2 + 2 = 5

[Kisama]

Ok, how about this:

Spoiler:

Given: 2 + 2 = 4

Now,
\begin{align}
-20 \ &= -20 \\
16 - 36 \ & = 25 - 45 \\
4^2 - 4 \times 9 \ & = 5^2 - 5 \times 9 \\
4^2 - 4 \times 9 + \frac{81}{4} \ & = 5^2 - 5 \times 9 + \frac{81}{4} \\
(4 - \frac{9}{2})^2 \ & = (5 - \frac{9}{2})^2 \\
4 - \frac{9}{2} \ & = 5 - \frac{9}{2} \\
4 \ & = 5
\end{align}

Since 2 + 2 = 4,

2 + 2 = 5

Done

[Actaeus]

Ok, how about this:

Spoiler:

Given: 2 + 2 = 4

Now,
\begin{align}
-20 \ &= -20 \\
16 - 36 \ & = 25 - 45 \\
4^2 - 4 \times 9 \ & = 5^2 - 5 \times 9 \\
4^2 - 4 \times 9 + \frac{81}{4} \ & = 5^2 - 5 \times 9 + \frac{81}{4} \\
(4 - \frac{9}{2})^2 \ & = (5 - \frac{9}{2})^2 \\
4 - \frac{9}{2} \ & = 5 - \frac{9}{2} \\
4 \ & = 5
\end{align}

Since 2 + 2 = 4,

2 + 2 = 5

Done

Found it!

Spoiler:

When you take the square root:
(1/2)^2=(-1/2)^2 (true)
1/2 = -1/2 (no)
It looks convincing, but you have to take the positive or negative square root of each side... not the +ve of one and -ve of the other.

[Kisama]

Ok, how about this:

Spoiler:

Given: 2 + 2 = 4

Now,
\begin{align}
-20 \ &= -20 \\
16 - 36 \ & = 25 - 45 \\
4^2 - 4 \times 9 \ & = 5^2 - 5 \times 9 \\
4^2 - 4 \times 9 + \frac{81}{4} \ & = 5^2 - 5 \times 9 + \frac{81}{4} \\
(4 - \frac{9}{2})^2 \ & = (5 - \frac{9}{2})^2 \\
4 - \frac{9}{2} \ & = 5 - \frac{9}{2} \\
4 \ & = 5
\end{align}

Since 2 + 2 = 4,

2 + 2 = 5

Done

Found it!

Spoiler:

When you take the square root:
(1/2)^2=(-1/2)^2 (true)
1/2 = -1/2 (no)
It looks convincing, but you have to take the positive or negative square root of each side... not the +ve of one and -ve of the other.

Aww was hoping to have at least one person befuddled But at least it's a "solution" that doesn't involve division by 0 this time.

[Axidos]

It makes the exact same division by zero error as every other proof like it.

Proof in case it's needed:

Spoiler:

\begin{eqnarray}
x &=& 1 \\
x-1 &=& 0 \\
x^2-1 &=& 0 \\
x^2-1 &=& x-1 \\
(x+1)(x-1) &=& x-1 \\
x+1 &=& 1 \\
x &=& 0 \\
1 &=& 0
\end{eqnarray}

To get from this point: (x+1)(x-1) = x-1
to the next line: x+1 = 1
- you have to divide by x - 1, which the second line stated was equal to 0.

Edit: Seems I didn't realise there were several more pages since this post...

[phlip]

These faulty proofs tend to fall in a handful of categories:
* Canceling a non-injective function (commonly: canceling multiplications without considering 0, canceling squares without adding a \pm)
* Taking an identity that only holds over the reals and trying to apply it to complex numbers (eg \sqrt x \sqrt y = \sqrt {xy})
* Introducing possible solutions, then treating those as if they're the only solutions (eg x^2 = -1 \to x^3 = -x \to x=1)
* Forgetting the constant of integration (eg \int f(x)dx = 1 + \int f(x)dx \to 0 = 1)

There are a few that don't, they're the more interesting ones, I find... like the occasional geometry one where the diagram doesn't exactly match the mathematical description, like the "all angles are equal" one...

[Invertin]

2 + 2 isn't 5.


It's 19.

