Difference between revisions of "Fallacious proof/2equals1"

Latest revision as of 07:29, 5 June 2013

The following proofs are examples of fallacious proofs, namely that $2 = 1$ .

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 $a^2-ab$ )

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.

$1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots$

$(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots$

$2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots$

$2 = 1$

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

@@ Line 15: / Line 15: @@
 === 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 [[zero (constant) | zero]] is undefined.
 == Proof 2 ==
@@ Line 29: / Line 29: @@
 === 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]]
+''[[Fallacious_proof#2_.3D_1 | Back to main article]]''