Difference between revisions of "Fallacious proof/2equals1"

(I read this "proof" in a magazine a long time ago)
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=== Explanation ===
 
=== Explanation ===
The first step never definitively ends at a certain number (it switches back and forth between 1 and 2). Thus, we can't equate it with itself while extending it [[infinite]]ly.
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The given series does not converge. Therefore, manipulations such as grouping terms before adding are invalid.
  
[[Fallacious proof | Back to main article]]
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''[[Fallacious_proof#2_.3D_1 | Back to main article]]''

Revision as of 15:04, 21 July 2009

The following proofs are examples of fallacious proofs, namely that $2 = 1$.

Proof 1

Let $a=b$.

Then we have

$a^2 = ab$ (since $a=b$)

$2a^2 - 2ab = a^2 - ab$ (adding $a^2-2ab$ to both sides)

$2(a^2 - ab) = a^2 - ab$ (factoring out a 2 on the LHS)

$2 = 1$ (dividing by $a^2-ab$)

Explanation

The trick in this argument is when we divide by $a^{2}-ab$. Since $a=b$, $a^2-ab = 0$, and dividing by zero is illegal.

Proof 2

$1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots$

$(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots$

$2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots$

$2 = 1$

Explanation

The given series does not converge. Therefore, manipulations such as grouping terms before adding are invalid.


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