Ruining the joke-

Spoiler:

I was going to do some complex sum instead of just "19", but then realized I'd have to work out the answer for the joke to work, and I've already reached the time of day when the brain decides to stop working.

[Carlington]

First time poster. Thought you'd like to know.

The way I see it, the mistake everyone else is making is that we are following convention. It is convention to use letters from the English alphabet as pronumerals, and not Hindu-Arabic numerals. However, if we free ourselves from such a convention, and use what we think of as definitive values, as pronumerals, we can solve this easily:

Let 2 = 2.5
2 + 2 = 2.5 + 2.5 = 5

QED.

P.S. How can I insert one of the "therefore" signs, the three dots in a triangle?

[MHD]

in quatum computing 2+2 is what you want (measure) it to be...

[eaglef2]

you could use a telescopic series to forge 1 out of 0.

Spoiler:

0 = 0 + 0+ 0...
0 = (1-1)+(1-1)+(1-1)+...
0 = 1 - 1 + 1 - 1...
0 = 1 + (-1+1) + (-1+1)...
0 = 1 + 0 + 0 + 0...
0 = 1

therefore 2 + 2 + 0 = 4 + 0
2 + 2 + 0 = 4 + 1
2 + 2 + 0 = 5
2 + 2 = 5

Of course this is all just illegal math that has been disproven.

[koogco]

Spoiler:

In Denmark this one is easy, a tax of 25% is applied to everything you buy, so obviously 2+2=5

[Messer11]

I personally Think there is one fundamental flaw to this whole deal. (it's on the spoiler)

Spoiler:

If you follow the false proof of division by zero. 2+2+0 = 2+2+1 =5 so 2+2+5 .. but this also assumes that because 0=1, 1=2 (2=2-0=2-1=1) 2=3 (via manipulations of 1 and 0) 3=4 ... and you get to Every Natural Number (with the addition of 0) to be equal to every other Natural Number, and as such The Natural Numbers is an identity relation onto it self. But that contradicts the Well ordering principle of the Natural Numbers ( that there exists an x in the Natural numbers that x is less than every other natural number).

[andy11235]

Spoiler:

... and you get to Every Natural Number (with the addition of 0) to be equal to every other Natural Number, and as such The Natural Numbers is an identity relation onto it self. But that contradicts the Well ordering principle of the Natural Numbers ( that there exists an x in the Natural numbers that x is less than every other natural number).

Spoiler:

No, you would find that the Natural Numbers become {0}, which is trivially Well Ordered.

[phlip]

Spoiler:

However, {0} doesn't satisfy the Peano axioms... specifically, S(0) is a natural number, but doesn't equal 0. So whether you could call that set the natural numbers is stretching the friendship a bit...

[dedalus]

there's a good proof for 0=1 by integrating by parts:

(let F(x,y) mean taking the integral from x to y)

F()1/x = lnx
F()1/x = x/x - F()x*-1/x^2 = 1+F()1/x = 1+lnx
lnx=1+lnx, therefore 1=0.
From whence, 2+2=5, not four, and freedom is inconsequential.

<ah the irony at my sig>

[Paper_Weasel]

Spoiler:

However, {0} doesn't satisfy the Peano axioms... specifically, S(0) is a natural number, but doesn't equal 0. So whether you could call that set the natural numbers is stretching the friendship a bit...

Ah, but S(0) = 1, and 1=0 by assumption!

[jestingrabbit]

But the axioms say that S(n) > n ie its strictly greater than, not equal to.

Seriously, expecting something sensible to come from an assumption that you can divide by 0 is folly.

[navigatr85]

is the reference to an episode of fairly odd parents? steven hawkings comes in and proves that 2+2=5.... but i doubt that any one has ANY idea what im talking about.

Haha....the fairly odd parents is a great show and I remember that episode. Timmy writes "2+2=5" on a math test, and the teacher makes fun of him for it. Then Timmy's rich friend Remy hires Stephen Hawking to come into the classroom and prove that 2+2 really does equal 5. The exact quote is:

Hawking: "There are certain scenarios under which the use of pre-transitive imaginary binomials might lead to the result described by Timmy Turner, 2 plus 2 equaling 5."

Of course it's complete nonsense just for comic effect. :) I suppose an "imaginary binomial" would be a binomial with imaginary coefficients, like 3ix³+4ix². But the word "pre-transitive" comes from graph theory and has nothing to do with polynomials.

