There are many examples of false mathematical proofs that are often presented to fool people with inadequate mathematical skills. Classic examples include the 1=2 "proof" and the 2^.5 = 2 "proof," both of which clearly use the same technique of many other false proofs. These techniques generally boil down to one of four different types, each of which is described below with a clear example.

Trick 1: Division by zero or infinity

This is perhaps the most popular form of trickery, because it is the simplest and the one that produces the most seemingly flagrant errors. This trick is used in the basic 1=2 "proof", an example of which is as follows.

```Step 1: Let a=b.
Step 2: Then , a^2 = a*b
Step 3: a^2 + a^2 = a^2 + a*b
Step 4: 2*a^2 = a^2 + a*b
Step 5: 2*a^2 - 2*a*b = a^2 + a*b - 2*a*b
Step 6: 2*a^2 - 2*a*b = a^2 - a*b
Step 7: This can be written as 2*(a^2 - a*b) = 1(a^2 - a*b)
Step 8: and cancelling the (a^2 - a*b) from both sides gives 1=2.
```

The divide by zero trick occurs in step 8. If a=b then a^2 - a*b = a^2 - a*a = a^2 - a^2 = 0. Thus, dividing both sides by zero results in an undefined answer, which if disguised using algebra, can produce the bogus result seen above. The actual truth of the matter is that starting with step 5, the equation boils down to 0 = 0; in fact, every step up to the final one is actually correct.

Other variations of this proof (and similar ones) use techniques that result in a division by infinity, a similarly messy trick.

Trick 2: Incorrect use of a square or a square root

This form is much more subtle than the above one and is often harder to pick out. However, it often results in similarly false statements, such as 1=2. Here's an example of the "imaginary number" version of the 1=2 "proof."

```Step 1: -1/1 = 1/-1
Step 2: Taking the square root of both sides gives (-1/1)^1/2 = (1/-1)^1/2
Step 3: Simplifying this gives (-1)^1/2 * (1)^1/2 = (1)^1/2 * (-1)^1/2
Step 4: In other words, i/1 = 1/i (i being the square root of -1)
Step 5: Therefore, i / 2 = 1 / (2i)
Step 6: i/2 + 3/(2i) = 1/(2i) + 3/(2i)
Step 7: i*(i/2 + 3/(2i)) = i*(1/(2i) + 3/(2i))
Step 8: (i^2)/2 + (3i)/(2i) = i/(2i) + (3i)/(2i)
Step 9: (-1)/2 + 3/2 = 1/2 + 3/2
Step 10: and this shows that 1=2
```

The incorrect use of a square root comes in step 3. There is no rule that guarantees that (a/b)^1/2 = (a)^1/2 * (b)^1/2, except if a and b are both positive. By using an imaginary number, the calculator here is assuming that a = -1, meaning that the above rule does not hold. If you see an imaginary number involved in such a bogus "proof," then the trick is usually to hide the fact that such a statement without any mathematical basis occurs.

Trick 3: Incorrect use of mathematical induction

Before I get into this, a review of mathematical induction is in order. Suppose you have a set of natural numbers (natural numbers are the numbers 1, 2, 3, 4, . . . ). Suppose that 1 is in the set. Suppose also that, whenever n is in the set, n+1 is also in the set. Then every natural number is in the set.

This is the basic idea of induction. A proof by induction must prove these two principles:

1. The given statement is true when n = 1.
2. If the given statement is true for n, then the given statement is true for n + 1.

Combining the two means that for every natural number, the given statement must be true.

My favorite example of this type of proof is as follows. It is commonly known as the "all of the people in the world are the same age" "proof."

Statement S(n): In any group of n people, everyone in that group has the same age.

Step 1: In any group of one person, everyone in the group has the same age.

Step 2: Therefore, statement S(1) is true.

Step 3: Now we must prove that whenever S(n) is true for one number (say n=k), it is also true for the next number (that is, n=k+1).

