Simple mathematical problem (?)

[RCNelson]

WOW!

600 posts in this thread discussing whether one equals one!

 

[BillyJoe]

That was the 600th post?

Aren't you proud though?
Where else but in the JREF Forum would such a topic even be started?

 

[Skeptoid]

BillyJoe,

I've seen this argument pop up on forums all over the net. I know the math newsgroups have had this discussion ad nauseum as well. There really is no end to it.

 

[BillyJoe]

Yeah well, apart from Maths Newsgroups, where else but.....

 

[LW]

It's transfinite. Infinite with certain properties that distinguish it from other infinites, such as the number of real numbers.

But how can that be? You yourself posted earlier in this thread that:

That only points to the fact that numbers have no "reality" or "truth" outside of real usage.

So, how can there an infinite number of numbers? It is perfectly clear that the vast majority of all "numbers" are so large that there is absolutely no way for humans to ever manipulate or even represent them in any way.

You accuse of others being incoherent when they say that 0.99... has an infinite number of nines as "In any real system, you will have a finite number of digits for the real answer.". Don't you realize that by this your own criterion you are yourself being incoherent when you claim that there are an infinite number of natural numbers since in a finite universe there exists some absolute upper limit for the size of a number that can be represented?

Think about this for a while.

 

[Suggestologist]

I'm not sure I understand what you're asking for. I did provide the definition. The definition of ".9999. . ." is<blockquote>the limit of the sequence .9, .99, .999, .9999, . . . .</blockquote>Or, if you prefer the general case, the definition of ".d₁d₂d₃d₄. . .", where the ds are decimal digits, is<blockquote>the limit of the sequence .d₁, .d₁d₂, .d₁d₂d₃, .d₁d₂d₃d₄, . . . .</blockquote>It really is as simple as that.

Please elaborate on how you turn this into the number line. How do you determine that number A is larger than number B in terms of their digital values?

 

[Suggestologist]

T'ai Chi

If .9~ = 1,
(.9~ + 1)/2 = 2/2 = 1

If .9~ != 1 then complete the equation....
(.9~ + 1)/2 = ?

BillyJoe
(Yeah, I saw the smilie )

Yes, so in integer math:

Since 3 DIV 2 = 1
And 2 DIV 2 = 1
Then 3 = 2.

 

[BillHoyt]

Yes, so in integer math:

Since 3 DIV 2 = 1
And 2 DIV 2 = 1
Then 3 = 2.

Still having problems understanding reality, Suggestologist?

 

[BillHoyt]

(.9~+1) / 2

Still having problems with math, T'ai? (yes, I saw the smilie, too. I think, though, we need a new, lame smilie for attempts at answers so lame they need to fly under the cover of smilies so nobody can challenge them. Eh?)

 

[Iconoclast]

Still having problems understanding reality, Suggestologist?

With all due respect, we're talking about pure maths here, it has little to do with reality. Now, does anyone know where I can get some lengths of 4" x 1.9999999999999..." lumber around here.

 

[BillHoyt]

With all due respect, we're talking about pure maths here, it has little to do with reality.

Yes, quite right. I should have been clearer; should have said "reality of mathematics."

Now, does anyone know where I can get some lengths of 4" x 1.9999999999999..." lumber around here.

That'll cost more than a few quid.

 

[xouper]

Suggestologist: How do you determine that number A is larger than number B in terms of their digital values?

Are you asking how we know that 3 is greater than 2? Is that your question?

 

[Iconoclast]

Yes, quite right. I should have been clearer; should have said "reality of mathematics."

I thought that's what you meant, but I'm a pedantic b*stard kind of mood tonight. Carry on.

 

[LW]

By the way, here's one more appeal to authority. In Martin Gardner's Six Sensational Discoveries (the April fool's article) one of the jokes is that it supposedly was proven that e^{pi*sqrt(163)} was exactly:

262,537,412,640,768,743.999...

about this he then comments that: "which, of course, makes that number integral".

