You Love Xei

Theorem: Everybody on DreamViews loves Xei

Proof:

Proposition: any group of n people on DV have the same opinion of me.

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

The proposition is clearly true for n = 1; any group of 1 person on DV has the same opinion of me as anybody else in the group, because a person must have the same opinion as themselves.

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

and hence, it must be true for everybody in the forum.

Now that I have proved the proposition that everybody on DV has the same opinion of me, all that remains is to observe that at least one person, namely myself, loves me; and hence I have proved my theorem that everybody on DreamViews loves Xei.

I'd really like to disprove you here, but I would have to be lying. :chuckle:

If everybody on DV loved you, then everybody on DV would love you. However n < total number of members. Noogah doesn't like you. Sorry. :(

Quote:

---

I'd really like to disprove you here, but I would have to be lying.

---

Spread the love!

I have a (very) similar proof that everybody loves dajo, in fact!

I've also discovered a proof that everybody thinks Noogah is an embarrassment to himself, which is again strikingly similar; but you can't argue with pure logic, after all.

Quote:

---

If everybody on DV loved you, then everybody on DV would love you. However n < total number of members. Noogah doesn't like you. Sorry.

---

But if there n - 1 other forum members who love me, and 1 Noogah, as I've proved above, it follows that n members love me, and hence Noogah does too!

my mind asplodededed from trying to compute this, but I still love you

[edit] Nevarmind that last part. I do not hate anybawdy.

Quote:

---

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

---

Those two groups would just have to happen to be sharing the same opinion. The opinion of people 1 through n-1 may not be the same as person number n, in which case you cannot apply your assumption to two groups.

Mm, that doesn't really make sense.

The assumption is that P(n - 1) is true, and P is specifically about any arbitrary group you choose.

So, if you have a population of people named as follows:

1, 2, 3, ..., n - 2, n - 1, n

Then it follows from P(n - 1) that persons 1 to n - 1 (A) all have the same opinion as each other, and that persons 2 to n (B) all have the same opinion as each other.

It cannot be true, as you say, that the opinion of people 1 through n - 1 need not be the same as person n, because the opinion of people 1 through n - 1 is the same as the opinion of people 2 through n - 1 (as they're in group A), and the opinion of person n is the same as the opinion of people 2 through n - 1 (as they're in group B), and if the opinion of people 1 through n - 1 is different from the opinion of person n, then this implies that the opinion of people 2 through n - 1 is different from the opinion of people 2 through n - 1, which is clearly a contradiction.

The only assumption is that P(n - 1) is true. It's shown above that this implies that P(n) is true. You are right to highlight that it is an assumption, but as I covered in my proof, it is in fact patently true for P(1), and hence true for P(2), P(3), and so on for any number.

Is this logic or psychology?

Oh, that makes sense.

But,

Quote:

---

Proposition: any group of n people on DV have the same opinion of me.

---

Is n people the total number of people on DV, or a number less than or equal to the total number?

Second, can you explain how using a proposition and an assumption in a proof makes sense?

I do not love you, no matter how much you say I do. Fail.

Quote:

---

Originally Posted by Xei

Theorem: Everybody on DreamViews loves Xei

Proof:

Proposition: any group of n people on DV have the same opinion of me.

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

The proposition is clearly true for n = 1; any group of 1 person on DV has the same opinion of me as anybody else in the group, because a person must have the same opinion as themselves.

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

and hence, it must be true for everybody in the forum.

Now that I have proved the proposition that everybody on DV has the same opinion of me, all that remains is to observe that at least one person, namely myself, loves me; and hence I have proved my theorem that everybody on DreamViews loves Xei.

---

Jesus. If there was ever anything I'd read, that made me understand how much I hate mathematics, this would be it. Lol.

But, is it not true that you didn't prove that "Every Person" on DV has the same opinion of you? Rather, you simply proved that every group of "n person" has the same opinion of you? And as far as I can see, you haven't proven n to equal anything more than 1. The technicality it looks like you are working off of is that each (individual) person does have "the same" opinion of you. (The "same" opinion as his/herself). So any entire group of just 1 person on DV, is going to be at a consensus.

Correct?

(Couldn't bear to read anymore than the OP, yet. My mind is already fried. And if I'm way off, just disregard it. I've been chiefing all morning. :chuckle:)

Quote:

---

Is n people the total number of people on DV, or a number less than or equal to the total number?

Second, can you explain how using a proposition and an assumption in a proof makes sense?

---

n is a symbol. The proof shows that the proposition is true if I replace n with any natural number.

Propositions and assumptions are made in proofs all the time.

Here I have proved the proposition by

1. Showing that if you assume the proposition is true for n - 1, it follows that it must be true for n.
2. Showing that the proposition is indeed true for 1; and hence by step 1., it must also be true for 2, and hence for 3, and so on, for any number.

This method is called the principle of mathematical induction.

For example, consider the statement 'the sum of all numbers from 1 to n is 1/2*n*(n + 1)', i.e.

