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1 + 1 = 3


Derpy.H

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hehe. Didn't think it was possible, eh?

Here. Proof.

1 + 1 = 3

If we let x = y,

 

x-y+y = y

x-y+y ÷ x-y = y ÷ x-y

 

If we simplify this we get:

 

x-y ÷ x-y + y ÷ x-y = y ÷ x-y

 

OR

 

1+y ÷ x-y = y ÷ x-y

1 = y ÷ x-y - y ÷ x-y (This means 1 = 0)

 

Now knowing x = y or 1=1 is true and that 1=0 (or 0=1), we show:

0 = 1,

1 = 1

and

1 = 1.

 

If we add all these equations we get:

0+1+1 = 1+1+1

 

or, fully simplified

 

1+1=3

 

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Yo Applejack...

 

Y'alright?

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Well, strictly speaking you cannot divide by zero, which is what pretty much all of these "proofs" have done. However...

 

The limit x->0+ (from the right side of the graph towards zero) f(x) = 1/x does equal infinity. So you "kinda" can in Calculus. Though f(0) is still technically undefined. It's just that it depends on where the function is approaching. lim x->0- (from the left side of the graph) would be minus infinity.

 

Anyway...the laws of mathematics are safe once again!

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Anyone think of this when they read the title of this thread?

 

 

Posted Image

 

 

(also I almost failed math so I am the worst person to handle this)

 

I was just about to post this:

 

 

1984 was a great book btw.


THIS SIGNATURE IS POTATOES

 

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hehe. Didn't think it was possible, eh?

Here. Proof.

1 + 1 = 3

If we let x = y,

 

x-y+y = y

x-y+y ÷ x-y = y ÷ x-y

 

If we simplify this we get:

 

x-y ÷ x-y + y ÷ x-y = y ÷ x-y

 

OR

 

1+y ÷ x-y = y ÷ x-y

1 = y ÷ x-y - y ÷ x-y (This means 1 = 0)

 

Now knowing x = y or 1=1 is true and that 1=0 (or 0=1), we show:

0 = 1,

1 = 1

and

1 = 1.

 

If we add all these equations we get:

0+1+1 = 1+1+1

 

or, fully simplified

 

1+1=3

Is there somewhere I went wrong? It took a while to work all of this out, so some feedback on this would help.

 

hehe. Didn't think it was possible, eh?

Here. Proof.

1 + 1 = 3

If we let x = y,

 

x-y+y = y

x-y+y ÷ x-y = y ÷ x-y

 

If we simplify this we get:

 

x-y ÷ x-y + y ÷ x-y = y ÷ x-y

 

OR

 

1+y ÷ x-y = y ÷ x-y

1 = y ÷ x-y - y ÷ x-y (This means 1 = 0)

 

Now knowing x = y or 1=1 is true and that 1=0 (or 0=1), we show:

0 = 1,

1 = 1

and

1 = 1.

 

If we add all these equations we get:

0+1+1 = 1+1+1

 

or, fully simplified

 

1+1=3

Is there somewhere I went wrong? It took a while to work all of this out, so some feedback on this would help.

 

Insane. I understand it.
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Did you know that 0.999 repeating is the same as 1?

 

0.333 repeating + 0.333 repeating + 0.333 repeating = 0.999 repeating.

 

1/3 + 1/3 + 1/3 = 3/3 = 1.

 

1/3 = 0.333 repeating.

 

Mind blown!

 

Also, 1/9=0.111111 repeating

2/9=0.2222222 repeating

3/9=0.3333333 repeating

etc...

 

So logicially, 9/9 would equal 0.999999 repeating. But 9/9=1

 

Problem, math?


Signature now 99% less edgy!

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Also, 1/9=0.111111 repeating

2/9=0.2222222 repeating

3/9=0.3333333 repeating

etc...

 

So logicially, 9/9 would equal 0.999999 repeating. But 9/9=1

 

Problem, math?

 

it's actually well-established that 0.99999(repeating infinitely) is equal to 1.

 

So nope. Math has no problem here.

 

If I'm not mistaken, it's because of this fact that you cannot induct over the set of real numbers.

Edited by RagingTwilight
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Also, 1/9=0.111111 repeating

2/9=0.2222222 repeating

3/9=0.3333333 repeating

etc...

 

So logicially, 9/9 would equal 0.999999 repeating. But 9/9=1

 

Problem, math?

 

That's right. That's why 0.999 repeating is 1. 9/9 is the same thing as 3/3, you just stretched it out.


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Alea Jacta Est

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