The Goldbach Conjecture:
Prove that every even number greater than 2 can be written as the sum of two primes.
Or prove that the above statement is not true. Either way.
Computer Science Canada Suggestions of mathematical theorems |
| Author: | mapleleafs [ Wed Nov 23, 2005 7:46 pm ] |
| Post subject: | Suggestions of mathematical theorems |
Can any of you guys suggest the names of some good (not too complicated), mathematical theorems? For example, I mean something along the lines of Pythagorean theorem, or Heron's theorem. I have this school assignment where I have to pick a theorem and prove it and so on, but I'm having a tough time just deciding which theorem to do. Suggestions much appreciated. |
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| Author: | Martin [ Wed Nov 23, 2005 8:03 pm ] |
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Fermat's last theorem: x^n + y^n = z^n has no integer solutions for all n > 2 and x,y,z != 0. |
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| Author: | Martin [ Wed Nov 23, 2005 8:05 pm ] |
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Prove that the number of primes are infinite. Easy, but cool. Or prove the chain rule for differential equations. EDIT: Better one. Prove that between any two non equal irrational numbers there is an irrational number, and between every two non-equal rational numbers, there is a rational number. EDIT: sorry, fixed that one ^^ |
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| Author: | mapleleafs [ Wed Nov 23, 2005 8:15 pm ] |
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Just for clarification, when you say "between" two such numbers, you mean the "difference between" the two numbers right? |
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| Author: | Martin [ Wed Nov 23, 2005 8:26 pm ] |
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Prove: If a and b are irrational, and a < b, then there exists an irrational c such that a < c < b If a and b are rational, and a < b, then there exists a rational c such that a < c < b |
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| Author: | Martin [ Wed Nov 23, 2005 8:32 pm ] |
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Proof of infinite primes: We know that the list of primes is not empty (since we know that 3, 5, 7, 11, 13, ... ) are prime. Let p be the set of all known primes. So assume that there are only n primes. Multiply all of these together creating M = p1 x p2 x ... x pn. We know that every prime number divides M. Add 1 to M, and now no prime number divides it (since dividing by any prime would give a remainder of 1). Therefore, M+1 is prime, or has prime factors not in P. Since this works for any n, there are infinite primes. |
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| Author: | Cervantes [ Wed Nov 23, 2005 8:51 pm ] |
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The Goldbach Conjecture: Prove that every even number greater than 2 can be written as the sum of two primes. Or prove that the above statement is not true. Either way. |
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| Author: | Hikaru79 [ Thu Nov 24, 2005 11:49 pm ] |
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You guys are evil Looks like fun.
Here's two more great ideas. Prove the Riemann Hypothesis or the Poincare Conjecture. If you do, you'll probably get a pretty good mark  |
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| Author: | Martin [ Fri Nov 25, 2005 1:18 am ] |
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On the more easy and serious side. Prove that 0.9... is exactly 1 and not just really close (it's true, I promise). |
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| Author: | Naveg [ Fri Nov 25, 2005 7:26 am ] |
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Martin wrote: On the more easy and serious side.
Prove that 0.9... is exactly 1 and not just really close (it's true, I promise). No it's not true, but i won't give away the solution. Once infinity is involved in a proof, the proof is pretty well null. This question must be expressed as a limit |
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| Author: | Martin [ Fri Nov 25, 2005 7:28 am ] |
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There's absolutely no question that it's true. EDIT: So as not to make 3 posts in a row, here are 54 proofs of the Pythagorean Theorem. http://www.cut-the-knot.org/pythagoras/index.shtml |
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| Author: | Martin [ Fri Nov 25, 2005 7:41 am ] |
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A bunch of proofs. 1 Let x = 0.9... 10x = 9.9... 10x - x = 9x = 9 .: x = 1 2 0.99... = 9/10 + 9/100 + 9/1000 + ... let s = 9 * (1/10 + 1/100 + ... + 1/(10^n) + ...) We know that the sum of a geometric series of the form S = a + ar + ar^2 + ar^3 + ... + ar^n + ... is S = a/(1 - r) Let a = 9/10 and r = 1/10 then S = 9/10 / (1 - 1/10) = 9/10 / 9/10 = 1. 3 1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999... I've got more if you're not convinced |
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| Author: | md [ Fri Nov 25, 2005 9:16 am ] |