P(n) <=> sum from 1 to n = 1/2*n*(n + 1)

if we assume P(n) is true, by adding n + 1 to both sides, we obtain

(sum from 1 to n) + (n + 1) = sum from n to n + 1
= 1/2*n*(n + 1) + (n + 1) = 1/2*(n + 1)(n + 2)

which is P(n + 1).

So, we have proved that if P(n) is true, P(n + 1) must be true.

Now consider P(1), which states that

The sum of all numbers from 1 to 1 = 1
= 1/2*1*(1 + 1) = 1

Which is true. So P(1) is true, and hence P(2) is true, and hence by induction P(n) is true for all n.

Quote:

---

But, is it not true that you didn't prove that "Every Person" on DV has the same opinion of you? Rather, you simply proved that every group of "n person" has the same opinion of you? And as far as I can see, you haven't proven n to equal anything more than 1. The technicality it looks like you are working off of is that each (individual) person does have "the same" opinion of you. (The "same" opinion as his/herself). So any entire group of just 1 person on DV, is going to be at a consensus.

Correct?

---

The key is that I've proved the truth of n implies the truth of n + 1.

In other words, because it is true for 1, which it patently is, it must be true for 2, 3, and so on; n can be any positive whole number, so there is no problem in making it the number of members on DV.

Quote:

---

I do not love you, no matter how much you say I do. Fail.

---

I didn't say it. I logically proved it.

You fail. :V

Quote:

---

Originally Posted by Xei

I didn't say it. I logically proved it. You fail. :V

---

Then your logic is fail, because as I said, I do not love you. Honestly all this is doing is proving that you cannot trust science and mathematics! Now my whole world is falling apart! :panic:

Why wouldn't your equation be true for everyone? So because your mathematical equation states that i love you means indeed the equation can't possibly be wrong? I think you should do some "field tests" on this and then report your findings...

...is this supposed to be a joke.

Xei, you seem to have the skills of a politician. I think you could use math fallacies like those and convince masses of people that you can end any social problem you bring up.

What fallacy? :0

The steps all seem to follow if you ask me.

Quote:

---

Honestly all this is doing is proving that you cannot trust science and mathematics!

---

No this is demonstrating the high degree of rigour which is required to be able to do mathematics.

I don't see any point in arguing to be honest; as you all have the same opinion as me, it follows that you all think my proof is infallible, so any objections you make are just you lot joking around.

Quote:

---

Originally Posted by Xei

Theorem: Everybody on DreamViews loves Xei

Proof:

Proposition: any group of n people on DV have the same opinion of me.

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

The proposition is clearly true for n = 1; any group of 1 person on DV has the same opinion of me as anybody else in the group, because a person must have the same opinion as themselves.

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

and hence, it must be true for everybody in the forum.

Now that I have proved the proposition that everybody on DV has the same opinion of me, all that remains is to observe that at least one person, namely myself, loves me; and hence I have proved my theorem that everybody on DreamViews loves Xei.

---

Theorem: Everybody on DreamViews does not love Xei

Proof:

Proposition: any group of n people on DV have the same opinion of Xei.

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of Xei.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of Xei.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

The proposition is clearly true for n = 1; any group of 1 person on DV has the same opinion of Xei as anybody else in the group, because a person must have the same opinion as themselves.

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

and hence, it must be true for everybody in the forum.

Now that I have proved the proposition that everybody on DV has the same opinion of Xei, all that remains is to observe that at least one person, namely Aquanina, does not love Xei; and hence I have proved my theorem that everybody on DreamViews does not love Xei.

Yeah I already did that joke.

Although to be honest I don't think either you or I can speak for Noogah. The only person whose opinions I am 100% sure of are mine, and I know for sure that I love myself, and hence by my flawless argument, everybody loves me.

An interesting corollary of this fact is that Noogah loves me too, so as I cautioned, you were indeed misplaced in your opinion of him.

Quote:

---

Originally Posted by Xei

Yeah I already did that joke.

Although to be honest I don't think either you or I can speak for Noogah. The only person whose opinions I am 100% sure of are mine, and I know for sure that I love myself, and hence by my flawless argument, everybody loves me.

An interesting corollary of this fact is that Noogah loves me too, so as I cautioned, you were indeed misplaced in your opinion of him.

---

I edited it because we have testimony from Aquanina. Your case has been dismissed, and you are hereby required to pay all court costs for the opposing parties.

She was lying.

I can't 100% trust anybody but myself, so it's only a proof when I apply my opinions.

And said proof implies that Nina does indeed love me. I guess she's just shy or something.

Quote:

---

Originally Posted by Xei

She was lying.

I can't 100% trust anybody but myself, so it's only a proof when I apply my opinions.

And said proof implies that Nina does indeed love me. I guess she's just shy or something.

---

Can you mathematically prove that I should trust your account of your own emotions more than I can trust her account of her own emotions?

Also, if n - 1 = 1, doesn't n = 2, which is 1 more than the number of people who love (or don't love) you?

