Mathematical fallacy: Information from Answers.com

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In mathematics, certain kinds of mistaken proof are often exhibited, and sometimes collected, as illustrations of a concept of mathematical fallacy. There is a distinction between a simple mistake and a mathematical fallacy in a proof: a mistake in a proof leads to an invalid proof just in the same way, but in the best-known examples of mathematical fallacies, there is some concealment in the presentation of the proof. For example the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented, it leads not only to an absurd result, but does so in a crafty or clever way.^[1] Therefore these fallacies, for pedagogic reasons, usually take the form of spurious proofs of obvious contradictions. Although the proofs are flawed, the errors, usually by design, are comparatively subtle, or designed to show that certain steps are conditional, and should not be applied in the cases that are the exceptions to the rules.

The traditional way of presenting a mathematical fallacy is to give an invalid step of deduction mixed in with valid steps, so that the meaning of fallacy is here slightly different from the logical fallacy. The latter applies normally to a form of argument that is not a genuine rule of logic, where the problematic mathematical step is typically a correct rule applied with a tacit wrong assumption. Beyond pedagogy, the resolution of a fallacy can lead to deeper insights into a subject (such as the introduction of Pasch's axiom of Euclidean geometry).^[2] Pseudaria, an ancient lost book of false proofs, is attributed to Euclid.^[3]

Mathematical fallacies exist in many branches of mathematics. In elementary algebra, typical examples may involve a step where division by zero is performed, where a root is incorrectly extracted or, more generally, where different values of a multiple valued function are equated. Well-known fallacies also exist in elementary Euclidean geometry and calculus.

Simple examples

The following example^[4] "proves" the absurd conclusion 1 = 0,

$\begin{align} x=0&\\ \therefore\quad x(x-1)=0&\\ \therefore \quad x-1=0&\\ \therefore \quad x=1&\\ \therefore \quad 1=0&\\ \end{align}$

The error here is in going from first to the second line. The assumption that 0 is the answer is incorrect; therefore, the second to the third line: if x = 0, then division by x make no sense, and should not be permitted. This division by zero fallacy has many variants.

A correct result obtained by an incorrect line of reasoning has a different status. This is the case, for instance, in the calculation

$\frac{16}{64} = \frac{16\!\!\!/}{6\!\!\!/4}=\frac{1}{4}.$

Although the conclusion 16/64 = 1/4 is correct, there is a fallacious invalid cancellation in the middle step. Bogus proofs constructed to produce a correct result in spite of incorrect logic are known as howlers.^[5]

Power and root

Fallacies involving disregarding the rules of elementary arithmetic through an incorrect manipulation of the radical are often of the following kind:^[6]

$1 = \sqrt{1} = \sqrt{(-1)(-1)} = \sqrt{-1}\sqrt{-1} = -1.$

The fallacy is that the rule $\sqrt{xy} = \sqrt{x}\sqrt{y}$ is generally valid only if at least one of the two numbers x or y is positive, which is not the case here.

Although the fallacy is easily detected here, sometimes it is concealed more effectively in notation. For instance,^[7] consider the equation

cos 2 x = 1 - sin 2 x

which holds as a consequence of the Pythagorean theorem. Then, by taking a square root,

cos x = (1 - sin 2 x) 1 / 2

so that

1 + cos x = 1 + (1 - sin 2 x) 1 / 2 .

Squaring both sides gives

$(1+\cos x)^2 = \left(1+(1-\sin^2x)^{1/2}\right)^2.$

But evaluating this when x = π implies

$(1-1)^2 = \left(1+(1-0)^{1/2}\right)^2$

0 = 4

which is absurd.

The error in each of these examples fundamentally lies in the fact that any equation of the form

x 2 = a 2

has two solutions, provided a ≠ 0,

$x=\pm a$

and it is essential to check which of these solutions is relevant to the problem at hand.^[8] In the above fallacy, the square root that allowed the second equation to be deduced from the first is valid only when cos x is positive. In particular, when x is set to π, the second equation is rendered invalid.

Another example of this kind of fallacy, where the error is immediately detectable, is the following invalid proof that −2 = 2. Letting x = −2, and then squaring gives

$x^2 = 4\,$

whereupon taking a square root implies

$x = \sqrt{4} = 2,$

so that x = −2 = 2, which is absurd. Clearly when the square root was extracted, it was the negative root −2, rather than the positive root, that was relevant for the particular solution in the problem.

Alternatively, imaginary roots are obfuscated in the following:

$\sqrt{-1} = (-1)^\frac{2}{4} = ((-1)^2)^\frac{1}{4} = 1^\frac{1}{4} = 1$

The error here lies in the equality, where we are ignoring the other fourth roots of 1,^[9] which are −1, i and −i (where i is the imaginary unit). Seeing as we have squared our figure and then taken roots, we cannot always assume that all the roots will be correct. So the correct fourth are i and −i, which are the imaginary numbers defined to be $\sqrt{-1}$ .

