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2 + 2 = 5? I will prove it.

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posted on Oct, 10 2009 @ 01:30 AM
Three boys decide to chip in and buy a bicycle.

They each put in $10, for a total of$30.

They goto the store and give the shopkeep $30 and take the bicycle home. The shopkeep decides he overcharged them. He gave$5 to the errand boy to take back to the boys.

The errand boy keeps $2 for himself, and gives the boys$3 back.

Each boy spent $10 but got$1 back, totalling for $9 spent each.$9 times 3 is $27, plus the$2 the errand boy kept adds up to $29. What happened to the last$1?

( Sorry, couldn't resist
)

EDIT:

Going back ontopic, I know a few programming terms such as:

2++ would equal 3.
2+=1 would 3.

In the first line, the first + stands for increment, as ++ side by side stands for increment by one. That being said, + does not equal to +1. If anything, I would say ++ would be equal to +1.

So if the statement would be 2 ++ 2, i would break it down to:

(2) (++) (2)
(2) (0+1) (2)
(2) (1) (2)
(2) (2)
(4)

OR:

(2++) (2)
(3) (2)
(6)

Hmm, I'm making up math I think. Time to goto sleep....

[edit on 10-10-2009 by Sevrens83]

posted on Oct, 10 2009 @ 09:47 AM
I tell you what, lend me 5 dollars.
I will pay you back 2 dollars tomorrow, and 2 dollars the day after that.
And by your reasoning, we will be quits !
Top tip: Don't do anything with real money for a living, cos you'll soon be broke.

posted on Oct, 10 2009 @ 03:14 PM
I'm surprised not one person thought of it this way.
If + sign equals 1, then would not the equation be 212=5?
Since you implied that it is a number then it would be added as a number which would make it 212, right?
You have removed the + sign and made it a whole number. You can not add the plus sign back into the equation again or it would be an infinite number. 2 (1) 2 = 4 because 2x1=2 therefore 2x2=4. All you have done is implied that the plus sign equals 1. You can imply that it is any number if you wanted. Not rocket science here.

posted on Oct, 10 2009 @ 07:00 PM
lol, I love threads like this.

2 + 2 can also equal 12.

If you write the digits in straight lines and count them.

I suppose any calculation has unlimited possibilities doesn't it?

Nice to read some of the posts here and to see some folks thinking with a 2 sided box.

posted on Oct, 10 2009 @ 07:33 PM

Originally posted by zazzafrazz
Stop using 'applied' mathematics, throw in the observer, and perhaps 2+2=5

Just saying, the observer makes the all the difference within infinity......

Forgot to add, technically everything equals infinity

posted on Oct, 11 2009 @ 11:26 AM
2 + 2 can equal 5, but it is for greater values of 2.

ok so we know that 2 = 2 but did you also know that 1.6 threw 2.5 also = 2 if we're not looking at the decimal, 1.6 = 2 because it is past the half way point; so if we was to round it off we would consider this a 2 and 2.5 is on the half way point, but not past it therefor 2 + 2 = 5 (2.5 + 2.5 = 5) but 2 + 2 can also equal 3.2 (1.6 + 1.6 = 3.2) so 2 + 2 can range from 3.2 to 5, if it was 2.4 + 2.6 then that would be 2 + 3 which would still equal 5. I can't remember the name of the knot but if you tied it in two knots then tie it again in two knots you will notice you have 5 knots even tho we only tied 4 knots. I will see if I can find out the name of this knot.

or we can say:

Which is the same as: 16-36 = 25-45
Which can also be expressed as: (2+2) 2 (9 X (2+2) = 52) 9 X 5
Add 81/4 to both sides: (2+2) 2 (9 X (2+2) + 81/4 = 52) 9 X 5 + 81/4
Rearrange the terms: ([2+2]) 9/2) 2 = (5-9/2) 2
Ergo: 2+2 - 9/2 = 5
Hence: 2 + 2 = 5

or we can also go into proof that x = y for any real x, y

Let x and y be any two numbers Then let a = x - y \ , \ b = (x + y)^2 \ , \ c = 4 (x^2 - x y + y^2)\, Let u = a \ , \ v = b - c\, Let's compute:

(u + \sqrt [v])^3 = u^3 + 3 u^2 \sqrt[v] + 3 u v + v \sqrt[v]
(u - \sqrt [v])^3 = u^3 - 3 u^2 \sqrt[v] + 3 u v - v \sqrt[v]

Replacing u = a \ , \ v = b - c, we get:

(a + \sqrt [b - c])^3 = a \ (a^2 + 3 b - 3 c) + (3 a^2 + b - c)\sqrt[b - c]
(a - \sqrt [b - c])^3 = a \ (a^2 + 3 b - 3 c) - (3 a^2 + b - c)\sqrt[b - c]

Let's compute 3 a^2 + b - c\, Replacing a = x - y \ , \ b = (x + y)^2 \ , \ c = 4 (x^2 - x y + y^2)\,:

3 a^2 + b - c = 3 (x^2 - 2 x y + y^2) + x^2 + 2 x y + y^2 - 4 (x^2 - x y + y^2) = 0\,

So:

(a + \sqrt[b - c])^3 = (a - \sqrt [b - c])^3\,
a + \sqrt[b - c] = a - \sqrt [b - c]\,
\sqrt [b - c] = 0\,
b - c = 0\,
b = c\,

Replacing b = (x + y)^2 \ , \ c = 4 (x^2 - x y + y^2)\,:

(x + y)^2 = 4 (x^2 - x y + y^2)\,
x^2 + 2 x y + y^2 = 4 (x^2 - x y + y^2)\,
-3 x^2 + 6 x y - 3 y^2 = 0\,
(x - y)^2 = 0\,
x - y = 0\,
x = y\,

The mistake here is that from z³ = w³ one may not in general deduce z = w (unless z and w are both real, which they are not in our case).

with that said I will leave on the note that ∞ = 1/4

Since an infinitely large plane has the coordinates of (-∞,∞) × (-∞,∞), this means that

∞ = [∞ - (-∞)]^2

Which can be simplified into

∞ = (2∞)^2

And finally

∞ = 4∞^2

Now combine the ∞'s:

1 = 4\frac[\∞^2][\∞]

This itself then simplifies into

1 = 4 \∞ \,

And finally, to find the value of ∞ itself,

\frac[1][4] = \∞

This can be checked by starting with the equation given in step 1,

∞ = [∞ - (-∞)]^2

Substitute in the above value of ∞ to see if it really works:

\frac[1][4] = \left[\frac[1][4] - \left(-\frac[1][4]\right)\right]^2

Which is then simplified to get

\frac[1][4] = \left[\frac[1][2]\right]^2

And that then simplifies into

\frac[1][4] = \frac[1][4]

One of the more unusual aspects of this type of invalid proof is that it can be checked, unlike many other invalid proofs, particularly ones which rely on division by zero. What is going on here is that in the extended real number system (real numbers with plus and minus infinity) the function of division, although extended, does not extend to the input (∞,∞). Using the Cauchy sequences characterization of real numbers, and the extended real number line, this is because the quotient of two sequences both tending to infinity could tend to any value from 0 to infinity, not just 1. But this is precisely what is being assumed when assuming that ∞/∞ is defined, and that ∞/∞=1. Similar problems arise if one chooses to work with another approach than the Cauchy sequences approach, making the desired proof that 1/4=∞ impossible to obtain.

[edit on 11-10-2009 by XeoniX]

posted on Oct, 11 2009 @ 01:12 PM

MISDIRECTION!

Here's the solution to the missing dollar problem.

We have 3 entities: The clerk, the bellhop, and the guests.

The initial payment of $30 is accounted for as the clerk takes$25, the bellhop takes $2, and the guests get a$3 refund. It adds up.