Quote:

---

Can you mathematically prove that I should trust your account of your own emotions more than I can trust her account of her own emotions?

---

Nah.

Quote:

---

Also, if n - 1 = 1, doesn't n = 2, which is 1 more than the number of people who love (or don't love) you?

---

I've proved that if P is true for n - 1, then P is true for n. In other words, if it's true for some number, then it is true for the next number.

I've also proved P is true for 1.

As we can put n - 1 = 1, then P is true for n - 1, and hence is true for n, which as you say is clearly 2.

You then iterate this process to generate any number you want.

Basic mathematical induction. (Check out my reply to Invader; also http://en.wikipedia.org/wiki/Mathematical_induction is quite good).

You defined n - 1 as the number of members who love you. Didn't you? Then n = 1 more than the number of members who love you. Is that correct?

If we accept your proposition that you love yourself, which we can only do with blind faith, then we arrive at the conclusion that n - 1 > 1. So far, we have no reason to conclude n - 1 is greater than 1, so we are only convinced that n - 1 = 1. In that case, n = 2, but n is just 1 more than n -1 and has no other significance. So how do we get to n - 1 = number of members?

This is where you touched on the topic...

Quote:

---

Originally Posted by Xei

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

---

Those are hypotheticals. You, with the factor of our blind faith involved, showed that n - 1 > 1 but only proved that n = 1. In other words, you proved that n - 1 could be greater than 1, not that it actually is. When you discussed how the statements would apply to 2 and 3 and so forth, you did not prove that n - 1 is 2 or more. You talked about how things would hypothetically work if it were the case. Thus, you did not prove that n = total number of members, only that it could be the case.

Here's the proof everyone on the forum loves Xei:

If you don't love him, I'll come after you during the night.

See? Instant success.

Quote:

---

Originally Posted by Xei

She was lying.

I can't 100% trust anybody but myself, so it's only a proof when I apply my opinions.

And said proof implies that Nina does indeed love me. I guess she's just shy or something.

---

Administer the polygraph! I'm ready! :shadewink:

Unfortunately there's no convincing evidence that polygraphs are much better than chance. :(

That's why it makes me a bit sick on talk shows where they force some guy to take a polygraph and it comes back with '99% certain you were cheating on your wife' and the guy is just like wtf no I didn't.

They're certainly not accepted in courts of law.

Quote:

---

You defined n - 1 as the number of members who love you. Didn't you? Then n = 1 more than the number of members who love you. Is that correct?

If we accept your proposition that you love yourself, which we can only do with blind faith, then we arrive at the conclusion that n - 1 > 1. So far, we have no reason to conclude n - 1 is greater than 1, so we are only convinced that n - 1 = 1. In that case, n = 2, but n is just 1 more than n -1 and has no other significance. So how do we get to n - 1 = number of members?

---

No, n can be any number. It is a dummy variable in the proposition. The proposition can be true or false depending on what value you assign n. The proof shows that if you assign n any positive whole number, the proposition is true. Therefore you can assign it the whole number of people on DV and it will still be true.

Quote:

---

Those are hypotheticals. You, with the factor of our blind faith involved, showed that n - 1 > 1 but only proved that n = 1. In other words, you proved that n - 1 could be greater than 1, not that it actually is. When you discussed how the statements would apply to 2 and 3 and so forth, you did not prove that n - 1 is 2 or more. You talked about how things would hypothetically work if it were the case. Thus, you did not prove that n = total number of members, only that it could be the case.

---

No, I proved that if P is true for some number n - 1 then P is true for the number n.

For example, if I could prove that P was true for -10, it would follow that P is true for -9. As P is true for -9, it would then follow that P is true for -8, and so on.

Of course, P is false for -10, because the phrase 'any group of -10 people' is meaningless, and therefore nothing is actually said about the truth of P for -9, or -8, etc.

However I did prove that P is true for 1, and then, given that P(n - 1) implies P(n), it is incontrovertible that P is true for 2, and hence true for 3, and hence true for any positive integer.

Quote:

---

Originally Posted by Aquanina

Administer the polygraph! I'm ready! :shadewink:

---

Quote:

---

Originally Posted by Xei

Unfortunately there's no convincing evidence that polygraphs are much better than chance. :(

That's why it makes me a bit sick on talk shows where they force some guy to take a polygraph and it comes back with '99% certain you were cheating on your wife' and the guy is just like wtf no I didn't.

They're certainly not accepted in courts of law.

---

Sodium thiopental.... that'll give you your answer

Quote:

---

Originally Posted by Xei

No, I proved that if P is true for some number n - 1 then P is true for the number n.

---

You did? Where? All I saw was that you said it must be true. Where was the proof again?

Quite clearly I don't just 'assert that it's true';

Quote:

---

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n. [So these groups share person 2 through to person n - 1].

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

---

:uhm: . . . . love xei?


:lipssealed:

:furious:

:listenup:

:madtongue:


















Okay I submit :teeth: (where's my prize?)