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The last proof certainly isn't true; 1/3 does not equal 0.33... it may be close but it's not equal. As for the other proofs, though they seem pretty convincing they only work because you're using an infinite number of 9's; and infinities are cheating. |
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| Author: | Martin [ Fri Nov 25, 2005 9:21 am ] |
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Infinity's not cheating. You don't believe in limits eh? Another one. Number theory tells us that there must be an infinite number of terms between 0.9... and 1, or they're equal. And these numbers are...? The last one is true, 1/3 is indeed equal to 0.333... |
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| Author: | Albrecd [ Fri Nov 25, 2005 10:53 am ] |
| Post subject: | Infiniti |
Quote: Once infinity is involved in a proof, the proof is pretty well null
It's used in converting repeating decimals into fractions in the same context as his first example (9.9... - 0.9...) |
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| Author: | Dan [ Fri Nov 25, 2005 12:41 pm ] |
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How did you get 9x = 9 in proof number 1? If x = 0.9999......... then it should be 0.999.......*9 saying that is = 9 is assuming what you are trying to prove. Also assuming you are assuming some things that may not be ture when saying 10X = 9.9999....... |
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| Author: | Martin [ Fri Nov 25, 2005 12:58 pm ] |
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so let x = 0.99... What is 10 times 0.999...? 9.99... Multiplying by 10 just shifts the decimal right a place. Subtract x, which we can since there are infinite decimal places, and you get 10x - x = 9x = 9, and therefore x = 1 and therefore 1 = 0.99... It's strange that our mathematical system requires a number of ways to write 1, but there are weirder things in math. This is one of the major reasons that the real numbers were so poorly understood by mathematicians up until suprisingly recently. It's a tough concept. EDIT: Here's another one. Now I'm going to bed. Look at it like this. 0.999... + 0.1 = 1 + 0.0999... > 1 0.999... + 0.01 = 1 + 0.00999... > 1 0.999... + 0.001 = 1 + 0.000999... > 1 0.999... + 0.0001 = 1 + 0.0000999... > 1 And so on. Each time, 0.999... + 0.000...1, for any number of 0's results in a number that's greater than 1. Therefore, 1 - 0.999 is less than any positive number and clearly not negative, so we can conclude that 0.99... = 1. |
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| Author: | Brightguy [ Fri Nov 25, 2005 3:43 pm ] |
| Post subject: | Re: Suggestions of mathematical theorems |
Assuming you accept the delta-epsilon definition of the limit, here's a formal proof. |
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| Author: | Dan [ Fri Nov 25, 2005 5:06 pm ] |
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Martin wrote: so let x = 0.99...
What is 10 times 0.999...? 9.99... 0.999......... is not a real number i do not think you can do mathical operations on it like that. Quote: Multiplying by 10 just shifts the decimal right a place. Subtract x, which we can since there are infinite decimal places, and you get 10x - x = 9x = 9, and therefore x = 1 and therefore 1 = 0.99... You are still saying 9x=9 with not justifaction and are assuming what you are trying to prove. By you reasning the falwoing whould also be a proof of it: X=0.99999.... (random whole number)X = (same reandom whole number) there for X = 1 If this was true you could prove that anything is anything with such a prof. |
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| Author: | Brightguy [ Fri Nov 25, 2005 5:50 pm ] |
| Post subject: | Re: Suggestions of mathematical theorems |
0.999... is real because it has a periodic decimal expansion (this can be proved). Although Proof #1 is not rigorous, the method does work because you can justify the multiplication using infinite series. The trick is to notice that multiplying by 10 just shifts the decimal point over (so you can't use any random number). In fact you can use this method to write any periodic decimal in rational form. |
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| Author: | Naveg [ Fri Nov 25, 2005 8:51 pm ] |
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Simple arithmetic cannot prove that 0.999... = 1 Use limits. Here is the only valid answer to that problem: lim (1-x) = 1 x->0 Since 1-0.1=0.9 1-0.01=0.99 1-0.001=.999 etc... |
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| Author: | Martin [ Fri Nov 25, 2005 9:03 pm ] |
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Yeah, alright, the first proof is kind of sketchy. True, but I should prove that the multiplication is valid first. 0.999... is a real number. It's 1. The definition of a real number is, according to mathworld, The field of all rational and irrational numbers is called the real numbers. |
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| Author: | Dan [ Fri Nov 25, 2005 9:43 pm ] |
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Martin wrote: Yeah, alright, the first proof is kind of sketchy. True, but I should prove that the multiplication is valid first.
