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Fallacious proof/2equals1: Difference between revisions

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I read this "proof" in a magazine a long time ago
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=== Explanation ===
=== Explanation ===
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 [[0|zero]] is undefined.


== Proof 2 ==
== Proof 2 ==
<center>
<math>1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots</math>


<math>(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots</math>
<cmath>1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots</cmath>
 
<cmath>(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots</cmath>
<math>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</math>
<cmath>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</cmath>
 
<cmath>2 = 1</cmath>
<math>2 = 1</math>
</center>


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


[[Fallacious proof | Back to main article]]
''[[Fallacy#2_.3D_1 | Back to main article]]''

Latest revision as of 13:04, 16 May 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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