p>

<p>…hate to say it, but i think you are wrong…sry
i mean, sooner or later i WILL ask you for the proof, so, does it matter when you send it to me? So what if i dont know integrals or infinite series? i still can keep it untill i think i would be able to understand it…Besides, the presense of it would stimulate my learning :wink:
he-he</p>

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<p>stop assuming things then. its a fact that 1/3 = .3333…
ok if you dont like decimals take a look at this 1/3 + 1/3 +1/3 = 1
OH MY GOD ITS A BREAKTHRU!!</p>

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<p>Hriundeli, I think the problem is is that you are getting confused with the concept of infinity. It’s a very difficult topic, but the point is that you have to have a good knowledge of infinite series and integrals for this. Actually, it isn’t that hard to learn the two, even if you just skim it.</p>

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<p>Hriundeli – <a href=“http://www.its.caltech.edu/~beng/decimalexpansions.pdf[/url]”>http://www.its.caltech.edu/~beng/decimalexpansions.pdf&lt;/a&gt; shows that each real number has at most two decimal expansions.</p>

<p>It is written in the terse style of analysis proofs, so you probably will need to have a bit more experience before reading it. I highly recommend the book Principles of Mathematical Analysis by Walter Rudin to help you get up to speed and able to read such things.</p>

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<p>Hriundeli – you need to understand that infinite sums – which is essentially what .999… is – have no meaning on their own. They have to be defined. In standard real analysis, if we have a sequence of numbers a<em>1, a</em>2, a_3,… then we define the partial sums</p>

<p>S<em>1 = a</em>1
S<em>2 = a</em>1 + a<em>2
S</em>3 = a<em>1 + a</em>2 + a<em>3
.
.
.
S</em>k = a<em>1 + a</em>2 + a<em>3 + … + a</em>k
.
.
.</p>

<p>The infinite sum a<em>1 + a</em>2 + a_3 + … is defined as </p>

<p>lim S_k
k->inf</p>

<p>if this limit exists.</p>

<p>It is easy to show that under this definition, .9999… (.9 + .09 + .009 + … ) is equal to 1. This is the standard framework of analysis. The reason we use it is because it works very well and gives a number of quite sensible, non-contradictory results.</p>

<p>If standard mathematics is unsatisfactory for you, you can guarantee whatever outcome you want by making your own definitions and axioms, but then you are not doing standard mathematics – in which the question is not up for debate – but something else, perhaps called Hriundeli-athics.</p>

<p>This is my last post in this discussion.</p>

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<p>1/3=0.333…is just approximation</p>

<p>^^This is where you went wrong. Infinite decimals are not approximations. To you, 0.333… is some sort of process of continually adding 3s. It is not. It is the number which consists of 0 followed by an infinite number of 3s. See what Ben said.</p>

<p>This is my last post also.</p>