(The joke is that actually the number doesn't contain an infinite number of nines but thirteenth digit is actually 2.)

 

[69dodge]

Originally posted by Suggestologist
Please elaborate on how you turn this into the number line. How do you determine that number A is larger than number B in terms of their digital values?

The only real numbers that have two decimal expansions are those whose decimals end in an infinite string of 0s or an infinite string of 9s. So we can simply pick one of those two forms as the canonical one, make sure that neither A nor B is in the noncanonical form by changing them to the canonical form if necessary, then compare the decimal expansions of A and B lexicographically, digit by digit, in the usual way.

This procedure assumes that we are given the numbers in a form that allows us to determine, in advance, whether they end in an infinite string of 0s or 9s. (The notation ".9999..." that we have been discussing in this thread is such a form, for example.) If, on the other hand, we are given the numbers one digit at a time, we must use a slightly different procedure. Of course, if the two digit-sequences that we've been given so far are identical, we can't yet decide which number is larger; we must ask for the next digit of each. But we must also allow for cases like .12399 and .12400. These two numbers may be equal or not, depending on the remaining digits; so in these cases too, we must delay our decision and ask for the next digit instead. If the two numbers differ, we will eventually find out. If they're equal, we never will know this with certainty. But that's not really surprising, and it's the case whether or not you think that .9999... = 1. How could we ever be sure that two number are equal if we've only seen a finite portion of them? They could always differ later on.

 

[69dodge]

Originally posted by Suggestologist
Yes, so in integer math:

Since 3 DIV 2 = 1
And 2 DIV 2 = 1
Then 3 = 2.

I do not understand your point.

Of course, 3 does not equal 2. So you have demonstrated that the following statement is false:<blockquote>For all integers n and m, if n DIV 2 = m DIV 2, then n = m.</blockquote>I agree. It is false. Now what?

 

[Suggestologist]

I do not understand your point.

Of course, 3 does not equal 2. So you have demonstrated that the following statement is false:<blockquote>For all integers n and m, if n DIV 2 = m DIV 2, then n = m.</blockquote>I agree. It is false. Now what?

My point is that numerical representational systems have limitations.

 

[xouper]

Suggestologist: My point is that numerical representational systems have limitations.

If you are suggesting that 0.999... is not equal to one because of some "representational limitation", then let's hear what you think that limitation is. Otherwise, this is just another one of your irrelevant points.

 

[Suggestologist]

The only real numbers that have two decimal expansions are those whose decimals end in an infinite string of 0s or an infinite string of 9s.

Where do you get this assumption about dual decimal expansions?

So we can simply pick one of those two forms as the canonical one, make sure that neither A nor B is in the noncanonical form by changing them to the canonical form if necessary, then compare the decimal expansions of A and B lexicographically, digit by digit, in the usual way.

And what is "the usual way"?

This procedure assumes that we are given the numbers in a form that allows us to determine, in advance, whether they end in an infinite string of 0s or 9s. (The notation ".9999..." that we have been discussing in this thread is such a form, for example.) If, on the other hand, we are given the numbers one digit at a time, we must use a slightly different procedure. Of course, if the two digit-sequences that we've been given so far are identical, we can't yet decide which number is larger; we must ask for the next digit of each. But we must also allow for cases like .12399 and .12400. These two numbers may be equal or not, depending on the remaining digits; so in these cases too, we must delay our decision and ask for the next digit instead. If the two numbers differ, we will eventually find out. If they're equal, we never will know this with certainty. But that's not really surprising, and it's the case whether or not you think that .9999... = 1. How could we ever be sure that two number are equal if we've only seen a finite portion of them? They could always differ later on.

 

[xouper]

69dodge: The only real numbers that have two decimal expansions are those whose decimals end in an infinite string of 0s or an infinite string of 9s.

Suggestologist: Where do you get this assumption about dual decimal expansions?

It's not an assumption. It's a fact. That is precisely the point of this whole thread.

0.24999... = 0.25000... for exactly the same reason that 0.999... = 1.000...