= 9/10+ lim n->inf sum[i,2,n] 910^i (extract bottom term from sum)
= 9/10 + (1/10) * (lim n->inf sum[i,1,n-1] 910^i) (scale all terms to shift sum range down)
= 9/10 + (1/10) * (lim n->inf sum[i,1,n] 9*10^i) (as n gets to infinity, n-1 is effectively n)
= 9/10 + (1/10) * x (substitute x back in)
so x = 9/10 + x/10
so 10 x = 9 + x
so 9x = 9
so x = 1
This only differs from the previous very simple ‘algebraic’ proof in that I use limits and summation notation (uh, subject to not drawing the sum symbol of course), so that no ‘multiplication of an infinite repeating decimal value’ was done. I just extracted and scaled terms in a sum. Admittedly an infinite sum, so Phage would still disagree, but its an alternate way of looking at it.
At the risk of (once more) revealing my ignorance, does the process involved in ascertaining the limit of .9 + .09 + .009 … not also depend on the Axiom of Choice?
William_Ashbless’s proof also assumes that the series converges, just like the algebraic proof. Seriously, I would consider that a big deficiency if I were grading the paper. You could find the limit better using a standard delta-epsilon proof.
KarlGauss, no, finding the limit does not require AC. I’m not sure how instructive that answer is for you, however; if you could mention how you think AC is being used, then maybe I can be of more help.
Thanks, Cabbage. I wondered if you needed AC in order to assume that all the terms of 9/10[sup]n [/sup] “already existed” (despite their infinitude) and hence that the limit also existed.
This thread’s bein’ hijacked all over the place
Ultrafilter, let me draw you out on this one -
I’m not sure I understand you. I don’t see why 1[sub]R[/sub] and 1[sub]Z[/sub] not “being in the same set” necessarily makes them different. Or, I don’t understand what you mean by “same type”.
Consider the Integers, and the even Integers. Two different sets, but no reason to think that 2[sub]Z[/sub] and 2[sub]even[/sub] are different objects.
And, from a definitional standpoint, there are often different, but compatible, definitions for the same thing without the idea that the things they define are different objects.
I see what you’re saying KarlGauss, most real numbers aren’t “describable” in mathematics. Most numbers will be of the nature:
.4356437985437…
with the digits just jumping around, having no pattern, and not generatable by any finite algorithm. To talk about such numbers we would need the axiom of choice (or at least a weaker form of it), just as we need at least a weak form of the axiom of choice to talk about non-measurable sets–otherwise, we just don’t have the power to describe such a bizarre set.
.9999… is simple, however, and easily described (just 9’s all the way across), and so AC isn’t necessary.
The crucial difference between the relationship between 1[sub]Z[/sub] and 1[sub]R[/sub] and that between 2[sub]Z[/sub] and 2[sub]even[/sub] is that the even integers are a subset of the integers, whereas the integers are not a subset of the reals.
This is going to be a bit complicated, so let me explain. I don’t know what your background in algebra is, so I’ll try to avoid too much jargon. If you want me to rewrite this using standard terminology, I can.
We say that two sets are isomorphic if, as far as the properties we’re interested in go, they can’t be distinguished. The integers are a ring, with all the properties that implies. There is a subset of the reals which is also a ring, and it works out that it’s isomorphic to the integers. You can get to this ring by replacing each integer z with the set {x | x is rational and x < z}. Given the definitions of real addition and multiplication, it works out that u + v ends up being replaced by {x | x is rational and x < u + v}, and uv ends up being replaced by {x | x is rational and x < uv}.
Refer to my earlier post for the precise definitions of N, Z, Q, and R. This is pretty obscure stuff, so don’t worry about exactly what it means–it’ll never make a difference to you unless you’re a mathematician, and even then, it probably doesn’t matter.
As for what exactly a type is…I can’t find a precise definition, but in general, two objects would have the same type only if they’re constructed from the set-theoretic axioms in the same manner. I’ll see if I can find something better.
… or to continue my analogy of strings and variables in programming, the set of even numbers is the set {“two”, “four”, “six”, …}, which is a subset of {“one”, “two”, “three”, …}. A member of the first set can be combined with a member of the second because they are both of the same type.
Remember, however, that the reals in my analogy are like the variable set {1, 2, 3, …}. You just can’t combine 1 with “one”.
ultrafilter’s isomorphism is like telling the programme “where you see “one”, replace it with 1”. We now have a way of linking the s strings with the v variables. However, it doesn’t get around the fact that they are still fundamentally different.
But if {“one”,“two”,“three”,…} is isomorphic with {1,2,3,…} aren’t they really just different representations of the same thing? To my intuition, if two sets are isomorphic, they really are the same fundamental “thing” underneath two different ways of writing them. After all, there’s no difference in the properties we’re interested in. (Maybe this is more philosopy than mathematics?)
I think the problem may be more with my analogy actually. Suppose your isomorphism instead is between {“one”, “two”, …} and {2 , 4 , 6, …} and remembering that the cardinality of the reals is greater than that of the integers, can you see where the problem arises? If {“one”, “two”, …} is somehow the “same” as {2 , 4 , 6, …} then how come you can do 1 + 2 but you can’t do 1 + “one”?
… of course, you can substitute 2 for “one” and claim that the answer is 3 in either case, but that still doesn’t get around the fact that “one” is fundamentally not 2. “One” is a string and 2 is a variable.
Same deal with 1.0000… in R and 1 in Z.
Come to think of it, that probably would still have worked with the original analogy. “one” was fundamentally not the same as 1 there either.
I don’t think I’m helping. Let’s go back to the discussion of ordered fields…
Ultrafilter, I understand your isomorphism discussion, but that’s not where my problem is. It’s the idea that Z is not a subset of R. I presume you’d also say that Q is not a subset of R.
I’ve been taking some classes towards a Master’s in Math. I’m sure that in one of my Analysis or Metric Spaces classes I’ve had an assignment to “prove that Q is dense in R”.
What kind of credit do you think I’d have gotten if I shot back “Q is not dense in R because there is no element of Q that even exists in R”?
I’m guessing that the distinctions you’re drawing are at a level of fomality that is above the most formal textbooks I have (so far, at least).
Q is isomorphic to a subset of R, and because of that, you can pretend that it is a subset of R. Most people do–you’re right that this discussion is beyond what even most professional mathematicians would deal with. Don’t concern yourself with it unless you start studying constructibility.
Let me use a simpler example: Z[sub]2[/sub] x Z[sub]3[/sub] is isomorphic to Z[sub]6[/sub] (both under addition). I would say that (1 1) is fundamentally the same thing as 1. But I still can’t ask what (1 1)+1 is. That operation is undefined. They are the same in the sense that they behave in identical ways in their respective sets. I’m only using different notations for the same entity.
So, in the sense that the subset of R {0,1,2,3,…} is isomorphic to Z, the real 1 is the same as the integer 1.
To further muddy the waters, the subsets of Z A={0,1,2,3,…} and B={0,2,4,6,…} are isomorphic. Thus 1 in A is the same as 2 in B. (“Same” meaning isomorphic, not equal.)