0.999... is a real number. It's 1. But see you can not use that in a proof that is trying to prove that fact. I am not saying that what you are clamining is not ture it is just that as Naveg was saying it can not be proven using the methods you used. I blive that acording to the rules of clacuces and limits it is ture, tho i do not blive such a number could exist in real life. |
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| Author: | Martin [ Fri Nov 25, 2005 9:59 pm ] |
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Nothing greater than 10^100 can exist in real life. The thing is, mathematics is an artificial system, and you're absolutely correct when you say that 0.99... can't exist in real life (as independent of 1). That's the point. Look at it like this. In Japanese, three is written as ミ for san. In English, we write 3. They mean the same thing. They're just different ways of writing the same thing. |
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| Author: | Naveg [ Fri Nov 25, 2005 11:29 pm ] |
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Hacker Dan wrote: Martin wrote: Yeah, alright, the first proof is kind of sketchy. True, but I should prove that the multiplication is valid first.
0.999... is a real number. It's 1. But see you can not use that in a proof that is trying to prove that fact. I am not saying that what you are clamining is not ture it is just that as Naveg was saying it can not be proven using the methods you used. I blive that acording to the rules of clacuces and limits it is ture, tho i do not blive such a number could exist in real life. Well yes and no. See, by the rules of limits it isn't true because of the reaon for using the limit in the first place. The whole point is that number never reaches 1, since, in the above limit, x never reaches 0. |
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| Author: | Naveg [ Fri Nov 25, 2005 11:36 pm ] |
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Martin wrote: Nothing greater than 10^100 can exist in real life.
The thing is, mathematics is an artificial system, and you're absolutely correct when you say that 0.99... can't exist in real life (as independent of 1). That's the point. Look at it like this. In Japanese, three is written as ミ for san. In English, we write 3. They mean the same thing. They're just different ways of writing the same thing. I'm beginning to see your point...and it makes very much sense. As you said, mathematics is an artificial system. However, if we're going to use it we must abide by its rules and limitations. Clearly, by mathematics 0.999.... is not equal to one. In a higher conceptual, non systematic sense yes it very well may be. |
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| Author: | Martin [ Sat Nov 26, 2005 1:01 am ] |
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No, in mathematics, 0.9... is exactly equal to 1 if not, there are infinite real numbers between 0.9... and 1. I challenge you to name one of them. |
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| Author: | Andy [ Sat Nov 26, 2005 1:38 am ] |
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yea martin's right, using archimedian's property, we know that between two rational numbers there must exist another rational number. and since u cant find that number for 0.999... and 1, that means the two are equal |
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| Author: | md [ Sat Nov 26, 2005 3:25 am ] |
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Andy wrote: yea martin's right, using archimedian's property, we know that between two rational numbers there must exist another rational number. and since u cant find that number for 0.999... and 1, that means the two are equal
While I'm willing to agree that the limit of 0.999... is 1; and thus 0.999... = 1; I disagree with the above statement. Proof: Assume that if there are no real numbers between two real numbers then they must be equal. Let a, b, and c be unique real numbers such that there is no number between a and b, and no number between b and c. By our assumption a = b, and b = c (because there are no number between them). Thus a = b = c, so a = c. But there is a unique number between a and c so they cannot be equal. This contradicts the assumption, which proves that if there is no number between two other numbers it does not mean they are equal. |
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| Author: | Martin [ Sat Nov 26, 2005 7:48 am ] |
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Cornflake wrote: Proof: Assume that if there are no real numbers between two real numbers then they must be equal.
Let a, b, and c be unique real numbers such that there is no number between a and b, and no number between b and c. By our assumption a = b, and b = c (because there are no number between them). Thus a = b = c, so a = c. But there is a unique number between a and c so they cannot be equal. This contradicts the assumption, which proves that if there is no number between two other numbers it does not mean they are equal. But if there are no real numbers between them, they can't be unique. All that this proves is that either your assumption or your conclusion is false. |
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| Author: | Paul [ Sat Nov 26, 2005 2:27 pm ] |
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Martin wrote: 3 1/3 = 0.333... 2/3 = 0.666... 3/3 = 0.999... I've always found this to be pretty cool, any fault in it? Either way, in real life applications I don't think there'd be any difference between 0.99999... and 1. |
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| Author: | md [ Sat Nov 26, 2005 2:53 pm ] |
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Martin wrote: Cornflake wrote: Proof: Assume that if there are no real numbers between two real numbers then they must be equal.
