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Simple mathematical problem (?)

jpowell said:
10 times 999 is 9990.

Therefore

x = 999
10x = 9990
10x -x = 9990-999
9x = 8991
x = 999

It is perfectly valid to use your method, but you need to know how to multiply numbers together in the first instance to get the right answer :P.
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Yep.

Vampyric:

Those multiplication tables you had to study in your last GED class weren't just for the test.

By the way, do you like the taste of shoe rubber?
 
LW said:
Oh dear, the Stars were Wrong and the Thread that Should Not Have Been Rose Again to Haunt the Mankind.



No. Because 0 is a real number and 0.000...01 is not. They can't be the same.
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They're both real numbers. They're both 0.
 
Errata said:
They're both real numbers. They're both 0.
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.0000...01 is a contradiction in terms

The "..." means continuing without end. If you add an end to it, you've contradicted the notation.

But I stand by my proof in post #705.
 
The ... means continuing without end, thus the 1 is meaningless, and the expression is just a stupid way of writing 0.
 
Errata said:
They're both real numbers. They're both 0.
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And 0 and 0.5 are both integers and they are both 0 since 0.5 is just a stupid way of writing 0, right?

Informally, integers are the whole numbers, rationals are the numbers that can be expressed in the form x/y where x and y are integers (and y is not zero), and real numbers are the numbers that can be written as x.y where x is an integer and y is an infinite sequence of digits. An infinite sequence followed by a digit has no place in here.

I could develop a number system where expression 0.00...01 had a well-defined meaning, but I wouldn't have real numbers anymore, just like 2/3 is well-defined but not an integer.
 
LW said:
I could develop a number system where expression 0.00...01 had a well-defined meaning, but I wouldn't have real numbers anymore, just like 2/3 is well-defined but not an integer.
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You'd have to change the meaning of ...
 
gnome said:
You'd have to change the meaning of ...
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Actually, you wouldn't; it makes sense as it is (though it wouldn't be a real number, of course).

To try to illustrate: Say I want to place a linear order (a well order, in fact) on the nonnegative integers in the following way:

1 < 2 < 3 < 4 < .... < 0

In other words, the positive integers have the usual order amongst themselves, but 0, under this order, has been placed larger than all of the (infinitely many) positive integers.

So you could think of the digits in 0.000...1 as being indexed by the nonnegative integers under this under: There's a "0" in each position corresponding to a positive integer, and there's a "1" corresponding to the ultimate position of the zero.

(I would be uncomfortable writing this expression as 0.000...01, since that notation implies to me there is a "0" immediately preceding the "1"; of course, that can't be. Just like in the above ordering of the nonnegative integers, there is no positive integer immediately preceding the final "0".)

Another order you could place on the natural numbers:

1 < 3 < 5 < 7 < ... < 2 < 4 < 6 < 8 <... < 0

In other words, the odd positive integers in their usual order, followed by the even positive integers, followed by 0.

If you use this set to index a string of digits, you could have something like:

0.000...1111...2

(Informally: infinitely many "0"s, followed by infinitely many "1"s, followed by a final "2").
 
LW said:
And 0 and 0.5 are both integers and they are both 0 since 0.5 is just a stupid way of writing 0, right?
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Nope. Congratulations on utterly failing to have a point.

Informally, integers are the whole numbers, rationals are the numbers that can be expressed in the form x/y where x and y are integers (and y is not zero), and real numbers are the numbers that can be written as x.y where x is an integer and y is an infinite sequence of digits. An infinite sequence followed by a digit has no place in here.
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Because an infinite sequence cannot be followed by another digit. If it could, it wouldn't be infinite, but rather it would just be very large. We're talking about "completed" infinity, not "potential" infinity.

I could develop a number system where expression 0.00...01 had a well-defined meaning, but I wouldn't have real numbers anymore, just like 2/3 is well-defined but not an integer.
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You could redefine anything you want, but under their commonly understood current meaning, that expression is an alternate way of representing the real number 0. You could also invent a number system where .9 equals 5, but we're not using that hypothetical system, and under the system we're using .9 is just another way of writing 1.
 
Most of this confusion seems to be over an incorrect concept of infinity. You can’t just tack a number at the end of an infinite string of them. .999…9995 isn’t possible if you take the .999… to be an infinite number of 9s, since there would be no end of the 9s to tack on the 5, hence the term infinite.

Placing another number at the end of an infinitely repeating number makes about as much sense as asking what happened before time began. Both are logically impossible by definition.
 
