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Difference between revisions of "Fallacious proof/2equals1"

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=== Explanation ===
 
=== Explanation ===
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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 undefined.
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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 [[0|zero]] is undefined.
  
 
== Proof 2 ==
 
== 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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<cmath>1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots</cmath>
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<cmath>(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots</cmath>
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<math>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</math>
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<cmath>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</cmath>
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<cmath>2 = 1</cmath>
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<math>2 = 1</math>
 
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</center>
 
  
 
=== Explanation ===
 
=== Explanation ===
 
The given series does not converge.  Therefore, manipulations such as grouping terms before adding are invalid.
 
The given series does not converge.  Therefore, manipulations such as grouping terms before adding are invalid.
  
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''[[Fallacy#2_.3D_1 | Back to main article]]''
  
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''[[Fallacious_proof#2_.3D_1 | Back to main article]]''
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Revision as of 12:50, 15 February 2025

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 undefined.

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.

Back to main article


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