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You are not logged in. #1 20091001 17:03:22
Is this cool with you?Hi; Last edited by bobbym (20091001 17:05:12) In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #2 20091002 17:15:16
Re: Is this cool with you?Hi; Last edited by bobbym (20091002 22:35:51) In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #3 20100101 13:48:34
Re: Is this cool with you?Hi; She manipulates this to: She says proved: Someone else says that you must use induction, presumably on the inequality. Who is right? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #4 20100102 02:34:48
Re: Is this cool with you?
Last edited by JaneFairfax (20100102 02:44:46) #5 20100102 02:47:27
Re: Is this cool with you?
That would be looking for a square between and . It does not imply that there is a square between and . Correct proof of #3. If the statement were false, there would exist a natural number and a natural number satisfying With some simple algebraic manipulation, we should arrive at It can be easily checked that there are no natural numbers satisfying for or and so we have a contradiction. Last edited by JaneFairfax (20100102 12:22:14) #6 20100105 01:01:09
Re: Is this cool with you?OK. Congratulations on your deep sense of knowing how to perplex me.
What did you do here? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #7 20100105 01:16:26
Re: Is this cool with you?She removed the n's by doubling the first inequality and combining them. Why did the vector cross the road? It wanted to be normal. #8 20100105 01:24:12
Re: Is this cool with you?Hi mathsyperson; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #9 20100109 22:17:48
Re: Is this cool with you?
The first step is invalid because in order to split the limit of a sum as the sum of limits, you have to make sure all the limits you are taking exist. In this case, none of the three limits exists. The rest of the proof is treating ∞ as a number, which it is not.
can be written since n is positive. There are n+1 integers from n to 2n inclusive and so there are at least n+1 integers between k² and (k+1)². Hence these two integers must differ by at least n+2: Combining with k² < n gives which is another contradiction. Last edited by scientia (20100110 00:53:46) #10 20100110 03:10:12
Re: Is this cool with you?Hi Jane and scientia;
The question was, is her manipulation a proof. Do you like where she stops? Do you find her proof valid? Is more required? When does algebraic manipulation constitute a proof? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #11 20100110 04:32:17
Re: Is this cool with you?
So you just want to prove for all natural numbers n (even though this is the wrong way to go about solving the original question of finding a square between n and 2n)? I don’t know what the person did, but if we want to use her method, we should proceed as follows. Start by noting that for all integers n ≥ 1, Hence for all natural numbers n, and the proof is completed by adding n² throughout. The wrong way to go about it is to start with [1] and get [2]. This would be assuming what you want to prove to be true already, and would therefore prove nothing – yet this is an all too common mistake among students (and sometimes even teachers). So what did the person do? Did she start from [1] and get [2] (which would be incorrect) or did she start with [2] and get [1]? Last edited by scientia (20100110 04:32:58) #12 20100110 06:00:18
Re: Is this cool with you?Hi scientia;
I can't say, I don't know.
I obviously forgot what you and Jane already know, that the first statement in her proof is wrong. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #13 20100127 19:24:14
Re: Is this cool with you?This prob came up somewhere else: Now sum that series: Set the sum to 2/5 and solve for p. Solve the equation any way you can, to get p = 1/3 Has B solved the problem? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #14 20100128 09:46:33
Re: Is this cool with you?Just want to point out that to prove #2 like that, you need to observe that the function is continuous. Why did the vector cross the road? It wanted to be normal. #15 20100128 13:09:04
Re: Is this cool with you?Hi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #16 20100202 03:26:39
Re: Is this cool with you?Hi; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #17 20100202 10:44:28
Re: Is this cool with you?I agree with B. Why did the vector cross the road? It wanted to be normal. #18 20100202 13:40:58
Re: Is this cool with you?Hi mathsyperson; In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #19 20100209 23:05:56
Re: Is this cool with you?How about this? How many integer values can a have? B says just substitute into the inequalities. Supposing a = 1 then 1 < b and b <= 1 + 3b is true. Supposing a = 2 then 2 < b and 2b <= 2 + 3b is true. Supposing a = 3 then 3 < b and 3b <= 3 + 3b is true. Supposing a = 4 then 4 < b and 4b <= 4 + 3b is not true when b>4. The same is true for a=5,6,7... So a can take positive values of 1,2,3 and nothing greater. Did you like this? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #20 20100217 08:06:20
Re: Is this cool with you?How many ways can you arrange 5 A's, 7 B's and 4 C's so there are 3 CA pairs? B) Says the magic number: 87120 ways. Who is right? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #21 20100219 06:26:59
Re: Is this cool with you?Exasperating: Solve these simultaneously by multiplying the top equation by 8 and subtracting the bottom. You get: plug back in: You get another 2x2: Solve for a: Plug into the a^2b^2+c^2 Which equals 1 so option b) 1 is the answer. A says start by just setting a=0 and then solve from there. Because there are 3 variables and two equations, there are an infinite number of ordered triplets (a,b,c) that can satisfy the two equations. However, while the variables change, the value of a^2b^2+c^2 stays constant. In one of these triplets, a=0. So if we set a to 0, we still get the same value of a^2b^2+c^2 as in any other triplet. Which method do you prefer? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #22 20100226 06:14:08
Re: Is this cool with you?Hi; B) says: What do you like? In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #23 20100226 07:55:05
Re: Is this cool with you?I like the value of the second answer, but I like the reasoning of the first. Why did the vector cross the road? It wanted to be normal. #24 20100227 02:58:23
Re: Is this cool with you?Hi mathsyperson; Each power of x represents how many non attacking rooks. The coefficient represents the number of ways. The coefficient of x^4 is 117600. That is the number of ways to place 4 non attacking rooks on an 8x8 board. Since the rook polys are used for derangements. I suspect that you could use the principle of inclusion and exclusion but that looks tedious. Claude Tardif on another forum solved it in this manner: (8*8*7*7*6*6*5*5)/4! = (117600) In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. #25 20100302 20:52:19
Re: Is this cool with you?Hi; has 4 real roots: A says when 27 < k < 5; B says when 5 < k < 27; C says when 5 < k < 0. D says never. In mathematics, you don't understand things. You just get used to them. I have the result, but I do not yet know how to get it. All physicists, and a good many quite respectable mathematicians are contemptuous about proof. 