you could use this thread to prove that everybody thinks so ;)
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you could use this thread to prove that everybody thinks so ;)
Heh... my attempt to do so would only prove that I can't do math! :panic:
Interesting observation. :V
Well, what I said can be generalised to 'if any group of n - 1 objects share some property then any group of n objects share that property'.Quote:
Our proofs used the same principle-- If n - 1 people do x, then n people do x.
Okay, true or false?... If n - 1 people do x, then n people do x.
And hence 'if any group of n - 1 people do the same thing, then any group of n people do the same thing'.
This is a bit different from your principle for a couple of (subtle but vital) reasons.
Firstly, you can't prove yours for the base case. In other words, you can't prove the statement that '1 person always does x'.
However, my statement in the base case, 'any group of 1 person does the same thing as everybody else in the group' is trivially always true.
That's why you can't prove your statement with my proof.
Secondly, your statement asserts the existence of n people (because if n people do x then at least n people exist).
However mine only asserts that if you can find a group of n people, then it follows that they all do the same thing; in other words it does not assert the existence of n people.
That's why your argument, if it were true, would (as you said) imply the existence of infinite Xeis; mine however only implies that all people who exist are Xei.
Something still isn't making sense. You're using 1 through n-1 to represent the number of people in a group. This number of people is the same as the sum of numbers 2 through n. The actual people in the groups however are denoted by the number (person 1, person 2, person 3 ... person n). The integers themselves are not arbitrary, they are referring to a person. So although you can say that the number in each of these groups is the same, you can't just shift your bounds over to include another person who's quality you're assessing may be unlike the shared quality in group 1 through n-1. Am I making sense? I mean, do you understand how I'm seeing this?
I don't really know what you mean. I've used '1 through n - 1' as shorthand for 'person 1 through to person n - 1' generally, I think. I don't get why you are referring to the 'sum of numbers 2 through n'; do you mean the number of numbers in 2 through n..?
I'm confused as to your general meaning but to try to address what I think may have been your main point: the proposition specifically applies to any arbitrary group of people, so there is no issue in applying it to any arbitrary group of people any number of times to establish further proofs, or 'shifting the bounds' as you say. If P(n - 1) is true, then it is true for person 1 to person n - 1, and it is also true for person 2 to person n (because P is specifically about any group of people of the correct size), and it follows that P(n) is true.
You said that you wanted to apply your proposal to two groups (1 through n-1 and 2 through n). I need to use an example..
Imagine that there are 1,000 users here, and that 400 of them love you. If we define that group of 400 as having people 1 through n-1 in it, n is equal to 401. Person one through person 400. If you want to apply it to a second group (person 2 through person n), then you are including a person that doesn't fit into the group who shared the same quality. The 401st person exists outside of group one. Each number is referring to one person in the group, see? I don't get how you can just apply your statement to a second group like this, obviously because I'm not getting the way your proof works.
When I said "sum of numbers.." I was referring to the size of the groups. Groups one and two have the same size but different members, and I thought that's where the confusion was coming into play.
well, shit. I'm terrible at math. I guess you're right, since there's no way I can prove you wrong...I love you.
now bend over. (unless you can come up with some fancy mathematical proof that love does not equal sex? ;D)
Dammit Nerve! You just killed my n theory!!
D:
Well yes, all you're doing there is pointing out that the proof is patently flawed, which we all knew anyway.Quote:
Imagine that there are 1,000 users here, and that 400 of them love you. If we define that group of 400 as having people 1 through n-1 in it, n is equal to 401. Person one through person 400. If you want to apply it to a second group (person 2 through person n), then you are including a person that doesn't fit into the group who shared the same quality. The 401st person exists outside of group one. Each number is referring to one person in the group, see? I don't get how you can just apply your statement to a second group like this, obviously because I'm not getting the way your proof works.
The question is why.
Why can't I apply the proposition to a second group? The proposition is specifically about any group of people you choose.
Please tell me which of these steps you don't understand or think is wrong, and why:
1. P(n - 1) implies P(n), in other words, if any group of n - 1 people you choose will have the same opinion as everybody else in the group, it follows that any group of n people you choose will have the same opinion as everybody else in that group.
2. P(1) is true, in other words, any group of 1 person you choose will have the same opinion as everybody else in the group.
3. Because P(1) is true, and because P(n - 1) implies P(n) (in other words the truth of P for any number implies the truth of P for that number plus 1), it follows that P(2) is true. Repeating this process, it follows that P(3) is true, and so on for any number.
