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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Correct!

Yes: The definition of a a convergent sequence (*a[sub]n[/sub]*) is:

Note that the ∃*L* comes *before* the ∀ε in other words, *L* *does not* depend on the choice of ε. This was not the case with the fake limit *a*[sub]*N*+1[/sub] in the false proof choose a different ε and the same *a*[sub]*N*+1[/sub] might not work at all.

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

I guess I'll post one now. I came up with this for a certain thread which I'm sure you all know of... I won't mention its name here as it has fallen dormant and it might be best for it to stay that way. But anyway, a bit after I wrote it I thought about it again and realized it had an error. Now it's your turn to find it!

Theorem: Let S = {0.9, 0.99, 0.999, ...}. Then 1 ∈ S.

Proof: First, note that S is closed, since its complement **R**\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open. Also, 1 is an accumulation point (limit point) of S, since any open ball B(1, r) contains some point in S that is not equal to 1. But S is closed, and hence contains all of its accumulation points. Then 1 ∈ S.

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**javacodeman****Member**- Registered: 2007-04-04
- Posts: 9

Zhylliolom wrote:

First, note that S is closed, since its complement R\S = (-∞, 0.9) U (0.99, 0.999) U ... is a union of open sets, and thus open

I'll take a stab that the error lies above. I'm not sure what a ball is (maybe you can expound), but S doesn't seem to be closed to me. If S were a finite set, then I think what you wrote above would work. But S is surely an infinite set (with a cardinality equal to the counting numbers). Each element of the set get's closer to 1, but never reaches it. You never address the numbers on the other side of the set, because that union would be [1, + ∞) .

.999999999999.......... is equal to one, but does your set contain this number?

That's all I've got folks.

java

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Yep. When I first wrote the proof I just said "**R**\S = (-∞, 0.9) U (0.99, 0.999) U ..." without thinking. The problem is that saying this actually is the same as saying 1 ∈ S, since I said each set in the union is open! So I proved my assertion by unknowingly assuming its truth. Getting excited about having an interesting proof to a classic problem and writing it quickly like that is how you accidentally make false proofs . Your turn to post one, or you can pass it on to whoever else wants to if you have nothing.

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**javacodeman****Member**- Registered: 2007-04-04
- Posts: 9

Okay. I'm new to the boards (so this may be older than dirt here and I wouldn't know it), but here's an easy one that might be good for some of the younger crowd (I'm sure most of you have seen it).

Theorem: 1 = 2

Proof:

Let A = B Then,

A² = AB

A² - B² = AB - B²

(A + B)(A - B) = B(A - B)

Divide both side by (A - B) gives

A + B = B

Since A = B

B + B = B

2B = B

2 = 1

edit: (You can use the commutative property of addition to switch 2 = 1 and 1 = 2 )

*Last edited by javacodeman (2007-04-12 14:41:18)*

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

I think it has been posted. But the error is in the part where you divide by A - B. A = B, so this is equivalent to division by zero, which we cannot do.

I'll let someone else post a erroneous proof next.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**adam_dsutton****Member**- Registered: 2007-09-12
- Posts: 10

what do you mean by "i"???...imaginary number? :S

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Unless you have a reason for e^pi = e^(ipi), I would say there is your error right there... no?

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

No, that bit's fine.

Why did the vector cross the road?

It wanted to be normal.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

Proof the square root of -1 is real:

sqrt(-1) = (-1)^(1/2), by the laws of fractions, 1/2 = 2/4, so (-1)^(1/2) = (-1)^(2/4), and by the laws of exponants,

= ((-1)^2)^(1/4), = 1^(1/4) = 1. But obviously 1*1 != -1. Where is the error?

A logarithm is just a misspelled algorithm.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I honestly don't know myself. Where did we break the rules? Its obvious that multiplying the exponent by 2/2 is what brought the trouble in, but if we had instead written it as ((-1)^(1/4))^2 we would have ended up with i, which is correct. So what rules did we break?

A logarithm is just a misspelled algorithm.

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

I heard that

ifbut i dunno why

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 22,548

mikau,

I guess the error crept in when deducing the result from the penultimate statement.

1^(1/4) = 1

For finding the fourth units of 1, lets assume a^4=1.

Therefore, a^4-1=0 or (a²-1)(a²+1) = 0

If a²-1 = 0, a²=1, a = +1 or -1.

If a²+1=0, a²=-1, a = +i or -i.

Therefore, there are four values of the fourth root of unity, viz. +1,

-1, +i, and -i.

In the instant case, both +i and -i satisfy the initial equation.

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

hmm.. i like it!

A logarithm is just a misspelled algorithm.

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