The_root_of_all_evil said:
Lukeje said:
(I've also noticed that you seem confused by the notion of mathematical equivalence. Saying that a is equivalent to b means that a=b if and only if b=a. Equivalence thus implies equality).
From what I can tell, the terminology has changed somewhat. My Maths training was back in the early 90s; pre-Windows 3.1. We had the three bar equals as equivalent, but it seems the approximation (curved upper bar) has taken over.
≡ equivalent to
≅ congruent with
≈ approximately equal to
These are three different concepts.
But .9rep is only equivalent to 1, where as 1 can be .9rep - the infinite repetition is a flag that it approximates for convenience - so it can be used but not without loss of accuracy.
Equivalent meaning the same, identical, exactly equal. Not approximate. The infinite repetition is a flag that it is an exact figure. If it was an approximation it would have to be a
finite repetition.
geizr said:
oktalist said:
f(x) = 1 - 10-x
limx→∞ f(x) = 1
Your function only approaches 1 in the limit, but it never actually gets there. This is because n/(infinity) only approaches zero in the limit; it never actually gets there. However, for purposes of convenience, we often ignore that subtlety because it is below our error tolerance, but, when we need to be more mathematically precise, we can not make that assertion.
I was simply demonstrating that the limits process does not end up with a 0.000...1 left over, when you actually do it abstractly as you suggested. Limits actually have nothing to do with the 0.999... = 1 thing, because when you write 0.333... = 1/3 you are saying that there is an actual infinite number of 3's after the decimal point, rather than it just tending to an infinite number of 3's.
oktalist said:
Infinity doesn't really have a size, as such.
Infinity very much has a size, and that size can be different in different cases. That is why you can take the limit of a numerator and denominator both going to infinity and, yet, obtain a finite ratio. It's because they are different sized infinities.
oktalist said:
There is no "at the end". Recurring decimals are endless.
It doesn't matter how endless they are, you can always add one more digit to make a larger infinity. That's what happens when you multiply by 10.
There are some transfinites that can be said to be smaller or larger than other transfinites, but you seem to be talking about their sizes in the same way as one might compare the sizes of real numbers. You can't get from one infinite to another by adding or multiplying by a finite number. If you add one more digit to an infinite, you get the same infinite.
m + 1 = m for any transfinite m
0.999... is not a set. It's a number.
The digits that we use to represent numbers are sets. Each element in the set represents a particular fraction multiplied by some factor. The arithmetic operations with which we are familiar perform transformations on those elements, which can themselves be sets. When you multiply x = 0.999rep by 10, you shift all the elements upward and then have to add an extra empty set element at the very end to represent the digit that was vacated as a result of the multiplication.
If you have to add an "empty set" "at the end" for the digit that was vacated after the multiplication, why was there not an "empty set" at the beginning for the digit that was vacant before the multiplication?
"Empty set" is in inverted commas because the digit zero is not the empty set. "At the end" is in inverted commas because there is no end to an infinite set. You are free to invent whatever mapping you wish in order to map the decimal representations of numbers onto sets, but this in no way shows that numbers should necessarily embody any of the properties of sets.
Sets are not ordered, and cannot contain the same entity twice. You could invent the notation D(d,p) to mean the digit d at position p, so that 123.456 can be represented with the set { D(1,2), D(2,1), D(3,0), D(4,-1), D(5,-2), D(6,-3) }, but that's just an alternative representation of a number.
Just because the set { D(9,-1), D(9,-2), D(9,-3) } differs from the set { D(9,-1), D(9,-2), D(9,-3), D(0,-4), D(0,-5), D(0,-6) }, does nothing to suggest that .999 differs in any way from .999000
EDIT: Also, if you add/remove any finite number of elements to/from a countably infinite set (like .999...), you still have a countably infinite set, meaning it has the same cardinality.