Division by zero

2 = 1

1. Let a and b be equal non-zero quantities

$a = b \,$

2. Multiply through by a

$a^2 = ab \,$

3. Subtract $b^2 \,$

$a^2 - b^2 = ab - b^2 \,$

4. Factor both sides

$(a - b)(a + b) = b(a - b) \,$

5. Divide out $(a - b) \,$

$a + b = b \,$

6. Observing that $a = b \,$

$b + b = b \,$

7. Combine like terms on the left

$2b = b \,$

8. Divide by the non-zero b

$2 = 1 \,$

Q.E.D.^[10]

The fallacy is in line 5: the progression from line 4 to line 5 involves division by a − b, which is zero since a equals b. Since division by zero is undefined, the argument is invalid. Deriving that the only possible solution for lines 5, 6, and 7, namely that a = b = 0, this flaw is evident again in line 7, where one must divide by b (0) in order to produce the fallacy (not to mention that the only possible solution denies the original premise that a and b are nonzero). A similar invalid proof would be to say that since 2 × 0 = 1 × 0 (which is true), one can divide by zero to obtain 2 = 1. An obvious modification "proves" that any two real numbers are equal.

Many variants of this fallacy exist. For instance, it is possible to attempt to "repair" the proof by supposing that a and b have a definite nonzero value to begin with, for instance, at the outset one can suppose that a and b are both equal to one:

$a = b = 1.\,$

However, as already noted the step in line 5, when the equation is divided by a - b, is still division by zero. As division by zero is undefined, the argument is invalid.

All numbers are equal to 1

^{[citation needed]}

Suppose we have the following system of linear equations:

$\left\{ \begin{matrix} c_1x_1 + c_1x_2 + \cdots + c_1x_n = c_1\\ c_2x_1 + c_2x_2 + \cdots + c_2x_n = c_2\\ \vdots \\ c_nx_1 + c_nx_2 + \cdots + c_nx_n = c_n \end{matrix}\right.$

Dividing the first equation by c₁, we get x₁ + x₂ + · · · + x_n = 1. Let us now try to solve the system via Cramer's rule:

$\begin{bmatrix} c_1 & c_1 & c_1 & \dots\\ c_2 & c_2 & c_2 & \dots\\ \vdots&\vdots&\vdots&\\ c_n & c_n & c_n & \dots \end{bmatrix} \begin{bmatrix} x_1\\ x_2\\ \vdots\\ x_n \end{bmatrix} = \begin{bmatrix} c_1\\ c_2\\ \vdots\\ c_n \end{bmatrix}$

Since each column of the coefficient matrix is equal to the resultant column vector, we have

$A_i = A \implies |A_i| = |A| \implies {|A_i| \over |A|} = 1 = x_i$

for all i. Substituting this back into x₁ + x₂ + · · · + x_n = 1, we get

$\sum_{i=1}^n 1 = 1$ .

Q.E.D.

This proof is fallacious because Cramer's rule can only be applied to systems with a unique solution; however, all the equations in the system are obviously equivalent, and thus are insufficient to provide a unique solution. The fallacy occurs when we try to divide |A_i| by |A|, as both are equal to 0.

Multivalued functions

2π = 0

^{[citation needed]}

$x = 2π$; $s i n (x) = 0$; $x = s i n - 1 (0)$; $x = 0$
$2π = 0$

Q.E.D

The problem is in the third step, where we take the arcsin of each side. Since the arcsin is an infinitely multivalued function, $x = s i n - 1 (0)$ is not necessarily true.

Calculus

0 = 1

^{[citation needed]}

Begin with the evaluation of the indefinite integral

$\int \frac{1}{x} dx$

Through integration by parts, let

$u=\frac{1}{x}$ and

d v = d x

Thus,

$du=-\frac{1}{x^2}dx$ and

v = x

Hence, by integration by parts

$\int \frac{1}{x} dx=\frac{x}{x} - \int \left ( - \frac{1}{x^2} \right ) x dx$

$\int \frac{1}{x} dx=1 + \int \frac{1}{x} dx$

$0 = 1 \,$

Q.E.D.

The error in this proof lies in an improper use of the integration by parts technique. Upon use of the formula, a constant, C, must be added to the right-hand side of the equation. This is due to the derivation of the integration by parts formula; the derivation involves the integration of an equation and so a constant must be added. In most uses of the integration by parts technique, this initial addition of C is ignored until the end when C is added a second time. However, in this case, the constant must be added immediately because the remaining two integrals cancel each other out.

In other words, the second to last line is correct (1 added to any antiderivative of 1/x is still an antiderivative of 1/x); but the last line is not. You cannot cancel $\int 1/x \,dx$ because they are not necessarily equal. There are infinitely many antiderivatives of a function, all differing by a constant. In this case, the antiderivatives on both sides differ by 1.