I am talking about everybody in a group (humans) being a Beatle, or doing stuff that qualifies them as Beatles. I don't need to prove that they always do it because you don't have to always do anything to be a Beatle except live.
Think of it as the group of humans who are Beatles. In the group of humans, if 1 person is a Beatle (as in, was a member at one time), then all other humans are Beatles. Paul McCartney is still alive, and there is tons of proof that he was a member of The Beatles when the band was together. For all we know, an infinite number of humans exist in the system of multiverses.
Also, you can use your reasoning to "prove" that every member here uses the screen name Xei.
(By the way, how do you pronounce your screen name? I have wondered that for a while.)
I concentrated on your conclusion for a bit and wrote some reductio ad absurdums of it, but now I am going to concentrate again on your base reasoning. What you have done is define n as the number of people in a group of members here who have the same opinion of you. However many people are in the group, n is that number. n - 1 is one less than that number. The trick you pulled is treating n - 1 as the least possible number of people in the group instead of treating n as that number. So, when you say you know that 1 person here loves you, you are really saying that n is at least 1, which means that n - 1 is at least 0. You did not prove that n - 1 is at least 1. By alternating the meaning of n repeatedly, you can make it look like the entire group of people here loves you.
lol ok.Quote:
I am talking about everybody in a group (humans) being a Beatle, or doing stuff that qualifies them as Beatles. I don't need to prove that they always do it because you don't have to always do anything to be a Beatle except live.
No what I mean is you can't prove every person (i.e. every group of 1 people) is a Beatle, which is required for your base case.
Well, if you add the premise that there are an infinite number of people, then yes my proposition would imply an infinite number of Xeis.Quote:
Think of it as the group of humans who are Beatles. In the group of humans, if 1 person is a Beatle (as in, was a member at one time), then all other humans are Beatles. Paul McCartney is still alive, and there is tons of proof that he was a member of The Beatles when the band was together. For all we know, an infinite number of humans exist in the system of multiverses.
I'm not sure which point you're making, it looks like we're back at just using my original proof to prove that everybody has attribute X.
"ξ". ;)Quote:
(By the way, how do you pronounce your screen name? I have wondered that for a while.)
The way a German would, basically.
No that's not the logic of the proof. I'm not really sure what your idea of the proof is, I think it is that I have tried to conflate n and n - 1. That is not what I did, or tried to do. In proof by induction n is just a dummy variable so that one may make essentially an infinite number of propositions concisely. One can then make statements like "P(n) implies P(n + 1)", which again represents an infinite number of propositions. Such a proposition on its own implies nothing about the truth of P for any value of n, only that if P is true for n (which it is when n = 1 in my case), it incontrovertibly follows that P is true for n + 1 (which I proved); and if P is not true for n, P(n + 1) may or may not be true.Quote:
I concentrated on your conclusion for a bit and wrote some reductio ad absurdums of it, but now I am going to concentrate again on your base reasoning. What you have done is define n as the number of people in a group of members here who have the same opinion of you. However many people are in the group, n is that number. n - 1 is one less than that number. The trick you pulled is treating n - 1 as the least possible number of people in the group instead of treating n as that number. So, when you say you know that 1 person here loves you, you are really saying that n is at least 1, which means that n - 1 is at least 0. You did not prove that n - 1 is at least 1. By alternating the meaning of n repeatedly, you can make it look like the entire group of people here loves you.
Please read post 12, it is a direct analogue of my proof.
I think a good idea would be to put a marker between the two lines in my proof where you think the false inference has occurred.
It looks like we posted at about the same time. Make sure you catch that last part I added.
Is that symbol a z? I was a baby when I lived in Germany, and I lived on a U.S. Army base any way.
It's the Greek 'Xi', in English I suppose 'Zye'. Germans pronounce 'ie' as 'e' and 'ei' as 'i'.
Note post 12 is a tiny bit wrong, where it says "= the sum from n to n + 1" it should clearly mean "= the sum from 1 to n + 1". It's too late to edit it.
The Wikipedia article's quite good.
Do I love Xei?
I don't know him well enough to say (damn... I thought that rhymed before you gave the pronunciation guide! :cheeky:)
But if you apply Schroedinger's Cat theory, then I love you and I hate you and I'm totally indifferent to you - all possibilities. So I guess that's a win for you.