In that scene, Hawking also quickly writes a "proof" of 2+2=5 on the chalkboard, which is also probably complete nonsense. I can't post a link, cuz I'm new here, but if you wanna see it, go to youtube and search for "remy rides again". (That's the title of the episode.) Once you're viewing the video of the full episode, jump ahead to about 3:15.

[eaglef2]

2 + 2 = 5

Math

It works bitches

That's all the proof I need

[dedalus]

no-one wants to comment on my proof.

[Kisama]

:( no-one wants to comment on my proof.

Haha, sorry, I promptly forgot integration by parts at the end of every semester of maths so I didn't actually read through your proof.

I do like your (very appropriate) sig though.

Oh, I see,

Spoiler:

you left out the constant of integration.

F()1/x = lnx + c

Spoiler:

In Denmark this one is easy, a tax of 25% is applied to everything you buy, so obviously 2+2=5

Wow that sucks.

Spoiler:

In South Africa it's 14% so 2+2= 4.56

This works:

Spoiler:

Code: Select all

    bool TwoPlusTwo = 2 + 2, Five = 5;
   
    printf("%s\n", (TwoPlusTwo == Five) ? "2 + 2 == 5" : "2 + 2 != 5");


[dedalus]

@Kisama nice job picking that up first try. Had a guy in a uni maths class try to fault my integration there, and refused to agree with me even after I told him where it was wrong.

[Preacha-Man]

let's make this up as I go along.

sin(0) = 1
sin(2π) = 1
0 = 2π
0 = 1 :: division by 2π

The sine of 0 is 0. And so is the sine of 2pi

[Random832]

let's make this up as I go along.

sin(0) = 1
sin(2π) = 1
sin(0)=sin(2π) ...incidentally you left this out
0 = 2π
0 = 1 :: division by 2π

Yeah, but you can't just eliminate a sine function from both sides, since it's not an injective function. (this is the same thing people do with the square function in some other proofs, incidentally)

[jestingrabbit]

If you guys are somehow surprised at the fact that all these "proofs" have issues then I don't think that there's any help that anyone here can offer you.

Yes, all these proofs are bull. This is as it should be. The moment we arrive at a proof that isn't bull someone is going to have to work out which peano/ZF axiom we need to do away with. I don't want to be at that meeting because I love them all equally. Except for Foundation, its crap.

[Random832]

If you guys are somehow surprised at the fact that all these "proofs" have issues then I don't think that there's any help that anyone here can offer you.

The fun is in finding it though.

[afarnen]

Since the current axioms cannot prove 2 + 2 = 5, then we need a new one, but one that doesn't contradict its entire system. So, we really need an entire, new arithmetic system: one with no contradictions, and one that can prove 2 + 2 = 5.

Of course the problem is now trivial. Here's one very simple proposed system:

1. There exists numbers 2 and 5
2. Two numbers connected by "+" are added to form a new number
3. Every number is equal to itself
4. 2 added to 2 is equal to 5

These axioms prove
2 = 2
5 = 5
2 + 2 = 5
2 + 2 + 2 = 5 + 2
2 + 2 + 2 + 2 = 5 + 5
and so on...

Assuming that we cannot rewrite arithmetic, I guess the best one can do is to trick some people for a minute until the error in their proof is obvious.

More fun than proving 2 + 2 = 5 is proving anything, with something false as a premise, as falsity implies anything.

[ChibiPez]

Spoiler:

X is any non zero number and y is any non zero number not equal to x
x/x = 1
y/y = 1
x = 1x
y = 1y
1y = 1x
y = x

therefor 2 is equal to any and all numbers it wishes to be to make 2+2 = 5

hah no zeros involved!

then again i'm totally relying on double stating an obvious..or else it should look like...

Spoiler:

x/x = 1
y/y = 1
x/x = y/y
really means xy = xy
1=1

edit: I had completely forgot about the rules of exponents! and as i was ironing my uniform for today this came to mind..

