Difference between revisions of "Fallacy"

Revision as of 11:45, 30 October 2007

A fallacious proof is a an attempted proof that is logically flawed in some way. The fact that a proof is fallacious says nothing about the validity of the original proposition.

Common false proofs

The fallacious proofs are stated first and then links to the explanations of their fallacies follow.

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 $a^2-ab$ )

Explanation

All horses are the same color

We shall prove that all horses are the same color by induction on the number of horses.

First we shall show our base case, that all horses in a group of 1 horse have the same color to be true. Of course, there's only 1 horse in the group so certainly our base case holds.

Now assume that all the horses in any group of $k$ horses are the same color. This is our inductive assumption.

Using our inductive assumption, we will now show that all horses in a group of $k+1$ horses have the same color. Number the horses 1 through $k+1$ . Horses 1 through $k$ must be the same color as must horses $2$ through $k+1$ . It follows that all of the horses are the same color.

Explanation

All Numbers are Equal

Proof: Choose arbitrary a and b, and let t = a + b. Then

$a + b = t$

$(a + b)(a - b) = t(a - b$ )

$a^2 - b^2 = ta - tb$

$a^2 - ta = b^2 - tb$

$a^2 - ta + \dfrac{t^2}{4} = b^2 - tb + \dfrac{t^2}{4}$

$(a - \dfrac{t}{2})^2 = (b - \dfrac{t}{2})^2$

$a - \dfrac{t}{2} = b - \dfrac{t}{2}$

$a = b$

QED

@@ Line 22: / Line 22: @@
 We shall prove that all horses are the same color by [[induction]] on the number of horses.
-First we shall show our base case, that all horses in a group of 1 horse have the same color, to be true.  Of course, there's only 1 horse in the group so certainly our base case holds.
+First we shall show our base case, that all horses in a group of 1 horse have the same color to be true.  Of course, there's only 1 horse in the group so certainly our base case holds.
 Now assume that all the horses in any group of <math>k</math> horses are the same color.  This is our inductive assumption.

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