Step 4: To prove this, we assume that in every group of k people, everyone has the same age, then deduce from it that in every group of k+1 people, everyone has the same age.

Step 5: Let G be an arbitrary group of k+1 people; to prove the statement, we just need to show that every member of G has the same age.

Step 6: To do this, we just need to show that, if P and Q are any members of G, then they have the same age. This must be true; if everyone has the same age, then choosing two random members of the group, the two chosen ones must have the same age as well.

Step 7: Consider everybody in G except P. These people form a group of k people, so they must all have the same age. Remember, we are assuming that in any group of k people, everyone has the same age (see Step 4).

Step 8: Consider everybody in G except Q. Again, they form a group of k people, so they must all have the same age.

Step 9: Let R be someone else in G other than P or Q.

Step 10: Since Q and R each belong to the group considered in Step 7, they are the same age.

Step 11: Since P and R each belong to the group considered in Step 8, they are the same age.

Step 12: Since Q and R are the same age, and P and R are the same age, it follows that P and Q are the same age.

Step 13: We have now seen that, if we consider any two people P and Q in G, they have the same age. It follows that everyone in G has the same age.

Step 14: The proof is now complete: we have shown that the statement is true for n=1, and we have shown that whenever it is true for n=k it is also true for n=k+1, so by induction it is true for all n.

A proof similar to this was used in an exam in a discrete mathematics course that I once took, where we were told to explain whether the induction proof was valid or not. Of course, it is not, but it uses a common trick used to exploit the idea in a subtle fashion.

The fallacy is in Step 9. With this assumption, we are assuming that G has a size of at least three (P, Q, and R, at least). We are not including the situation where G is of size 2; without that crucial part, we can't prove the second part of the induction and thus it is false.

The common proof for 2^.5 = 2 uses this technique in a clever fashion, as do similar arguments for proving that all males are female and so forth.

Trick 4: False use of the English language

This one is perhaps the sneakiest of all. A clever trickster will often turn the problem into a word problem to disguise the inherent logical flaw in their argument. Another clever example of how this proof can trick you comes from the following false proof, easily the most subtle one given here.

Every natural number can be unambiguously described in fourteen words or less.

The proof of this statement is as follows:

Step 1: Suppose there is some natural number which cannot be unambiguously described in fourteen words or less.

Step 2: Then there must be a smallest such number. Let's call it n.

Step 3: But now n is "the smallest natural number that cannot be unambiguously described in fourteen words or less".

Step 4: This is a complete and unambiguous description of n in fourteen words, contradicting the fact that n was supposed not to have such a description.

Step 5: Since the assumption (step 1) of the existence of a natural number that cannot be unambiguously described in fourteen words or less led to a contradiction, it must be an incorrect assumption.

Step 6: Therefore, all natural numbers can be unambiguously described in fourteen words or less.

It turns out that it is Step 4 of the proof that is at fault. It mistakes the self-inconsistent nature of S with a mathematical contradiction arising from the existence of n. In essence, it assumes a mathematical contradiction because of the apparent self-contradicting statement of Step 3.

It turns out, though, that although Step 3 is a true statement, it is not a complete and unambiguous description of n. The statement refers to itself in a logically nonsensical manner. If you try to apply the description S to a number, then it ends up stating that S does not apply to that number. Thus, there is in fact no contradiction and thus no basis for the proof. It trys to slip this idea by using verbiage, but a clever mind can catch it.

The results of all of this?

If someone claims something to be true that is obviously false and offers a "proof," more often than not the proof falls into one of these logical holes. Be warned, however; a truly clever prankster might also incorporate laws of physics into their fallacies in order to confuse you! A careful examination of the so-called "proof" will bring this trick and the ones presented here to light very quickly.