After the refund has been applied, we only have to account for a payment of $27. Again, the clerk keeps$25 and the bellhop gets $2. This also adds up. If someone says, "Okay, but it's still true that the bellhop's$2 plus the guest's $27 equals$29. How does that add up?" The answer is that the bellhop's $2 was part of the$27 payment, leaving the $25 kept by the clerk. There is no reason to add the$2 and $27, and no reason to expect a particular result. This becomes clearer when we write the initial and net payments as simple equations. The first equation shows what happened to the initial payment of$30:

$30 (initial payment) =$25 (to clerk) + $2 (to bellhop) +$3 (refund)

The second equation shows the net payment after the refund is applied (subtracted from both sides):

$27 (net payment) =$25 (to clerk) + $2 (to bellhop) Both equations make sense, with equal totals on either side of the equal sign. The correct way to get the bellhop's$2 and the guests $27 on the same side of the equal sign ("The bellhop has$2, and the guests paid $27, how does that add up?") is to subtract, not add, and when you do --$27 (final payment) - $2 (to bellhop) =$25 (to clerk)

-- that makes sense too.

The Missing Dollar Problem

posted on Oct, 11 2009 @ 01:59 PM

Anyway, I think the OP needs to provide more proof for his/her claims.

posted on Oct, 11 2009 @ 02:06 PM

What proof???

2 AND 2 combined is still 4 in ANY universe.

posted on Oct, 11 2009 @ 03:04 PM
reply to post by Deaf Alien

Well if you have to say what proof then I take it you couldn't understand what I said. Let's put it this way: when you have $2.50 and$2.50 do you have $4 or do you have$5. Like I said it's for greater values of 2; since 2.5 is a value of 2or is 2.5 not a number now? If so you can send me the amount greater then \$4 since it don't exist to you. If your on this site; shouldn't you know how to look outside the box? I take it that time goes in a straight line in ANY universe too. sbuıɥʇ ʇɐ ʞoo1 noʎ ʎɐʍ ǝɥʇ ǝbuɐɥɔ

[edit on 11-10-2009 by XeoniX]

posted on Oct, 11 2009 @ 03:08 PM

LOL the reply was for Unlimitedpossibilities, not for you. Take a look at my post again.

posted on Oct, 11 2009 @ 03:13 PM

Oh, you are talking about rounding off the real numbers.

2.0 to 2.49999... rounded is 2.

so let's say you have 2.25 and 2.40 rounded.

It's 2 + 2 = 4.

If you have 2.65 and 2.85 then it's

3 + 3 = 6.

You are just simply talking about what is CLOSEST to an integer.

posted on Oct, 11 2009 @ 03:23 PM

sbuıɥʇ ʇɐ ʞoo1 noʎ ʎɐʍ ǝɥʇ ǝbuɐɥɔ

LOL I like that one. Yes, since I am on ATS, I do think outside the box.

HOWEVER.

This is a counting problem.

The symbol 2 is taken to mean that you count TWO things.

I.e. I count the asterisks (* *) to be TWO.

The plus symbol (+) is taken to mean "to put together" or something to that effect.

So therefore, 2 + 2 MEANS

* * AND * *

then

* * * *

I count that to be FOUR.

This is true in ANY universe.

If you want to discuss real numbers, then I'm game.

posted on Oct, 11 2009 @ 03:29 PM
reply to post by Deaf Alien

In a lot of ways yes. In reality there is always exception to everything.
Most the time with equations like this they call them "invalid proof".
Like us saying the sky is "blue", that too is not truly true, and is not an absolute answer (as it has a lot of factors which can change the color of the sky) , we too can say the same about 2 + 2 = 4 as it too has a lot of factors which can change it's value enough within it's own range to conclude a different end result.

posted on Oct, 11 2009 @ 03:35 PM

But how can the OP say that when you have this

* * * *

You count it and it comes to * * * * *?