(And I see Nina really wasn't joking! :eek:) :lol:
Before I do this, I need to know two things:
1. When you talk about n representing a group, you mean that the group is some number of members equal to or less than the number of people on DV. The group you are referring to is not essentially all of DV, is it?
2. I thought you could not apply your premise to two groups and relate the groups to each other because each group may be sporting a different opinion. The people in group 1 will have some opinion A. You can say your premise applies to group 2, who will share opinion B and not necessarily A. It is for that reason I thought the argument was flawed.
I saw it like this: People 1 through n-1 have opinion A. People 2 through n have opinion B. So the application works for both groups, but both groups are not relateable. Person 1 only has opinion A, person n only has opinion B, and people 2 through n-1 have both opinions A and B. So, what's the deal?
n is a dummy variable and not fixed, soQuote:
1. When you talk about n representing a group, you mean that the group is some number of members equal to or less than the number of people on DV. The group you are referring to is not essentially all of DV, is it?
P(5) means 'every person in any group of 5 people you choose on DV will have the same opinion as every other person in that group',
and so on.
The statement only applies to the whole of DV when n = whatever the number of people is on DV; but the proof establishes that P is true for any positive whole number, which 'whatever the number of people is on DV' clearly is.
The deal is that the logical argument was only ever made in terms of P, and P is not 'this group of people of size n - 1 has opinion A and this other group of people size n - 1 has opinion B'; P says nothing about if the opinions of any two groups are different, or the same. It only asserts that the opinions within the group are the same (and nothing about the inter-group relationships), and goes on to prove that this implies P is also true for n.Quote:
I saw it like this: People 1 through n-1 have opinion A. People 2 through n have opinion B. So the application works for both groups, but both groups are not relateable. Person 1 only has opinion A, person n only has opinion B, and people 2 through n-1 have both opinions A and B. So, what's the deal?
What you are saying is right in the sense that you are stating 'there is something wrong with the argument because it isn't true', but you're not actually saying what's wrong with the argument itself.
I thought you were "proving" that P is true for every positive integer with the group relation though. I'm not understanding why you would apply it to two groups at all then. D:
If the proof does not specify that the collective opinion of group one is the same as the collective opinion of group 2, how can we conclude that P is true for all n? @_@Quote:
Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.
I'm saying that P(n) is true if n is a positive integer.Quote:
I'll try interpreting that another way. Are you saying that no matter what value you give n - 1, the statement is true?
In other words, P(1) is true, P(2) is true, P(3) is true, and so on.
There are some values of n for which P(n) might not be true, like 0 or negative numbers or fractions, but I'm not bothered about them because they're not important.
P(1) is true because I proved that separately. P is true for all positive integers because I proved that if P is true for some n then it also true for n + 1.
If you still don't get this method of proof then perhaps try some of the questions here:
http://www.hsc.csu.edu.au/maths/ext1.../mathinduc.htm
The first one, prove 1 + 3 + 5 + . . . + (2n - 1) = n^2 when n is a positive integer, is quite good.
Induction can be a bit weird at first but it all follows logically.
Again I think it would be a good idea to indicate which line of the stated proof is incorrect.
The proof deduces that their collective opinions must be the same if P is true.Quote:
If the proof does not specify that the collective opinion of group one is the same as the collective opinion of group 2, how can we conclude that P is true for all n? @_@
Assume P(n) is true, for instance.
Then take a group of n + 1 people.
Because we have taken P(n) to be true, we know that people 1 to n have the same opinion as each other. We also know that people 2 to n + 1 have the same opinion as each other. Nowhere does the proof assert that these two opinions are the same. However we can then go on to deduce that they must be the same, because if they were not, people 2 to n, who are in both groups, would not have the same opinions as themselves, which is impossible.
Hence P(n + 1) is true if P(n) is.
<3
<3 <3 <3
Well, what you say is true, although it's more of the result of the proof than the proof itself.Quote:
Well, yeah, I meant for any value that can actually be a number of people. Aside from that technicality, did I characterize it correctly?
n - 1 is only really used in the inductive step of said proof.
The statement which is actually proved would be more concisely stated as "P(n) is true for all n"; n - 1 isn't important at this point.
Then that brings up the next question. How does it mean that it is legitimate to give n - 1 any value other than 0 (with n being 1)? You told us you (1 person) love yourself, which makes n = 1 and n - 1 = 0. Although your assessment of how n and n - 1 are related when n equals other numbers is true, you didn't give a reason for us to believe n should be anything other than 1 for your particular situation. Your point was about hypothetical situations and not the actual situation.