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05-14-2003, 06:40 PM | #201 | |
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Here we go again... "God said it, I believe it, and that settles it," right yguy? You are so wrong in the above analogy. I can *prove* that 1 + 1 is equal to 2 (in a Euclidean space, anyway). One husky plus one weimeraner equals two dogs. what plus what equals one God? Hmm.... I guess, using your worldview, you could say that we are both atheists... I just happen to believe in one less god than you do. Tenspace |
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05-14-2003, 06:50 PM | #202 | |||
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(Oh, no, it couldn't.) Oh, yes, it could. Jesus, Yguy, this is a bloody pantomime. The universe can exist quite happily without a god--and it does. Quote:
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05-14-2003, 06:50 PM | #203 | |
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05-14-2003, 06:53 PM | #204 | |
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05-14-2003, 06:59 PM | #205 | |
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But, I would still like the answers to the following questions that you removed when quoting earlier posts: 1. Outside of the Bible, where do you find knowledge of the Judeo-Christian God? 2. How does being created in an image equal free will? 3. quote: -------------------------------------------------------------------------------- As for whether He actually parted the Red Sea and all that stuff, that is belief. -------------------------------------------------------------------------------- What do you mean by this? That each believer is open to interpret whether to believe or not believe sections of the Bible? What basis is used to determine if a story is Word, parable, or fable? 4. What about the order of creation? Do you believe in the sequence of events as spelled out in Genesis I? 5. How can one argue for the existence of God before arguing for the infallibility of the Bible? You answered, "I do it all the time." I'm still awaiting the "How". Tenspace |
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05-14-2003, 06:59 PM | #206 | |
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numbers N. N is the smallest set satisfying these postulates: P1. 1 is in N. P2. If x is in N, then its "successor" x' is in N. P3. There is no x such that x' = 1. P4. If x isn't 1, then there is a y in N such that y' = x. P5. If S is a subset of N, 1 is in S, and the implication (x in S => x' in S) holds, then S = N. Then you have to define addition recursively: Def: Let a and b be in N. If b = 1, then define a + b = a' (using P1 and P2). If b isn't 1, then let c' = b, with c in N (using P4), and define a + b = (a + c)'. Then you have to define 2: Def: 2 = 1' 2 is in N by P1, P2, and the definition of 2. Theorem: 1 + 1 = 2 Proof: Use the first part of the definition of + with a = b = 1. Then 1 + 1 = 1' = 2 Q.E.D. Note: There is an alternate formulation of the Peano Postulates which replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the definition of addition to this: Def: Let a and b be in N. If b = 0, then define a + b = a. If b isn't 0, then let c' = b, with c in N, and define a + b = (a + c)'. You also have to define 1 = 0', and 2 = 1'. Then the proof of the Theorem above is a little different: Proof: Use the second part of the definition of + first: 1 + 1 = (1 + 0)' Now use the first part of the definition of + on the sum in parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D. ***** Check out The Math Forum for further details. |
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05-14-2003, 07:42 PM | #207 | |
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05-14-2003, 07:46 PM | #208 | ||
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To the point, your statement is nothing more than x=x. It's void of all meaningful context, whether you actually understand this or not. Quote:
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05-14-2003, 07:50 PM | #209 | ||
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The rest of the questions were answered previously. |
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05-14-2003, 07:54 PM | #210 | ||
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