It's counting.

If the OP want to redefine the plus symbol (+), then fine. It still won't change the counting system.

posted on Oct, 11 2009 @ 03:39 PM

Originally posted by oozyism
People always try to see if an impossible logic is possible, here I will demonstrate that 2 + 2 can actually = 5.

How is it possible you may ask? Well all you do is change the definition of '+'. If + was defined as +1, we have succeeded.

This way all numbers can live in harmony. Every number can equal to another number without destroying our logic.

If that makes sense then good, if it doesn't I apologize because I don't know how else to explain it.

oozy

Are you serious or is this a joke thread?

posted on Oct, 11 2009 @ 03:43 PM

IT IS COUNTING FOR GOODNESS SAKE!!!

1 2 3 4 5 6 7 8.........

PLUS SIGN (+) MEANS "AND"

* * * + * *

Count it!

* * * * *

That is FIVE!

Maybe the OP came from a demented universe? Maybe he's living in an Orwellan universe where 2 + 2 does equal to 5?

I am not understanding this.

posted on Oct, 11 2009 @ 03:47 PM
This dispute has been going on for over 800 years I doubt we're gonna put an end to it here

"First and above all he was a logician. At least thirty-five years of the half-century or so of his existence had been devoted exclusively to proving that two and two always equal four, except in unusual cases, where they equal three or five, as the case may be."

- Jacques Futrelle, "The Problem of Cell 13"

Most mathematicians are familiar with -- or have at least seen references in the literature to -- the equation 2 + 2 = 4. However, the less well known equation 2 + 2 = 5 also has a rich, complex history behind it. Like any other complex quantity, this history has a real part and an imaginary part; we shall deal exclusively with the latter here.

Many cultures, in their early mathematical development, discovered the equation 2 + 2 = 5. For example, consider the Bolb tribe, descended from the Incas of South America. The Bolbs counted by tying knots in ropes. They quickly realized that when a 2-knot rope is put together with another 2-knot rope, a 5-knot rope results.

Recent findings indicate that the Pythagorean Brotherhood discovered a proof that 2 + 2 = 5, but the proof never got written up. Contrary to what one might expect, the proof's nonappearance was not caused by a cover-up such as the Pythagoras attempted with the irrationality of the square root of two. Rather, they simply could not pay for the necessary scribe service. They had lost their grant money due to the protests of an oxen-rights activist who objected to the Brotherhood's method of celebrating the discovery of theorems. Thus it was that only the equation 2 + 2 = 4 was used in Euclid's "Elements," and nothing more was heard of 2 + 2 = 5 for several centuries.

Around A.D. 1200 Leonardo of Pisa (Fibonacci) discovered that a few weeks after putting 2 male rabbits plus 2 female rabbits in the same cage, he ended up with considerably more than 4 rabbits. Fearing that too strong a challenge to the value 4 given in Euclid would meet with opposition, Leonardo conservatively stated, "2 + 2 is more like 5 than 4." Even this cautious rendition of his data was roundly condemned and earned Leonardo the nickname "Blockhead." By the way, his practice of underestimating the number of rabbits persisted; his celebrated model of rabbit populations had each birth consisting of only two babies, a gross underestimate if ever there was one.

www.kfunigraz.ac.at...

[edit on 11-10-2009 by XeoniX]

posted on Oct, 11 2009 @ 03:47 PM

How is it possible you may ask? Well all you do is change the definition of '+'. If + was defined as +1, we have succeeded.

Ok let's see hmm

2 + 2 =
2 +1 2 =
2 +11 2 =
2 +111 2 =
2 +1111 2 =

and so on

logic? pffft

posted on Oct, 11 2009 @ 03:55 PM

Even this cautious rendition of his data was roundly condemned and earned Leonardo the nickname "Blockhead."

I am not surprised.

I am not understanding the OP's point.

[edit on 11-10-2009 by Deaf Alien]

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