Two more methods of incorrect proof, somewhat less commonly seen than those previously addressed, but also in some ways more subtle:

Trick: Use of Ill-Defined Expressions/Misapplication of Definitions

We will prove that 1=2 using calculus and the definition of x2:

(*) First note that x2 is equal to x*x, that is, x added to itself x times:

x2 = x + x + ... + x (done x times)

Take the derivative of each side, on the left using the power rule, and on the right using the sum rule:

2*x = 1 + 1 + ... + 1 (done x times)

Now 1 added to itself x times is simply x:

2*x = x

Divide each side by x...

2 = 1

And QED! A falsehood is proved, mathematics as we know it is inconsistent, the stars fall into themselves, and the universe ends. Uh-oh.

The main flaw in this proof is that it relies on our alternate expression of x2 above at (*). This is a prefectly adequate definition for x2, but only if x is a positive integer! Our definition does not hold if x is any other real number; in particular using this definition to create the function y=x2 does not give us a continuous function. So its derivative is undefined, and the second step of our proof is not correct. Since we are trying to perform a continuous operation (differentiation) on something defined discretely, we are committing a major error.

Trick: Poorly Constructed or Misleading Diagram

It is difficult to node about geometry, since one's graphical abilities are limited in this medium, but I shall do my best to give a written explanation.

The canonical example of false proofs of this type are puzzles where one dissects a square into various pieces and reassembles them into a rectangle of an apparently different area than the original. See a similar puzzle at the disappearing area problem.

A neater example of this trick, given explicitly in proof form, involves "proving" that a particular triangle has more than 180 degrees in its three angles, a violation of the triangle sum rule. Although it relies on extensive use of diagrams that cannot be duplicated here, I will do my best to accurately describe the process by which this so-called proof does its work. Once again, the key here is to draw this diagram poorly:

Disproof of the Triangle Sum Rule:

Begin by drawing two circles which intersect each other at two points. Call the two points of intersection A and B. From A, draw a diameter of each circle.

Call the other endpoints of the diameters P and Q. Now our two diameters are AP and AQ. Connect P and Q to form triangle APQ. Now PQ will intersect each of the two circles, call these two points of intersection H (on the same circle as P) and K (on the same circle as Q).

Now draw in AH and AK. Consider angles AHP and AKQ. Since each of these angles subtends a diameter, each angle's measure is 90 degrees. But this means that triangle AHK has two 90 degree angles, an impossibility, since the third angle is yet unaccounted for and every triangle must have exactly 180 degrees by the Triangle Sum Rule. What has gone wrong?

The trick here relies on a sloppy diagram. If you construct your diagram carefully, you will see that PQ actually passes through point B. This means that H and K are actually the same point (point B), and that there isn't really a triangle AHK--it's just a line segment, AB.

This type of trick is particularly devious since when one is reading through a proof, one is often depending on the author's own diagrams--and the author can slightly skew the diagrams however he or she sees fit to mislead you.

A similar false proof may be found at Proof that all triangles are isosceles, once again relying on a misleading diagram (or in this case an assumption about the positioning of elements in a description of a diagram).

1=12=(-1)2=√1, yes? Are we all familiar with that? (What? You're not? Oh, go take a math class.)
Therefore:
√12=√(-1)2
=1  =-1 (since square roots always cancel out square powers)
Therefore 1=-1 and 2=0 (adding 1 to both sides).

So what's the problem? Well, the problem is: square roots do not cancel out square powers, although they certainly appear to. Ever heard of BODMAS or PEMDAS? Same thing, different acronym. Evaluate 12 or (-1)2 first, then find the root of the answer. It ends up coming out to 1 both times.

Also: ever heard of the modulus or piecewise function? It's defined as √x2 or |x|, and its graph looks like

```          y
\        |        /
\       |       /
\      |      /
\     |     /
\    |    /
\   |   /
\  |  /
\ | /
\|/
---------------------x
|
|
```

Note that y=x looks like:

```         y
|        /
|       /
|      /
|     /
|    /
|   /
|  /
| /
|/
---------------------x
/|
/ |
/  |
/   |
/    |
```

and nothing like our friend |x|.

God, I love proving stuff.

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