Spoiler:

2 +2 = 5 -> 2^1 + 2^1 = 2^2 + 2^0
1 + 1 does indeed equal 2 + 0. there for 2 + 2 = 5

[Alkalannar]

But surely 0 = 1 is a special case! Here's the general case for n = 0:

Spoiler:

e^2*pi*i = 1
ln |e^2 * pi* i| = ln | 1|
2*pi*i ln|e| = ln|1|
2*pi*i*1 = 0
i = 0
1 = 0
m = 0

Of course there is a flaw in this, but it's a more interesting one than usual, yes?

[Carlington]

Does no-one like my proof. I thought it was good use of lateral thinking. As my maths teacher says: "In Mathematics, when we don't like the question, we change it." Of course, he's referring to inverting the second fraction when dividing algebraic fractions, e.g.

2a+2/3 divided by 4a+5/5
=2a+2/3 multiplied by 5/4a+5
etc.

[Koyaanisqatsi]

Let me show you a neat little trick,
To make two and two make five,
Llamapalooza may be a video game in Jimmy Neutron,
I will kill you.
I can do this;
for I need not follow the Chinese scheme of prose if I don't want to.

It's easy. All I have to do is redefine the meaning of the Arabic numerals.

The following is 4 I's: IIIII
The following is 5 I's: IIII
The following is 2 I's: II

By the above definitions, 2 + 2 = 5.

[Qaanol]

More fun than proving 2 + 2 = 5 is proving anything, with something false as a premise, as falsity implies anything.

Indeed, there is a famous proof attributed to G. H. Hardy that, taking “2 + 2 = 5” as a premise, shows McTaggart is the pope.

[MrJohnnyBEE]

What if I wrap the 5 around a sphere, then view it from a very long ways off? If I did that, then while I would think that 2+2=4, I would in fact be wrong, because my perspective doesn't allow me to see the extra bits that are wrapped around...

[The Chosen One]

If I have 2 knots each in 2 distinct ropes, and I tie the ropes together, I have 5 knots!

[TheGrammarBolshevik]

More fun than proving 2 + 2 = 5 is proving anything, with something false as a premise, as falsity implies anything.

A contradiction implies anything. Falsity, not so much.

[KrO2]

Well, if you know that the statement is false then you can use its negation as another premise. Then you have your contradiction. So it's still correct to say that given the truth of a false statement anything else can be proven.

[taggedjc]

I'm not so sure about that.

Let's say that 2 + 2 = 5. Now, with that as a premise, can you prove that unicorns exist?

[Adacore]

Only if you do something silly like say: the number of unicorns standing in front of me is n=0, use 2+2=5 to prove that 0=1, therefore n=1, so there is a unicorn standing in front of me and unicorns must therefore exist.

[taggedjc]

Pretty good case! Nice way to think about it, though you are obviously correct that it is silly (as silly as 2+2=5 coming out as being true somehow!). Although this actually comes as a result of the typical proof by contradiction: The number of unicorns standing in front of you is 1, except that one of your previous assumptions was that the number of unicorns standing in front of you was 0. Thus, either that premise is wrong (ie there is actually a unicorn standing in front of you and therefore unicorns exist) or another premise is wrong, which we know is actually the assumption that 0=1. Either way, if you stop immediately at the "therefore", you can see that there's a little " ... or my other premise is not true, or both" at the end of the sentence.. Much like many of our current proofs! Haha.

While it's true that assuming something like 2+2=5 can arrive at such a proof for the existence of unicorns, at that point it's easy (for us) to look at the assumed premises and see which ones may actually be the contradictory ones and which ones are obviously observably true premises. I wonder if it's possible to construct something that isn't a proof by contradiction using something like 2+2=5, though.

[Qaanol]

I wonder if it's possible to construct something that isn't a proof by contradiction using something like 2+2=5, though.

Adacore already gave one. Here is another.

Suppose that 2+2=5. We also know that 2+2=4. By the reflexive property of equality we see that 2+2=2+2. Substituting on each side reveals 4=5. Subtracting 4 from both sides yields 0=1. We know from observation that there exists a horse with 0 horns. And since 0=1, we conclude that there exists a horse with 1 horn.

[TheGrammarBolshevik]

Well, if you know that the statement is false then you can use its negation as another premise. Then you have your contradiction. So it's still correct to say that given the truth of a false statement anything else can be proven.

In the same way that everything true can be proved given nothing at all, sure. But that's not what "implies" means in the sense that people talk about the principle of explosion.