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 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 [[ | + | 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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− | < | + | <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> | ||
+ | <cmath>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</cmath> | ||
+ | <cmath>2 = 1</cmath> | ||
− | + | === Explanation === | |
+ | The given series does not converge. Therefore, manipulations such as grouping terms before adding are invalid. | ||
− | + | ''[[Fallacy#2_.3D_1 | Back to main article]]'' | |
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− | + | {{delete|unnecessary page}} |
Latest revision as of 11:50, 15 February 2025
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 undefined.
Proof 2
Explanation
The given series does not converge. Therefore, manipulations such as grouping terms before adding are invalid.
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