Quote:

---

Originally Posted by Xei

Quite clearly I don't just 'assert that it's true';

---

Quote:

---

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

---

I must be pretty dense then, because it really seems like this is just an assertion.:? I don't deny the possibility but could you explain to me how it clearly follows that person 1 through to person n must have the same opinion of you? I know you showed that person n-1 when n=2 must have the same opinion (as themself) of you, that was pretty simple to understand. Where is the proof for the rest of it?

I feel like you learned this whole thing in class today and you have missed some vital piece of information. Despite the slightly esoteric language used, it still just seems like a baseless assertion to me, and this opinion is supported by the fact that I know the logic must be false, since someone somewhere in the world doesn't love someone else despite the fact that identical reasoning could be applied to any person.

It's right there in my post above, and I was pretty meticulous about it.

What you just quoted is only an assertion because you removed the whole of the preceding proof. :l

We assume that P(n - 1) is true.

In other words, it is true that you can take any group of n - 1 people and they will all have the same opinion as each other.

We now take two groups with some people in common; let's say we have a population of people,

1, 2, 3, ..., n - 2, n - 1, n.

and we take the groups 1 to n - 1, and 2 to n.

Because these are groups of n - 1 people, P(n - 1) tells us that within each group, all the people have the same opinion. As the two groups share people 2 to n - 1, it follows that the whole population, 1 to n, has the same opinion.

Hence the truth of P(n - 1) implies the truth of P(n).

For example, let's assume P(100) is true; so, any group of 100 people I choose have the same opinion as each other.

Now consider the population

1, 2, 3, ..., 99, 100, 101.

By P(100), we know that people 1 to 100 all have the same opinion as each other, and that people 2 to 101 all have the same opinion as each other, and hence people 1 to 101 all have the same opinion as each other, hence if P(100) is true, P(101) must also be true.

Quote:

---

Originally Posted by Universal Mind

If everybody on DV loved you, then everybody on DV would love you. However n < total number of members. Noogah doesn't like you. Sorry. :(

---

Noogah = 0

Right, I got that; I just don't understand how you get from, "We assume that P(n - 1) is true." to "hence it must be true for everyone on the forum."

Your proof is based on an assumption that has no logical basis. It seems so obvious that this is true that I feel like I'm either a crazy person or that is the entire point of the thread, which is why I originally asked if this was a logical problem or a psychological one.

Are you trying to show that people base too much on assumptions? That it is easy to be mislead by authoritative sounding rhetoric? What is the point exactly?

Quote:

---

Originally Posted by Xei

No, I proved that if P is true for some number n - 1 then P is true for the number n.

---

How exactly did you prove that?

There were 4 members of The Beatles during the time they were recording albums (if you don't count their sound mixing producer or any of their guest studio musicians). If we define the number of Beatles as n - 1, then there were therefore n Beatles? n - 1 = 4, and there had to have been n Beatles, so there were 5 Beatles? Hey, George Martin really was the 5th Beatle. So now n - 1 = 5. Because there were of course n Beatles if there were n - 1, there were 6. Would that be Billy Preston or Eric Clapton? Following the reasoning further, there was an infinite number of them. We are all Beatles.

What about the Rolling Stones?

You can use the same idea to prove that there is more than 1 of you. If n - 1 = the number of Xeis, there are n Xei's, so there are 2, thus 3, thus 5, etc. An infinite number of Xeis exist.

Are you The Beatles?

Quote:

---

Right, I got that; I just don't understand how you get from, "We assume that P(n - 1) is true." to "hence it must be true for everyone on the forum."

Your proof is based on an assumption that has no logical basis. It seems so obvious that this is true that I feel like I'm either a crazy person or that is the entire point of the thread, which is why I originally asked if this was a logical problem or a psychological one.

Are you trying to show that people base too much on assumptions? That it is easy to be mislead by authoritative sounding rhetoric? What is the point exactly?

---

It was partly for fun, partly be because I thought people would be interested; because it does seem to be a logical argument.

The assumption has a firm logical basis, which is P(1).

Because the truth of P(n - 1) implies the truth of P(n), the truth of P(1) implies the truth of P(2), and the truth of P(2) implies the truth of P(3), and so on.

It's a standard case of mathematical induction. Nothing is wrong with the above step.

If you don't understand mathematical induction then please refer to post 12, and the relatively good Wikipedia article,

http://en.wikipedia.org/wiki/Mathematical_induction.

Quote:

---

There were 4 members of The Beatles during the time they were recording albums (if you don't count their sound mixing producer or any of their guest studio musicians). If we define the number of Beatles as n - 1, then there were therefore n Beatles? n - 1 = 4, and there had to have been n Beatles, so there were 5 Beatles?

---

I never made this argument; if I had I would clearly be wrong. Saying the number of Beatles is n - 1, and then asserting the number of Beatles is also n, is clearly baseless.

If you could somehow prove that there being n - 1 Beatles implied that there were n Beatles then the argument would be correct, but you didn't prove this.

