Difference between revisions of "Fallacious proof/2equals1"

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(I read this "proof" in a magazine a long time ago)
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=== 2 = 1 ===
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The following proofs are examples of [[fallacious proof]]s, namely that <math>2 = 1</math>.
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== Proof 1 ==
 
Let <math>a=b</math>.
 
Let <math>a=b</math>.
  
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The trick in this argument is when we divide by <math>a^{2}-ab</math>. Since <math>a=b</math>, <math>a^2-ab = 0</math>, and dividing by [[zero (constant) | zero]] is illegal.
 
The trick in this argument is when we divide by <math>a^{2}-ab</math>. Since <math>a=b</math>, <math>a^2-ab = 0</math>, and dividing by [[zero (constant) | zero]] is illegal.
  
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== Proof 2 ==
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<center>
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<math>1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots</math>
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<math>(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots</math>
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<math>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</math>
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<math>2 = 1</math>
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</center>
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=== Explanation ===
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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.
  
 
[[Fallacious proof | Back to main article]]
 
[[Fallacious proof | Back to main article]]

Revision as of 22:15, 15 February 2007

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 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 infinitely.

Back to main article

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