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1 + 1 = 3 can someone prove this? (1 Viewer)

Mark Giles

Second Unit
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Aug 30, 2002
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There is a way. This is not a riddle. Purely mathematical. Wont give any hints yet, but curious who can prove this first. Good luck!
 

Rob Gardiner

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If I remember correctly, the solution to this puzzle requires dividing by zero, which is a big no-no.
 

ThomasC

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I saw this on a professor's door. The logic is flawed, though. "Proof" that 1 = 2. Let a = 1 and b = 1.

a = b

a^2 = ab

a^2 - b^2 = ab - b^2

(a + b)(a - b) = b (a - b)

a + b = b

Unfortunately, (a - b) = 0, negating (a + b) and b.
 

Bill Williams

Screenwriter
Joined
May 28, 2003
Messages
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Here's one way of looking at it...

Husband - 1
Wife - 1
Husband and wife have baby
Husband, wife, and baby makes 3!

It takes two to make three!
 

Haggai

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Rob, you remembered right, as ThomasC's post explains in the spoiler.

Here's a fallacy that illustrates a different concept. Can you spot the error in this "proof" that 1 = 0? The ... in each line here means to repeat the pattern on into infinity.

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

Danny Tse

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I had a math professor who did this on the board one day and it was quite an eye-opener.

Now that I am a government analyst, I do see this type of "new math" used regularly.
 

RobertR

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You left out the first 1 in line 3, Haggai :)

Here's the "proof" that girls are evil:

girls = time * money

There's an old saying that time is money, so

girls = money * money

Another saying is that money is the root of all evil:

money = sqrt(evil)

substituting, we get

girls = sqrt(evil) * sqrt(evil), or

girls = evil
 

ThomasC

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And that knowledge is evil as well:

Knowledge is power
Power corrupts
Corruption is evil
Thus, knowledge is evil
 

Haggai

Senior HTF Member
Joined
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Messages
3,883


Nope, that wasn't the intention. The formatting here isn't ideal, but look at the difference between lines 2 and 3, the intent there is that I've just re-arranged the parentheses from line 2, all of them being shifted one spot to the left. I'll change it to make the # of 1's match up, now there are 7 of them in both line 2 and in line 3:

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

Of course, that step between lines 2 and 3 is where the fallacy/error is...any takers?
 

RobertR

Senior HTF Member
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But that was what I was getting at. In line 2, it's of the form 1 + n(1-1), where n can be infinite. In line 3, it's just n(1-1), which isn't the same thing.
 

Holadem

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Messages
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Yeah well, the question is how a seemingly correct operation using the associative property of addition produced such a result.

I can only say that the associative property doesn't apply to infinite sums/series?

--
H
 

Haggai

Senior HTF Member
Joined
Nov 3, 2003
Messages
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You're getting close...that property does apply to certain kinds of infinite sums, but not to others. Full explanation soon, unless someone tries to kill me first. :D
 

Haggai

Senior HTF Member
Joined
Nov 3, 2003
Messages
3,883
OK, I'll go ahead and fill in all the details.

Using the associative property (i.e., moving parentheses around without changing the value of anything) for infinite sums only works if the series, when written without any parentheses, converges to a finite number (and in a specific way, to boot, convergence alone isn't always enough, but I'll leave that alone for now). For instance, the infinite sum:

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + ...

...equals 1 when summed to infinity. You can re-arrange the terms any way you like, group them in any way with any pattern of parentheses, and the infinite sum will still converge to the value of 1. But let's see what the infinite sum from my previous posts looks like without any parentheses:

1 - 1 + 1 - 1 + 1 - 1 + 1 - ....

And so forth. Does this infinite sum have a finite value? No, it diverges. Certain sums are obviously divergent because they keep getting bigger in an unstoppable way, i.e. 1 + 2 + 3 + 4 + ..., or 1 + 1 + 1 + 1 + ..., for instance. The series in question here doesn't do that, as each successive 1 is cancelled out by the next -1, but it diverges because it never approaches one particular value in an inexorable way. If you stop after finitely many steps, you get a sum of either 1 or 0, depending on where you stop.

So you can't re-arrange any parentheses in this case because the infinite sum does not actually represent a finite value in and of itself.
 

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