I however did prove that for my proposition the truth of n - 1 implied the truth of n, and to address your query,

Quote:

---

How exactly did you prove that?

---

I went into obscene detail in post 33.

Quote:

---

Originally Posted by Xei

If you could somehow prove that there being n - 1 Beatles implied that there were n Beatles then the argument would be correct, but you didn't prove this.

---

It only works with DV members?

I did read the wiki. Now having reread it, it appears that the disproof to "Pólya's proof that there is no 'horse of a different color'" also applies to this situation, as P(n-1) is false in the case of n=1, since 0 people cannot have the same opinion of you. Let me know if I've misunderstood again.

Quote:

---

It only works with DV members?

---

You were trying a totally different (mock) proof by basically asserting the inductive step. I however proved the inductive step.

Quote:

---

I did read the wiki. Now having reread it, it appears that the disproof to "Pólya's proof that there is no 'horse of a different color'" also applies to this situation, as P(n-1) is false in the case of n=1, since 0 people cannot have the same opinion of you. Let me know if I've misunderstood again.

---

lol, this is indeed a bastardization of Polya's horses. However, the problem is not with P(n - 1) when n = 1, in other words P(0), because I never asserted P(0), only P(1) and upwards. Try a couple of other special cases of n - 1 though and you'll have it (I find it much more rewarding if you get the trick yourself).

Quote:

---

Originally Posted by Xei

I however proved the inductive step.

---

I then applied it to my proof.

But they were different propositions.

My proposition was that any group of n people has the same opinion of me.

Your proposition was that there are n Beatles.

How was the proof adapted?

Quote:

---

Originally Posted by Xei

(I find it much more rewarding if you get the trick yourself).

---

Can't tell if you intended this or not but this sounds like condescension. Thanks professor Xei. :roll::P

Of course not. You know where the mistake is, I'm just saying that if you work it out yourself then you'll have understood it, whereas if I tell you then there's not much point. Just like any other puzzle.

Yeah I know. I threw in so many smileys to try to make it obvious that I wasn't offended.

:cheeky::panic::):D:shadewink::lol:;):P:banana:

Quote:

---

Originally Posted by Xei

But they were different propositions.

My proposition was that any group of n people has the same opinion of me.

Your proposition was that there are n Beatles.

How was the proof adapted?

---

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.

I can't follow the math at all... just seeing it hurts my head... but I find it very interesting that you chose the letter N, and the only people we know for a fact DON'T love you are Nina and Noogah! :cheeky:

(Though I suspect Nina is just in denial)

Hm, to use n here would be pretty standart..

... and yet... funny! :banana:

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:

Quote:

---

Originally Posted by Darkmatters

(Though I suspect Nina is just in denial)

---

Men are so stupid. :P

Quote:

---

Originally Posted by Darkmatters

I can't follow the math at all... just seeing it hurts my head... but I find it very interesting that you chose the letter N, and the only people we know for a fact DON'T love you are Nina and Noogah! :cheeky:

(Though I suspect Nina is just in denial)

---

Interesting observation. :V

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.

---

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'.

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.

Quote:

---

Originally Posted by Darkmatters

I can't follow the math at all... just seeing it hurts my head... but I find it very interesting that you chose the letter N, and the only people we know for a fact DON'T love you are Nina and Noogah! :cheeky:

(Though I suspect Nina is just in denial)

---

Lol! That's hilarious! :D The power of N!

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:

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.

---

Well yes, all you're doing there is pointing out that the proof is patently flawed, which we all knew anyway.

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.

Quote:

---

Originally Posted by Xei

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.

---

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.

Quote:

---

Originally Posted by Xei

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.

---

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.)

Quote:

---

Originally Posted by Xei

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

---

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.

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.

---

lol ok.

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.

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.

---

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.

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.

Quote:

---

(By the way, how do you pronounce your screen name? I have wondered that for a while.)

---

"ξ". ;)

The way a German would, basically.

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.

---

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.

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:

Quote:

---

Originally Posted by Xei

Please tell me which of these steps you don't understand or think is wrong, and why:

---

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?

Quote:

---

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?

---

n is a dummy variable and not fixed, so

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.

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?

---

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.

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.

Quote:

---

Originally Posted by Xei

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

---

I'll try interpreting that another way. Are you saying that no matter what value you give n - 1, the statement is true? If so, how does that mean that it is legitimate to give it any value other than 0 (with n being 1)?

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:

Quote:

---

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

---

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:

---

I'll try interpreting that another way. Are you saying that no matter what value you give n - 1, the statement is true?

---

I'm saying that P(n) is true if n is a positive integer.

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.

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? @_@

---

The proof deduces that their collective opinions must be the same if P is true.

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.

Quote:

---

Originally Posted by Xei

I'm saying that P(n) is true if n is a positive integer.

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.

---

Well, yeah, I meant for any value that can actually be a number of people. Aside from that technicality, did I characterize it correctly?

<3

<3 <3 <3

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?