Let a, b, and c be unique real numbers such that there is no number between a and b, and no number between b and c. By our assumption a = b, and b = c (because there are no number between them). Thus a = b = c, so a = c. But there is a unique number between a and c so they cannot be equal. This contradicts the assumption, which proves that if there is no number between two other numbers it does not mean they are equal. But if there are no real numbers between them, they can't be unique. All that this proves is that either your assumption or your conclusion is false. Bah, your probably right... this is why I hate getting into math arguments... stupid reals... they break everything! |
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| Author: | Martin [ Sat Nov 26, 2005 3:32 pm ] |
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Haha, it's alright. I'm here to keep you in line |
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| Author: | MysticVegeta [ Sat Nov 26, 2005 9:33 pm ] |
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I have to agree with Martin because our teacher did this proof. But there is one thing that bothers me. if 10x = 9.999.... and x = 0.999.... Then how can you subtract infinite from infinite? Since infinite is nothing but just a concept, I dont think this operation will be valid, but I assume its true since I have been taught to. You guys have some suggestions? |
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| Author: | md [ Sat Nov 26, 2005 9:58 pm ] |
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the 10x = 9.999... proof isn't a valid proof; there are valid proofs using limits though. (sorry martin ) |
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| Author: | Paul [ Sun Nov 27, 2005 12:50 pm ] |
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MysticVegeta wrote: I have to agree with Martin because our teacher did this proof. But there is one thing that bothers me. if 10x = 9.999.... and x = 0.999.... Then how can you subtract infinite from infinite? Since infinite is nothing but just a concept, I dont think this operation will be valid, but I assume its true since I have been taught to. You guys have some suggestions?
Not only can you do that, there are also different levels of infinity. For example, there are more real numbers between 1 and 2, then there are all integers. If there are an infinite number of integers, then the real numbers between 1 and 2 are represented by a higher level of infinity. |
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| Author: | Dan [ Sun Nov 27, 2005 1:32 pm ] |
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Paul wrote: MysticVegeta wrote: I have to agree with Martin because our teacher did this proof. But there is one thing that bothers me. if 10x = 9.999.... and x = 0.999.... Then how can you subtract infinite from infinite? Since infinite is nothing but just a concept, I dont think this operation will be valid, but I assume its true since I have been taught to. You guys have some suggestions?
Not only can you do that, there are also different levels of infinity. For example, there are more real numbers between 1 and 2, then there are all integers. If there are an infinite number of integers, then the real numbers between 1 and 2 are represented by a higher level of infinity. You can't do it with out assuming the facts you are trying to prove in this case tho.... |
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| Author: | Martin [ Sun Nov 27, 2005 6:16 pm ] |
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Paul wrote: MysticVegeta wrote: I have to agree with Martin because our teacher did this proof. But there is one thing that bothers me. if 10x = 9.999.... and x = 0.999.... Then how can you subtract infinite from infinite? Since infinite is nothing but just a concept, I dont think this operation will be valid, but I assume its true since I have been taught to. You guys have some suggestions?
Not only can you do that, there are also different levels of infinity. For example, there are more real numbers between 1 and 2, then there are all integers. If there are an infinite number of integers, then the real numbers between 1 and 2 are represented by a higher level of infinity. There aren't higher levels of infinity. Infinity isn't a number, just a concept. What you're talking about is how sparse a set is. From 1 - 5, there are 5 integers but infinite real numbers (including 1, 2, 3, 4 and 5). Therefor we say that the real numbers are more dense than the integers. And the 10x = 9.999... is also valid, I promise. |
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| Author: | Paul [ Sun Nov 27, 2005 6:32 pm ] |