I less than three logic said:
Placing another number at the end of an infinitely repeating number makes about as much sense as asking what happened before time began. Both are logically impossible by definition.
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But it does make sense in the manner I described in my previous post. The underlying ideas are the ordinal numbers:
Perhaps a clearer intuition of ordinals can be formed by examining a first few of them: as mentioned above, they start with the natural numbers, 0, 1, 2, 3, 4, 5, … After all natural numbers comes the first infinite ordinal, ω, and after that come ω+1, ω+2, ω+3, and so on.
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( from http://en.wikipedia.org/wiki/Ordinals )
 
No Cabbage. It doesn't matter that there are higher ordinal infinities. You still can't count to ω, so its impossible to ever place digits after the repeating sequence.

I Less Than Three Logic, that was pretty much my point in raising the tangent, before people went off on it. The point is that the digits after the repeating sequence just don't exist, so you can write them if you really want, but the value is just the same as if you hadn't, because they don't exist. The concept was raised as my version of what people who think 1!=.9 must think 1 - .9 is, but because the series is actually infinite and not merely very long, that is actually 0.
 
Errata said:
No Cabbage. It doesn't matter that there are higher ordinal infinities. You still can't count to ω, so its impossible to ever place digits after the repeating sequence.
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You can't "count" to ω in a step by step process where you just add one to the previous integer to get the next one (e.g., 0,1,2,3,4,...), but ω is still there (as well as all the ordinals beyond ω). In the same way, you can have a string of digits indexed by any arbitrary ordinal. The important thing to note is that not every digit will have an immediate predecessor (just as ω has no immediate predecessor), but it's still perfectly OK to talk about such strings of digits.

What I'm really getting at is that you can have a function f:alpha -> {0,1,2,3,4,5,6,7,8,9}, where alpha is any arbitrary ordinal. You can think of this as a string of decimal digits indexed by the arbitrary ordinal alpha (just as you can think of a real number's decimal expansion as a string of decimal digits indexed by the particular ordinal ω).

Now, just to be clear, I'm not claiming there is a natural way to define algebraic operations such as addition or multiplication on decimal strings indexed by an arbitrary ordinal (and I'm not claiming there isn't a natural way either, I simply haven't thought about it much either way). I'm only claiming that such strings do make sense mathematically.
 
I less than three logic said:
Most of this confusion seems to be over an incorrect concept of infinity. You can’t just tack a number at the end of an infinite string of them.
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A somewhat analogous situation: How many positive real numbers are less than 1? Isn't 1 greater than all of them?

("Greater than" in the analogy corresponds to "to the right of" in the string of digits.)
 
69dodge said:
A somewhat analogous situation: How many positive real numbers are less than 1? Isn't 1 greater than all of them?

("Greater than" in the analogy corresponds to "to the right of" in the string of digits.)
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2 is a positive real number greater than one

so is sqrt(13).

I'm not sure what you're getting at.
 
Wait, I do now. You're referring to the infinite number of positive reals that happen to be less than one.

However, there is no list that contains them in any infinite series, that it makes sense to include "1" in the list at all. "1" would have to be part of a completely different series.
 
Seems like this is simply a convoluted way of writing R x {0,1,2,3,4,5,6,7,8,9}, since, absent any algebraic structure, .999...1 would, in your formulation, function identically to (.999... , 1). One infinite list of digits "followed" by another infinite list of digits is the same as simply having an ordered pair, each element of which is one infinite list of digits.

I also disagree that your formulation follows naturally from the current definition of ".999...". It is a valid extension, true, but it requires stating the defintion of ".999..." in a particular way so that it can be extended. When you speak of one infinite list following another infinite list, "follow" cannot be anything but an abstract term, as the usual meaning does not allow for such a use.
 
Cabbage said:
You can't "count" to ω in a step by step process where you just add one to the previous integer to get the next one (e.g., 0,1,2,3,4,...), but ω is still there (as well as all the ordinals beyond ω). In the same way, you can have a string of digits indexed by any arbitrary ordinal. The important thing to note is that not every digit will have an immediate predecessor (just as ω has no immediate predecessor), but it's still perfectly OK to talk about such strings of digits.
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ω isn't there though. The number of digits (not necessarily the value of the whole string of them) is clearly a natural number, so any specific quantity of digits is < ω, and an infinite number of digits is mappable to ω. For higher ordinals of infinity to apply, you'd have to be talking about a set that cannot be mapped to the natural numbers, which is not the case here. No matter how many sets of natural numbers you combine, you've still got a set which is mappable to the natural numbers. You need something else entirely before you can talk about the next higher ordinal infinity.
 
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