---

Well, what you say is true, although it's more of the result of the proof than the proof itself.

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.

Quote:

---

Originally Posted by Xei

<3 <3 <3

Well, what you say is true, although it's more of the result of the proof than the proof itself.

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.

n and n - 1 are all just dummy variables.

The inductive statement "P(n - 1) implies P(n) for all n - 1" is completely equivalent to "P(n) implies P(n + 1) for all n" or "P(x) implies P(x + 1) for all x".

n can be given any value without contradiction. If n - 1 = 5, the inductive step "P(n - 1) implies P(n)" says that P(5) implies P(6).

This says nothing about the truth of P(5).

However, because P(1) is true, we know that P(2) is true. And because P(2) is true, we know that P(3) is true. And so on.

Wouldn't you have to define n + 1 as the number of people in the group for n to prove n + 1?

What is the significance of P(2) when you have only proven P(1)? I know the statement you are using would work with a value of 2 if you had proven P(2), but you never proved P(2).

I don't love you.

Wrong.

Quote:

---

Wouldn't you have to define n + 1 as the number of people in the group for n to prove n + 1?

What is the significance of P(2) when you have only proven P(1)? I know the statement you are using would work with a value of 2 if you had proven P(2), but you never proved P(2).

---

Because I proved that for all positive integers n if P(n) is true then P(n + 1) is true.

1 is a positive integer and P(1) is true hence P(1 + 1) = P(2) is true.

Quote:

---

Proposition: any group of n people on DV have the same opinion of me.

---

^You are proving this, in the proof. "Any group" can mean 5 people, or everyone.

Quote:

---

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

---

^You use this as a part of the proof. This can also mean 4 people, or every member minus 1 member. But this alone isn't proven either.

I don't think that P(n-1) is implying P(n). I think P(n) is implying P(n-1). Any group of people that has the same opinion will have a group inside that group with the same opinion.

By applying your original assumption to "two overlapping groups," you were doing nothing more than taking one group of 1 through n people and labeling them '1 through n-1' and '2 through n', which is a clever way of making it look like the shared opinion of a smaller group is implying that this total larger group shares the same opinion, when in fact the whole group would already have to have had the same opinion to begin with in order for any of this to work.

Am I getting warmer? I haz headache.

P(n) implies P(n-1), and not vice-versa.

You're in the right ballpark, ish.

P(n) does imply P(n - 1), this is true, but it isn't particularly important.

The trick is not wrong for the reason you give; there is nothing wrong with the idea that if all groups of, say, 5 people have the same opinion, then it follows, by taking two groups of n - 1 people with 4 people in common, that every group of 6 people must have the same opinion.

You are right that 'in real life' this is clearly absurd because the two groups don't have to have the same opinion, but we already knew the proof was absurd. This doesn't show why it's absurd. So far the logical argument hasn't been contradicted and so we must accept its truth.

Quote:

---

there is nothing wrong with the idea that if all groups of, say, 5 people have the same opinion

---

Why is there nothing wrong with this? You're trying to prove that a group of people has the same opinion by stating that any group of people will already have the same opinion.

I think the logic is wrong in that P(1) does not imply (P2). In your example you use two groups with enough people that some of the members in each group will be in both groups. If you only have two people, however, you can't do this. The argument is built on the fact that we can know that a group of one people will have the same opinion as everyone in the group. The flaw has to come in implying that this will suggest the same for a group of two people.

So let's apply it to two groups of one people. Person 1 through person n-1, and person 2 through person n, when n=2. Group one will have person 1 through 1, and group two will have person 2 through 2. The member of neither group exists in the other group, so no relationship can be drawn, and P(1) can therefor not imply P(2).

Quote:

---

Why is there nothing wrong with this? You're trying to prove that a group of people has the same opinion by stating that any group of people will already have the same opinion.

---

There can't be anything wrong with a statement which just says 'if X'.

Quote:

---

I think the logic is wrong in that P(1) does not imply (P2). In your example you use two groups with enough people that some of the members in each group will be in both groups. If you only have two people, however, you can't do this. The argument is built on the fact that we can know that a group of one people will have the same opinion as everyone in the group. The flaw has to come in implying that this will suggest the same for a group of two people.

So let's apply it to two groups of one people. Person 1 through person n-1, and person 2 through person n, when n=2. Group one will have person 1 through 1, and group two will have person 2 through 2. The member of neither group exists in the other group, so no relationship can be drawn, and P(1) can therefor not imply P(2).

---

:V

Yeah that's the answer essentially.

P(n) does imply P(n + 1), except when n = 1, because in the proof that P(n) implies P(n + 1), the assertion that there is an intersection of two groups is not founded when n = 1.

Google for Polya's horses for more info. You win the prize.

yeah i didn't read this, but i loooove you. let's be fest briends.

also i'm kidna drunk.

If Xei gets laid because of a flawed mathematical proof, I am going to be pissed.

Too late, you've all had sex with me.

(Want a proof?).

Many of you were sleeping though.

