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October 3rd, 2018, 07:52 AM   #91
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Quote:
 Originally Posted by AplanisTophet $0-----0.0_2 = \frac{0}{2^1}$ $1-----0.1_2 = \frac{1}{2^1}$ $2-----0.01_2 = \frac{1}{2^2}$ $3-----0.11_2 = \frac{3}{2^2}$ $4-----0.001_2 = \frac{1}{2^3}$ . . . The list contains only dyadic rationals of the form $\frac{a}{2^b}$ where $a$ is an integer and $b$ is a natural number. As a result, your list is a mere subset of the rational numbers. It also contains no irrational numbers. Thatâ€™s like saying you can You now need to prove that each real number in [0,1) can be expressed in the form $\frac{a}{2^b}$ as a dyadic rational. Alternatively, you can get that sailboat.
Please give an example of a real number [0,1) not in the list. Note the list contains ALL binary and decimal sequences.

October 3rd, 2018, 07:58 AM   #92
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Quote:
 Originally Posted by zylo Every entry in the list is finite but the list has no end.
I'm glad we can finally agree - every entry of the list is finite. (Of course the list has no end, this was never in question.) If you add a binary radix point in front of a finite string of ones and zeros, you get a rational number. So no irrational number arises from your list. Not even all the rational numbers in [0,1) show up (see AplanisTophet's comment to see which ones do).

Quote:
 Originally Posted by zylo Please give an example of a real number [0,1) not in the list. Note the list contains ALL binary and decimal sequences.
Any irrational number in [0,1) or any rational number not expressible in the form $\frac{a}{2^b}$ with $a, b$ natural will do. As you admitted, every element of your list is finite. So no infinite string is in your list.

Last edited by skipjack; October 3rd, 2018 at 09:15 AM.

October 3rd, 2018, 08:08 AM   #93
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Quote:
 Originally Posted by zylo Please give an example of a real number [0,1) not in the list. Note the list contains ALL binary and decimal sequences.
The list contains only dyadic rationals so $\frac{1}{3}$ isn't in the list.

If you can show me an element of the list that is not a dyadic rational, I'll give you my sailboat.

PS - Get a sailboat.

 October 3rd, 2018, 08:37 AM #94 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 What you say is true IF the list is finite (ends) it doesn't. .33333..... is a nested sequence which by a theorem of analysis zeroes in on one point of the real line. Want to guess what it is? You're spending too much time with the sailboat.
October 3rd, 2018, 08:53 AM   #95
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Quote:
 Originally Posted by zylo What you say is true IF the list is finite (ends) it doesn't.
Yes, and it's also true if the list is infinite (which it is). If $x$ is a binary string in your list (yes, any of the infinitely many of them), then its corresponding number in [0,1) must be rational. I have proven this by induction.

With not much extra work, you could prove by induction that every element of your list actually corresponds to a dyadic rational number.

Quote:
 Originally Posted by zylo .33333..... is a nested sequence which by a theorem of analysis zeroes in on one point of the real line. Want to guess what it is?
For a string to correspond to $\frac{1}{3}$, it would have to be infinitely long. However, as you've said, all the strings in your list are finite. Therefore no string in your list can correspond to $\frac{1}{3}$.

Also, what you've said here is nonsense. $.3333...$ is not a sequence, but is instead a real number. It's by definition the limit of the sequence $.3, .33, .333, ...$.

Last edited by cjem; October 3rd, 2018 at 09:16 AM.

October 3rd, 2018, 09:32 AM   #96
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Quote:
 Originally Posted by cjem 1) With not much extra work, you could prove by induction that every element of your list is actually dyadic rational number. 2) For a string to correspond to $\frac{1}{3}$, it would have to be infinitely long. However, as you've said, all the strings in your list are finite. Therefore no string in your list can correspond to $\frac{1}{3}$. 3) Also, what you've said here is nonsense. $.3333...$ is not a sequence, but is instead a real number. It by definition the limit of the sequence $.3, .33, .333, ...$.
1) and 2): But the list doesn't end.
3) It defines a sequence of closed nests each one containing the subsequent one, and there is only one number common to all.

Apparently there is the inability to comprehend that any particular natural number is "finite," but the list of natural numbers doesn't end. If you have a problem with that, you will just keep repeating the same thing over and over and over again, and compensate for your inability to comprehend with insults.

October 3rd, 2018, 10:36 AM   #97
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Quote:
 Originally Posted by zylo 1) and 2): But the list doesn't end.
I agree. Fortunately my proof works even though the list doesn't end. That's the power of induction for you!

Quote:
 Originally Posted by zylo 3) It defines a sequence of closed nests each one containing the subsequent one, and there is only one number common to all.
Sure. I was just disagreeing with your claim that $.333.....$ is itself a sequence.

Quote:
 Originally Posted by zylo Any particular natural number is "finite," but the list of natural numbers doesn't end. If you have a problem with that, you will just keep repeating the same thing over and over and over again
I have no problem with this whatsoever. I wholeheartedly agree.

Last edited by cjem; October 3rd, 2018 at 11:04 AM.

October 3rd, 2018, 10:55 AM   #98
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Quote:
 Originally Posted by zylo To answer Maschke's interesting question, offhand, an infinite power series has an infinite number of roots unless it converges.
Care to give an example?

October 3rd, 2018, 11:41 AM   #99
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Quote:
 Originally Posted by zylo Skipjack and CJEM. Every entry in the list is finite, but the list has no end.
If you agree that every entry is finite, please explain your remark that the list contains ALL imaginable binary sequences, given that anyone can imagine an infinitely long string of digits.

If you disagree that every entry is finite, please ignore the previous request and instead give an example of an entry that isn't finite (preferably the first such entry in the list), stating where exactly it occurs in the list.

October 3rd, 2018, 11:47 AM   #100
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Quote:
 Originally Posted by Maschke Care to give an example?
I can't. I thought I could get there from an nth degree polynomial has n roots for all n. I can't. e^z was a mistake.

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