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Member (Idle past 2325 days) Posts: 2870 From: Limburg, The Netherlands Joined: |
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Author | Topic: 0.99999~ = 1 ? | |||||||||||||||||||||||||||
Jon Inactive Member |
lyx2no writes: To prove that 0.999 is a real number one reads the dictionary. Jon writes: I just pray to God that the definition that gets used for 'real' number be not so pitiful that anything qualifies. I guess my prayers were not answered. Can you provide a definition of 'real number' so that I do not have to look for one myself? [O]ur tiny half-kilogram rock just compeltely fucked up our starship. - Rahvin
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
LOL. Again, you base your conclusion on the premisethough, I give you extra credit for more cleverly disguising it this timethat .9999| is a real number. Which it is by definition. BTW, I have obviously not attempted to disguise the fact that we are reasoning about real numbers and not about omelets or geraniums. Your admiration of how "clever" I am in doing so, though flattering, is therefore misplaced.
Your flow can just as easily be reworded as follows ... No it can't.
{ABE: I noticed you edited your post after I started typing my reply; worry not, though; italicizing your conclusion doesn't affect its truth value} That was not the change I made. If you will look at my edit and compare it to the original, you will see what change I made. The purpose of it was to prevent you from reading it as though being "distinct" could be the property of one number, or of two numbers separately, rather than of a pair of numbers. Edited by Dr Adequate, : No reason given.
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
DP.
Edited by Dr Adequate, : No reason given.
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Rrhain Member Posts: 6351 From: San Diego, CA, USA Joined: |
Jon writes:
quote: Well, the problem is that a "Real" number isn't nearly as easily defined as other kinds of numbers. For example, there are the "Natural" numbers: 1, 2, 3, .... Notice that "0" is not among them. If you add 0 to it, you get "Whole" numbers. Add the negatives, and you have the "Integers." A "Rational" number is any number that can be expressed as a/b where a and b are Integers. Real numbers, though, start to get complicated. One of the simplest way of constructing the "Reals" is through decimal expansion. For any number, you can converge upon its decimal representation: 2, 2.7, 2.71, 2.718, 2.7182, etc. will converge upon e, a Real number. Rrhain Thank you for your submission to Science. Your paper was reviewed by a jury of seventh graders so that they could look for balance and to allow them to make up their own minds. We are sorry to say that they found your paper "bogus," specifically describing the section on the laboratory work "boring." We regret that we will be unable to publish your work at this time. Minds are like parachutes. Just because you've lost yours doesn't mean you can use mine.
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Jon Inactive Member |
No it can't. LOL. I'm too busy for malarkey. My debate is with lyx2no, now. If you want to join, you can start by supporting this latest claim. Jon [O]ur tiny half-kilogram rock just compeltely fucked up our starship. - Rahvin
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
I guess my prayers were not answered. Can you provide a definition of 'real number' so that I do not have to look for one myself? Well, try the following definition: --- Let R be the set of Cauchy sequences of rational numbers. That is, sequences x1,x2,x3,... of rational numbers such that for every rational ε > 0, there exists an integer N such that for all natural numbers m,n > N, |xm-xn|<ε. Here the vertical bars denote the absolute value. Cauchy sequences (x) and (y) can be added, multiplied and compared as follows:* (xn) + (yn) = (xn + yn) * (xn) (yn) = (xn × yn) * (xn) ≥ (yn) if and only if for every rational ε > 0, there exists an integer N such that xn ≥ yn - ε for all n > N. Two Cauchy sequences are called equivalent if and only if for every rational ε > 0, there exists an integer N such that |xn -yn|<ε for all n > N. This does indeed define an equivalence relation, it is compatible with the operations defined above, and the set R of all equivalence classes can be shown to satisfy all the usual axioms of the real numbers. --- Note that the sequence 0.9, 0.99, 0.999, 0.9999 ... is indeed a Cauchy sequence, since this is a sequence of rational numbers and since for any rational ε > 0 there exists some N for which ε > 10-N. Indeed, one can also immediately prove from the definition of the equivalence relation that 0.999999... = 1. Edited by Dr Adequate, : No reason given.
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
LOL. I'm too busy for malarkey. Your posts belie this claim.
My debate is with lyx2no, now. This declaration will not, of course, prevent me from pointing out that you are wrong.