Quote:

---

Originally Posted by Xei

Yeah that's the answer essentially.

---

I would like to thank Xei for this prestigious award, whom I can now say I do not love.

I am also grateful that I can sleep in peace now. *sigh of relief*

Quote:

---

Google for Polya's horses for more info.

---

Will do. :)

I read the Wikipedia page on Polya's horse, and it cleared up my confusion. I missed the word "any" in the original propostion and thought n represents a number of members who love you though n does not necessarily represent the total number. The whole time, I kept thinking, "Uh, n people love Xei, and perhaps more than that. And how do we get to the conclusion?"

Here's the key Wikipedia excerpt on Poly'a horse.

The Argument
The flawed argument claims to be based on mathematical induction, and proceeds as follows:
Suppose that we have a set of five horses. We wish to prove that they are all the same color. Suppose that we had a proof that all sets of four horses were the same color. If that were true, we could prove that all five horses are the same color by removing a horse to leave a group of four horses. Do this in two ways, and we have two different groups of four horses. By our supposed existing proof, since these are groups of four, all horses in them must be the same color. For example, the first, second, third and fourth horses constitute a group of four, and thus must all be the same color; and the second, third, fourth and fifth horses also constitute a group of four and thus must also all be the same color. For this to occur, all five horses in the group of five must be the same color.
But how are we to get a proof that all sets of four horses are the same color? We apply the same logic again. By the same process, a group of four horses could be broken down into groups of three, and then a group of three horses could be broken down into groups of two, and so on. Eventually we will reach a group size of one, and it is obvious that all horses in a group of one horse must be the same color.
By the same logic we can also increase the group size. A group of five horses can be increased to a group of six, and so on upwards, so that all finite sized groups of horses must be the same color.

Explanation
The argument above makes the implicit assumption that the two subsets of horses to which the induction assumption is applied have a common element. This is not true when n = 1, that is, when the original set only contains 2 horses.
Indeed, let the two horses be horse A and horse B. When horse A is removed, it is true that the remaining horses in the set are the same color (only horse B remains). If horse B is removed instead, this leaves a different set containing only horse A, which may or may not be the same color as horse B.
The problem in the argument is the assumption that because each of these two sets contains only one color of horses, the original set also contained only one color of horses. Because there are no common elements (horses) in the two sets, it is unknown whether the two horses share the same color. The proof forms a falsidical paradox; it seems to show something manifestly false by valid reasoning, but in fact the reasoning is flawed. The horse paradox exposes the pitfalls arising from failure to consider special cases for which a general statement may be false.

I like you already. Let me give this a crack. So, we've got the set of all of Dream Viewers, P; elements of that set (people), with a limiting behavior as x approaches P_max, n, n-1, n-2... Based on the assumption that the worst case scenario one person may not like Xei, n-1, and if P = n, and P = n-1 (in order for everyone to like Xei not matter what), then n = n-1. Saddly, 1 ≠ 0. This is also assuming, however, a static emotional field, where as an emotional field in flux may change over time.

Algebraically, this might be represented with big O notation as O[P(n)] < P as n → ∞. Too funny.

Quote:

---

Originally Posted by Xei

Too late, you've all had sex with me.

(Want a proof?).

Many of you were sleeping though.

---

Must not have been that good then.

Actually I can prove that you all loved it.

Quote:

---

I like you already. Let me give this a crack. So, we've got the set of all of Dream Viewers, P; elements of that set (people), with a limiting behavior as x approaches P_max, n, n-1, n-2... Based on the assumption that the worst case scenario one person may not like Xei, n-1, and if P = n, and P = n-1 (in order for everyone to like Xei not matter what), then n = n-1. Saddly, 1 ≠ 0. This is also assuming, however, a static emotional field, where as an emotional field in flux may change over time.

---

Is this supposed to be another proof or a restatement of mine?

pics or it didn't happen

My intent was another rigorous, highbrow counter-proof. Then again, I am a little late to the party.

Arcanum doesn't love Xei, Arcanum loves everyone. Therefore, everyone doesn't love Xei, but Xei is loved.

Quote:

---

Originally Posted by ArcanumNoctis

Arcanum doesn't love Xei, Arcanum loves everyone. Therefore, everyone doesn't love Xei, but Xei is loved.

---

I.e. You Love Xei

Quote:

---

Originally Posted by Xei

Theorem: Everybody on DreamViews loves Xei

Proof:

Proposition: any group of n people on DV have the same opinion of me.

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

The proposition is clearly true for n = 1; any group of 1 person on DV has the same opinion of me as anybody else in the group, because a person must have the same opinion as themselves.

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

and hence, it must be true for everybody in the forum.

Now that I have proved the proposition that everybody on DV has the same opinion of me, all that remains is to observe that at least one person, namely myself, loves me; and hence I have proved my theorem that everybody on DreamViews loves Xei.

---

Dammit to late to spoil the fun.

Induction isn't that tricky.

