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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− | === | + | The following proofs are examples of [[fallacious proof]]s, namely that <math>2 = 1</math>. |
+ | |||
+ | == 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. | ||
+ | == 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> | ||
+ | |||
+ | <math>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</math> | ||
+ | |||
+ | <math>2 = 1</math> | ||
+ | </center> | ||
+ | |||
+ | === 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. | ||
[[Fallacious proof | Back to main article]] | [[Fallacious proof | Back to main article]] |
Revision as of 21:15, 15 February 2007
The following proofs are examples of fallacious proofs, namely that .
Contents
[hide]Proof 1
Let .
Then we have
(since )
(adding to both sides)
(factoring out a 2 on the LHS)
(dividing by )
Explanation
The trick in this argument is when we divide by . Since , , and dividing by zero is illegal.
Proof 2
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.