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Difference between revisions of "Fallacy"
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A '''fallacious proof''' is a an attempted [[proof]] that is [[logic]]ally flawed in some way.  The fact that a proof is fallacious says nothing about the validity of the original proposition.
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A '''fallacy''' is a step an attempted [[proof]] that is [[logic]]ally flawed in some way.  The fact that a proof is fallacious says nothing about the validity of the original proposition.
  
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== Common false proofs ==
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== False proofs ==
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The fallacious proofs are stated first and then links to the explanations of their fallacies follow.
 
  
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=== 2 = 1 ===
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Some common fallacious proofs follow
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Let <math>a=b</math>.
 
  
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Then we have
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=== <math>2 = 1</math> ===
  
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<math> a^2 = ab </math> (since <math>a=b</math>)
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==== Proof 1 ====
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Let <math>a=b</math>. Then we have
  
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<math> 2a^2 - 2ab = a^2 - ab </math> (adding <math>a^2-2ab</math> to both sides)
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<cmath>a^2=b^2</cmath>
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<cmath>a^2 = ab</cmath>
  
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<math> 2(a^2 - ab) = a^2 - ab </math> (factoring out a 2 on the [[LHS]])
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<math>2a^2 - 2ab = a^2 - ab</math> (adding <math>a^2-2ab</math> to both sides)
  
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<math> 2 = 1 </math> (dividing by <math>a^2-ab</math>)
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<math>2(a^2 - ab) = a^2 - ab</math> (factoring out a 2 on the [[LHS]])
  
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[[Fallacious proof/2equals1 | Explanation]]
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<math>2 = 1</math> (dividing by <math>a^2-ab</math>)
  
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=== All horses are the same color ===
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===== Explanation =====
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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 [[0|zero]] is undefined.
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==== Proof 2 ====
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<cmath>1 + 1 - 1 + 1 - 1 \ldots = 1 + 1 - 1 + 1 - 1 \ldots</cmath>
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<cmath>(1 + 1) + (-1 + 1) + (-1 + 1) \ldots = 1 + (1 - 1) + (1 - 1) \ldots</cmath>
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<cmath>2 + 0 + 0 \ldots = 1 + 0 + 0 \ldots</cmath>
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<cmath>2 = 1</cmath>
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===== Explanation =====
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The given series does not converge.  Therefore, manipulations such as grouping terms before adding are invalid.
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=== Polya's Proof That All horses Are the Same Color ===
 
We shall prove that all horses are the same color by [[induction]] on the number of horses.
 
We shall prove that all horses are the same color by [[induction]] on the number of horses.
  
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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.
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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.
 
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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Using our inductive assumption, we will now show that all horses in a group of <math>k+1</math> horses have the same color.  Number the horses 1 through <math>k+1</math>.  Horses 1 through <math>k</math> must be the same color as must horses <math>2</math> through <math>k+1</math>.  It follows that all of the horses are the same color.
 
Using our inductive assumption, we will now show that all horses in a group of <math>k+1</math> horses have the same color.  Number the horses 1 through <math>k+1</math>.  Horses 1 through <math>k</math> must be the same color as must horses <math>2</math> through <math>k+1</math>.  It follows that all of the horses are the same color.
  
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[[Fallacious proof/all horses are the same color | Explanation]]
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==== Explanation ====
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Our base case is not the appropriate base case: if one could show that every pair of horses has the same color (the result for <math>n = 2</math>), the fact that all horses have the same color would follow.  Unfortunately, the case <math>n = 2</math> does not follow from the case <math>n = 1</math>. The first horse is the same color as itself, and so is the second horse, but there is no overlap and so we cannot deduce that the two horses are the same color
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=== All numbers are equal ===
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Consider arbitrary reals <math>a</math> and <math>b</math>, and let <math>t = a + b</math>. Then
 +
<cmath>a + b = t</cmath>
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<cmath>(a + b)(a - b) = t(a - b)</cmath>
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<cmath>a^2 - b^2 = ta - tb</cmath>
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<cmath>a^2 - ta = b^2 - tb</cmath>
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<cmath>a^2 - ta + \dfrac{t^2}{4} = b^2 - tb + \dfrac{t^2}{4}</cmath>
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<cmath>\left(a - \dfrac{t}{2}\right)^2 = \left(b - \dfrac{t}{2}\right)^2</cmath>
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<cmath>a - \dfrac{t}{2} = b - \dfrac{t}{2}</cmath>
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<cmath>a = b</cmath>
 +
 
 +
==== Explanation ====
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 +
It does not follow in general from <math>x^2 = y^2</math> that <math>x = y</math>.  Instead, we have either that <math>x = y</math> or <math>x = -y</math>.  In the given situation, we have that <math>a - \frac{t}{2} = \frac{a - b}{2} = -\left(\frac{b - a}{2}\right) = - \left(b - \frac{t}{2}\right)</math>, and so this was the fallacious step.
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=== Bread ===
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Nothing is better than happiness.
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A few small crumbs of bread are better than nothing.
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Thus, a few small crumbs of bread are better than happiness.
  
 
== See also ==
 
== See also ==
 
* [[Proof writing]]
 
* [[Proof writing]]
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 +
{{stub}}

Latest revision as of 12:47, 15 February 2025

A fallacy is a step 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.

False proofs

Some common fallacious proofs follow

$2 = 1$

Proof 1

Let $a=b$. Then we have

\[a^2=b^2\] \[a^2 = ab\]

$2a^2 - 2ab = a^2 - ab$ (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.

Polya's Proof That 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

Our base case is not the appropriate base case: if one could show that every pair of horses has the same color (the result for $n = 2$), the fact that all horses have the same color would follow. Unfortunately, the case $n = 2$ does not follow from the case $n = 1$. The first horse is the same color as itself, and so is the second horse, but there is no overlap and so we cannot deduce that the two horses are the same color

All numbers are equal

Consider arbitrary reals $a$ and $b$, and let $t = a + b$. Then \[a + b = t\] \[(a + b)(a - b) = t(a - b)\] \[a^2 - b^2 = ta - tb\] \[a^2 - ta = b^2 - tb\] \[a^2 - ta + \dfrac{t^2}{4} = b^2 - tb + \dfrac{t^2}{4}\] \[\left(a - \dfrac{t}{2}\right)^2 = \left(b - \dfrac{t}{2}\right)^2\] \[a - \dfrac{t}{2} = b - \dfrac{t}{2}\] \[a = b\]

Explanation

It does not follow in general from $x^2 = y^2$ that $x = y$. Instead, we have either that $x = y$ or $x = -y$. In the given situation, we have that $a - \frac{t}{2} = \frac{a - b}{2} = -\left(\frac{b - a}{2}\right) = - \left(b - \frac{t}{2}\right)$, and so this was the fallacious step.

Bread

Nothing is better than happiness. A few small crumbs of bread are better than nothing. Thus, a few small crumbs of bread are better than happiness.

See also

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