Quote:

---

Originally Posted by Xei

Theorem: Everybody on DreamViews loves Xei

Proof:

Proposition: any group of n people on DV have the same opinion of me.

Let's proceed by first assuming that the proposition is true for n - 1; i.e., any group of n - 1 people on DV have the same opinion of me.

Apply this to two groups, for example, person 1 through to person n - 1, and person 2 through to person n.

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

In other words, if the proposition is true for n - 1 people, it must be true for n people.

The proposition is clearly true for n = 1; any group of 1 person on DV has the same opinion of me as anybody else in the group, because a person must have the same opinion as themselves.

Hence, as the truth of n - 1 implies the truth of n, as it is true for 1, it must be true for 2, and hence it must be true for 3, and hence...

and hence, it must be true for everybody in the forum.

Now that I have proved the proposition that everybody on DV has the same opinion of me, all that remains is to observe that at least one person, namely myself, loves me; and hence I have proved my theorem that everybody on DreamViews loves Xei.

---

Yeah... so if 1 person loves you, 2 people love you, therefore 3 people love you, therefore

WHAT THE HELL ARE YOU SMOKING LOL

no I don't. you are the most closed minded ass in all of ED.

Go back to hyperspace.

Quote:

---

Originally Posted by Dreams4free

Yeah... so if 1 person loves you, 2 people love you, therefore 3 people love you, therefore

WHAT THE HELL ARE YOU SMOKING LOL

---

Dunno, what am I smoking? Point out the flawed step in the argument and maybe I will tell you.

Quote:

---

Originally Posted by Man of Shred

no I don't. you are the most closed minded ass in all of ED.

---

You have zero comprehension of this thread. Go away.

That made absolutely no sense to me. O__e

Feel my love, Xei <3 HUGS

I'm glad for the resurgence of this mighty thread! I love you all too (and I can prove it), my devoted followers!

Quote:

---

Originally Posted by Xei

Dunno, what am I smoking? Point out the flawed step in the argument and maybe I will tell you.


You have zero comprehension of this thread. Go away.

---

You are assuming that 1 person "loves" you, sure I will give you that.

But then... You essentially define every n group and n-1 group as a variation of the same person.
So now we are looking at groups of 1, 2, 3 that all consist of this same person multiplied, who happens to love you.

And as we know, groups of 1,2,3 people have different individuals in them (even groups of 1 person).

So what happens when N transforms from a constant into a variable, you have this^ dilemma.

If we are talking about static groups and arrays, you might have more flexibility.

But here on DV group n doesn't constitute every group n. because groups of (1) and groups of (2) are variable not constant.

You see, we could count 1,000,000 people are 1,000,000 groups of (1), and label all groups of (1) = n.

that is essentially what you just did.

I hope this was a joke :D.

Hypothetically, probabilistic theory and limits would be a more sound path to take to prove everyone loves you.

Quote:

---

Originally Posted by Dreams4free

But then... You essentially define every n group and n-1 group as a variation of the same person.

---

I never defined a group as anything. A group is a group.

Can you point to the actual flawed step in the argument I wrote?

Yeah, this is a bit of a joke, and only numpties have thought it was serious and that somebody could possibly be so deluded and arrogant (obviously I can use the same theorem to 'prove' that every colour is blue or something), but it is an interesting logical exercise.

I'm a little late to the party and I have only skimmed this thread, but here is my explanation as to why the proof is flawed.

There is a flaw in your proposition.

Your Proposition:

Quote:

---

Proposition: any group of n people on DV have the same opinion of me.

---

You are trying to proof it by induction on n, i.e. by showing the following two things:

(1) The proposition is true for n = 1.
(2) If the proposition is true for n - 1, it is also true for n (assuming n > 1).

Your reasoning for (1) is sound. However, there is a problem with the way you are trying to proof (2).

Given n people, you are forming two groups:
A = {1 … n-1} and
B = {2 … n}

Clearly A and B both contain n-1 people and we are assuming that the proposition is true for n-1.

You say:

Quote:

---

It clearly follows that all of these people, person 1 through to person n, must have the same opinion of me.

---

Wrong!

The fundamental idea here is that if you have two sets of objects (in this case, people) and the sets were formed in such a way that all the objects within are equal in one particular aspect (in this case, their opinion of Xei) and – this is the third and important part of the premise – those sets intersect, then you can conclude that all the objects within the union of those sets are also equal in this particular aspect.

But again, this only works if the sets intersect. And as for the problem at hand, there is a case where those two groups of people you have formed, A and B, do not intersect, namely the case n = 2.

Let's examine the case n = 2. The groups of people are:
A = {1 … n-1} = {1} and
B = {2 … n} = {2}

As you can see, A and B are disjoint, therefore you can not conclude that the two groups share the same opinion of you. Person 1 might like you and Person 2 might hate you.

For the reasons outlined above, your induction is flawed and thus your proof is too.

Yes, that's exactly correct. It's a famous 'paradox' called Polya's Horses (which 'proves' all horses are the same colour).