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I'm just conveying what my math teacher taught me, I dunno if he is wrong. But right now I'm more inclined to believe him, as he is a former NASA astrophysicist, and semi-famously named around here "the black hole man", and I've never seen him to be wrong on any mathematical subject before. He told me that there are different levels of infinity. There is a number between 1 and 2 that corresponds with all the integers to infinity, AND there is more. So if the integers go to infinity, for every integer, there is a corresponding number between 1 and 2, PLUS there are more, it'd be logical to say that there are more numbers between 1 and 2 then there are in infinite integers. He did say it doesn't have many practical applications, but the higher math goes, the more fields there are, and this is one of the fields. |
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| Author: | Martin [ Sun Nov 27, 2005 7:06 pm ] |
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He's talking about density, not infinity. It's like: Are there more rational or irrational numbers? Also, for example, look at 1/3 = 0.3333.... However, 1/3 in base 9 is simply 0.3 Some more logic too. Let's assume that there are two real numbers a and b such that a < b and there are no real numbers between a and b. If we subtract a positive real number c from both a and b, we end up with two smaller numbers between which there are no numbers. Similarly, if we add a positive real number to each, we end up with two larger numbers between which there are no numbers. Now, here's some more math. x - y is the distance between two numbers on a real number line. It's easy to see that 10 * (x - y) is just 10 times the distance between two numbers. 100 * (x - y) is 100 times this distance. And so on. However, if there are no numbers between x and y, then 10 * (x - y) can't ever be a number other than 0. 100 * (x - y) = 0 and in general, n * (x - y) = 0. How is this possible if n is a huge positive number? |
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| Author: | Martin [ Sun Nov 27, 2005 9:37 pm ] |
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First, we find the sum of a geometric series (1) s = a + ar + ar^2 + ar^3 + ... + r^n multiplying both sides by r, we get (2) rs = ar + ar^2 + ar^3 + ... + r^(n+1) (2) - (1) = s - rs = a + ar - ar + ar^2 - ar^2 + ... + r^n - r^n + 0 - r^(n+1) Factoring the s on the left side we get (1 - r)s = a - r^(n+1) Divide both sides by 1 - r to get s = [a - r^(n+1)]/(1-r) Which is the sum of a geometric series of n terms, for all +n. Now, we look at the series representing the decimal expansion of 0.999... 0.99... = 0.9 + 0.09 + 0.009 + ... Factoring out a 9 from the right hand side gives: 0.99... = 9 * (0.1 + 0.01 + 0.001 + ...) Now we multiply both sides by 10. 10 * 0.99... = 9 * 10 * (0.1 + 0.01 + 0.001 + ...) Because multiplication is distributive, we can say that: 10 * 0.99... = 9 * (10 * (0.1 + 0.01 + 0.001 + ...) 10 * 0.99... = 9 * (10 * 0.1 + 10* 0.01 + 10 * 0.001 + ...) (3) 10 * 0.99... = 9 * (1 + 0.1 + 0.01 + ...) (1 + 0.1 + 0.01 + ...) a geometric series with a = 1 and r = 1/10 Therefore, the sum is (a - r^(n+1)) / (1-r) = (1 - (1/10)^(n+1)) / (9/10) This series converges (has a sum) as n -> infinity Since 10^(n+1) -> infinity as n -> infinity, 1 / (10 ^ (n+1)) -> 0 So the series converges to is 1 / (9/10) = 10/9 So going back to (3), we get. 10 * 0.99... = 9 * (10/9) And we can therefore conclude that 10 * 0.99.. = 10 (and therefore that 10 * 0.99.. = 9.99..) |
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| Author: | Brightguy [ Sun Nov 27, 2005 11:26 pm ] |
| Post subject: | Re: Suggestions of mathematical theorems |
Martin wrote: Nothing greater than 10^100 can exist in real life.
Forget about the googol, what about the goololplex? You might be interested in this page I came across some time ago...  Yeah...
Cornflake wrote: the 10x = 9.999... proof isn't a valid proof; there are valid proofs using limits though. (sorry martin
)See the proof I posted earlier. Gotta love ε-δ. Paul wrote: For example, there are more real numbers between 1 and 2, then there are all integers. If there are an infinite number of integers, then the real numbers between 1 and 2 are represented by a higher level of infinity.
He's talking about the sizes of infinite sets. For example, the set of irrational numbers between 1 and 2 is "larger" than the entire set of rational numbers. The set of rational numbers is "countable", that is, you can count them while putting them in some order. The set of real numbers is "uncountable". |
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| Author: | Martin [ Mon Nov 28, 2005 3:56 am ] |
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Another perspective. 1/9 = 0.1111... In base 9, 1/9 = 0.1 Also, in base 9, 9 base 10 = 10 base 9. So: 0.9999... = 0.1111... * 9 (base 10) = 0.1 * 10 (base 9) = 1 (base 9) = 1 (base 10) Q.E.D. |
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| Author: | bugzpodder [ Mon Nov 28, 2005 10:46 am ] |
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open your math textbook. if you've grasped every theorem, try the net. for starters, www.cut-the-knot.com |
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