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Jon Inactive Member |
This declaration will not, of course, prevent me from pointing out that you are wrong. LOL. Yes; too busy for malarkey, but I can make time for bullshit. If you wish to be obtuse, care to point out where in my rewording the meaning of your original was lost? Edited by Jon, : Mr. T [O]ur tiny half-kilogram rock just compeltely fucked up our starship. - Rahvin
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Jon Inactive Member |
Rrhain,
Thank you for actually participating with reason and evidence.
Rrhain writes: For any number, you can converge upon its decimal representation: 2, 2.7, 2.71, 2.718, 2.7182, etc. will converge upon e, a Real number. My math is a little rusty; can you explain this one to me? What does it mean for numbers to converge upon e? Thanks,Jon [O]ur tiny half-kilogram rock just compeltely fucked up our starship. - Rahvin
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
LOL. Yes; too busy for malarkey, but I can make time for bullshit. So I see.
If you wish to be obtuse, care to point out where in my rewording the meaning of your original was lost? But what if I do not wish to be obtuse?
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Jon Inactive Member |
I did not think 'rational' and 'real' numbers to be the same. One's a subset of the other, no?
... the usual axioms of the real numbers Which are...? [O]ur tiny half-kilogram rock just compeltely fucked up our starship. - Rahvin
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
My math is a little rusty; can you explain this one to me? What does it mean for numbers to converge upon e? A sequence of numbers x1, x2, x3 ... converges on e if for every ε > 0 there is some N such that for all i > N, |xi - e| < ε.
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
I did not think 'rational' and 'real' numbers to be the same. You're right. Why do you mention it?
Which are...? A model for the real number system consists of a set R, two distinct elements 0 and 1 of R, two binary operations + and * on R (called addition and multiplication, respectively), a binary relation ≤ on R, satisfying the following properties. 1. (R, +, *) forms a field. In other words, * For all x, y, and z in R, x + (y + z) = (x + y) + z and x * (y * z) = (x * y) * z. (associativity of addition and multiplication)* For all x and y in R, x + y = y + x and x * y = y * x. (commutativity of addition and multiplication) * For all x, y, and z in R, x * (y + z) = (x * y) + (x * z). (distributivity of multiplication over addition) * For all x in R, x + 0 = x. (existence of additive identity) * 0 is not equal to 1, and for all x in R, x * 1 = x. (existence of multiplicative identity) * For every x in R, there exists an element −x in R, such that x + (−x) = 0. (existence of additive inverses) * For every x ≠ 0 in R, there exists an element x-1 in R, such that x * x-1 = 1. (existence of multiplicative inverses) 2. (R, ≤ ) forms a totally ordered set. In other words, * For all x in R, x ≤ x. (reflexivity)* For all x and y in R, if x ≤ y and y ≤ x, then x = y. (antisymmetry) * For all x, y, and z in R, if x ≤ y and y ≤ z, then x ≤ z. (transitivity) * For all x and y in R, x ≤ y or y ≤ x. (totalness) 3. The field operations + and * on R are compatible with the order ≤. In other words, * For all x, y and z in R, if x ≤ y, then x + z ≤ y + z. (preservation of order under addition)* For all x and y in R, if 0 ≤ x and 0 ≤ y, then 0 ≤ x * y (preservation of order under multiplication) 4. The order ≤ is complete in the following sense: every non-empty subset of R bounded above has a least upper bound. In other words, * If A is a non-empty subset of R, and if A has an upper bound, then A has a least upper bound u, such that for every upper bound v of A, u ≤ v. Edited by Dr Adequate, : No reason given. Edited by Dr Adequate, : No reason given.
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Jon Inactive Member |
How is this different from saying:
Real numbers are numbers that meet definition XReal numbers are defined by X All real numbers are Real What purpose does the Real/Non-real distinction serve? Does it actually differentiate, or is the distinction merely made up?
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Dr Adequate Member (Idle past 315 days) Posts: 16113 Joined: |
How is this different from saying: Real numbers are numbers that meet definition X Clearly the only way to define the real numbers is to supply a definition of the real numbers.
What purpose does the Real/Non-real distinction serve? Does it actually differentiate, or is the distinction merely made up? Clearly there are structures which are not the real numbers. Such as the complex numbers. Or quaternions. Or, for that matter, asparagus.
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