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Deeper than primes

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The Man said:
Again, not my "maneuvers" Doron just your own assertions about what you claim your “AB means” and if you can't even agree with just yourself on that, how can anyone even possibly agree with you on that.



Unless of course we just disagree with you, which would be agreeing with your disagreement with yourself. Hey,…

looks like we've had this "OM" stuff down pat from the start. The only way to agree with Doron and his “OM” is to disagree with him and his “OM” as that is all he and his "OM" do, just disagree.
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http://www.internationalskeptics.com/forums/showpost.php?p=7405615&postcount=16077

Still poor The Man can't distinguish between non-strict AB and strict A or strict B.
 
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doronshadmi said:
jsfisher, you still do not understand that the members of a given set are its organs, where a set is an organism.

If you take them out of their organism, they are not its organs anymore.

For example, {{N},{M}} is an organism, and this organism is the union of its organs.

By {N}U{M} = {N,M} you get another organism, which its organs are different than the organs of {{N},{M}} organism, exactly because {N} or {M} are not organs of {{N},{M}} organism.
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:dl:
 
The Man said:
Well when you do finally agree with at least yourself about what your "non-strict AB" means, you be sure to let us know.
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When you finally open your mind in order to get the difference between AB and A,B , let me know.
 
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doronshadmi said:
I do I do, can I?:idea:
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Apparently, you cannot.

{2}U{{1},2}={{1},2} exactly because all the involved objects are sets' members.
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No, {2} U {{1}, 2} = {{1}, 2} because of how the set union operator is defined. It's operands, by the way, need to be sets. Just about everything is a member of some set, so the "objects are sets' members" is trivial and irrelevant.

2U{{1},2} is gibberish
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Not necessarily. Although the notation suggests otherwise, the operand, 2, could be a set. By some conventions, it would be the set {0, 1}. Under that convention, 2 U {{1}, 2} = {0, 1, 2, {1}}.

...exactly because object 2 is not a member.
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(1) It is a member (of many, many sets), and (2) the property is irrelevant; only that it be a set is necessary (and sufficient).
 
doronshadmi said:
Wrong.
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So what is the "strictly non-wrong option?"

Is the identity

0.333... [base10] = .1 [base 3]

wrong too?

Your inequality in the decimal base

1/3 > 0.333...

is still waiting for the proof. See, the problem is that multiplication and division in any number base don't require actual computation when numbers which are not integers are multiplied or divided by the number which denotes the base. Instead, the separation point (in base 10 it is called the decimal point) is shifted to the right or left. For example, if you multiply binary number 101.011 by binary 1000 then

101.011 * 1000 = 101011

The number was multiplied by 8[base 10], which is 23 and so the point that separates the integer part from the fractional part has been shifted 3 (the exponent) places to the right. This shift avoids the incompleteness of multiplication of numbers with infinite fractional part. So in base 10, multiplying or dividing by 10 means shifting the decimal point by one place respectively.

x = 0.3333...

10x = 3.333...

10x - x = 3 (the infinitely long fractional parts are being subtracted)

9x = 3

x = 3/9 = 1/3

0.3333... = 1/3

That means the expression 0.333... is equal to the expression 1/3. At the same time, 0.333... is getting more and more precise as the number of the decimal digits is growing without bound. That means the number is getting "infinitely precise." :confused:What is the lowest number that 0.333... approaches but never reaches?:confused: Is it number expressed as 1/3? According to the above equations, it cannot be. So your inequality calls for 1/3 to be converted to the number with a radix (approximate format) to make a comparison other than shown above.

We can prove that UP = DOWN. Since UP and DOWN are opposites, we prove that opposites are equal and demonstrate it on the opposites APPROXIMATE and EXACT:

Approximate = Exact

t = 0.25000... = 1/4

As the number of zeroes grows without bound, the magnitude of t doesn't change. Since any expression that is free of the radix (the decimal point) is called exact, like 1/4, the assumption is correct, and by analogy, STUPID = SMART, or ATHEIST = THEIST.
 
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jsfisher said:
Not necessarily. Although the notation suggests otherwise, the operand, 2, could be a set. By some conventions, it would be the set {0, 1}. Under that convention, 2 U {{1}, 2} = {0, 1, 2, {1}}.
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Thank you for supporting me, because all you did is to translate 2 into some set ({0,1}, in this case) in order to avoid 2 U {{1}, 2} gibberish.
 
jsfisher said:
No, {2} U {{1}, 2} = {{1}, 2} because of how the set union operator is defined.
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And it is defined between members of sets, for example:

{1,2,3,4} can be, for example, the result of the union of the members of set {1,2} and the members of set {2,3,4}.
 
doronshadmi said:
And it is defined between members of sets, for example:

{1,2,3,4} can be, for example, the result of the union of the members of set {1,2} and the members of set {2,3,4}.
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No. {1, 2, 3, 4} is identical to the union of the sets, {1, 2} and {2, 3, 4}, but certainly not the union of the members of those sets.

You really, really need to review the meaning of the set-union operation. It would be a good time to brush up on set member, too.
 
jsfisher, according to your "reasoning" we can strip off also {} and get ,such that U {{1}, 2} is not gibberish by your "reasoning",
exactly as 2 U {{1}, 2} is not gibberish by your "reasoning".

Shell we :clap: for your "reasoning"?
 
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jsfisher said:
No. {1, 2, 3, 4} is identical to the union of the sets, {1, 2} and {2, 3, 4}, but certainly not the union of the members of those sets.

You really, really need to review the meaning of the set-union operation. It would be a good time to brush up on set member, too.
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No. {1, 2, 3, 4} is defined, in this case, by the union of the members of sets {1, 2} and {2, 3, 4}.

You really, really need to get out of your :dig:
 
doronshadmi said:
jsfisher, according to your "reasoning" we can strip off also {} and get ,such that U {{1}, 2} is not gibberish by your "reasoning"
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Got not argument so you just invent one? You are the one who advocated bracket stripping. I opposed it as meaningless gibberish. But now you try to project your failing on to me.

Well, at least you seem willing now to admit your original strip-the-parenthesis plan was nonsense.

...exactly as 2 U {{1}, 2} is not gibberish by your "reasoning".
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Not necessarily gibberish. Not necessarily. Your reading comprehension skills fail you get again.

Shell we :clap: for your "reasoning"?
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Yes, that would be appropriate.
 
doronshadmi said:
No. {1, 2, 3, 4} is defined, in this case, by the union of the members of sets {1, 2} and {2, 3, 4}.
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You never meant a definition you couldn't ignore, have you, doron?

The members of {1, 2} are 1 and 2. The members of {2, 3, 4} are 2, 3, and 4. The union of the members of those two sets, then, is clearly:

1 U 2 U 2 U 3 U 4

As you have already observed, this union is undefined since the operands fall outside the domain of the set union operator (unless, of course we adopt a special convention that has those numbers representing actual sets). Special convention or not, we don't get {1, 2, 3, 4} from the union.
 
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