This problem can be avoided if we use definite integrals (i.e. use bounds). Then in the second to last line, 1 would be evaluated between some bounds, which would always evaluate to 1 - 1 = 0. The remaining definite integrals on both sides would indeed be equal.

1 = 0

^{[citation needed]}

Take the statement

$x = 1\,$

Taking the derivative of each side,

$\frac{d}{dx}x = \frac{ d}{dx}1$

The derivative of x is 1, and the derivative of 1 is 0. Therefore,

$1 = 0\,$

Q.E.D.

The error in this proof is it treats x as a variable, and not as a constant as stated with x = 1 in the proof, when taking its derivative. Taking the proper derivative of x yields the correct result, 0 = 0.

Infinite series

0 = 1

^{[citation needed]}

Start with the addition of an infinite succession of zeros

$0 = 0 + 0 + 0 + \cdots$

Then recognize that $0 = 1 - 1$

$0 = (1 - 1) + (1 - 1) + (1 - 1) + \cdots$

Applying the associative law of addition results in

$0 = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + \cdots$

Of course $- 1 + 1 = 0$

$0 = 1 + 0 + 0 + 0 + \cdots$

And the addition of an infinite string of zeros can be discarded leaving

$0 = 1 \,$

Q.E.D.

The error here is that the associative law cannot be applied freely to an infinite sum unless the sum is absolutely convergent (see also conditionally convergent). Here that sum is 1 − 1 + 1 − 1 + · · ·, a classic divergent series. In this particular argument, the second line gives the sequence of partial sums 0, 0, 0, ... (which converges to 0) while the third line gives the sequence of partial sums 1, 1, 1, ... (which converges to 1), so these expressions need not be equal. This can be seen as a counterexample to generalizing Fubini's theorem and Tonelli's theorem to infinite integrals (sums) over measurable functions taking negative values.

In fact the associative law for addition just states something about three-term sums: $(a + b) + c = a + (b + c)$ . It can easily be shown to imply that for any finite sequence of terms separated by "+" signs, and any two ways to insert parentheses so as to completely determine which are the operands of each "+", the sums have the same value; the proof is by induction on the number of additions involved. In the given "proof" it is in fact not so easy to see how to start applying the basic associative law, but with some effort one can arrange larger and larger initial parts of the first summation to look like the second. However this would take an infinite number of steps to "reach" the second summation completely. So the real error is that the proof compresses infinitely many steps into one, while a mathematical proof must consist of only finitely many steps. To illustrate this, consider the following "proof" of $1 = 0$ that only uses convergent infinite sums, and only the law allowing to interchange two consecutive terms in such a sum, which is definitely valid:

$\begin{align}1&=1+0+0+0+0+\cdots\\ &=0+1+0+0+0+\cdots\\ &=0+0+1+0+0+\cdots\\ &=0+0+0+1+0+\cdots\\ &=0+0+0+0+1+\cdots\\ &\vdots\\ &=0+0+0+0+0+\cdots\\ &=0\end{align}$

The sum of all positive integers is negative

^{[citation needed]}

Define the constants S and A by

$S=\sum_{k=1}^{\infin} k=1+2+3+4+\cdots \quad A=\sum_{k=1}^{\infin} k(-1)^{k+1}=1-2+3-4+\cdots$ .

Therefore

$S-A=(1-1)+(2-(-2))+(3-3)+(4-(-4))+\cdots=4+8+12+16+\cdots=4S$

$S+A=(1+1)+(2+(-2))+(3+3)+(4+(-4))+\cdots=2+6+10+14+\cdots=4S+(2+2+2+2+\cdots)\; .$

Adding these two equations gives

$2S = 8S + (2+2+2+2+\cdots)$

$\Rightarrow S = -\left(\frac{1}{6}\right)(2+2+2+2+\cdots)\; .$

Therefore, the sum of all positive integers is negative.

The error in this proof is that it assumes that divergent series obey the ordinary laws of arithmetic.

The sum of all positive integral powers of 2 is -1

^{[citation needed]}

Let $x=1+2+4+8+16+\cdots$ .

Then, $2x=2+4+8+16+\cdots$ .

Subtracting these two equations produces $2x-x=x=(2+4+8+16+\cdots)-(1+2+4+8+16+\cdots)=-1$ .

Therefore, $1+2+4+8+16+\cdots=-1$ .

(see also p-adic numbers)

Extraneous solutions

−2 = 1

^{[citation needed]}

Start by attempting to solve the equation

$\sqrt[3] {1-x} + \sqrt[3] {x-3} = 1\,$

Taking the cube of both sides yields

$(1-x) + 3 \sqrt[3]{1-x} \ \sqrt[3]{x-3} \ (\sqrt[3]{1-x} + \sqrt[3]{x-3}) + (x-3) = 1\,$

Replacing the expression within parenthesis by the initial equation and canceling common terms yields

$-2 + 3 \sqrt[3]{1-x} \ \sqrt[3]{x-3} = 1\,$

$\sqrt[3]{1-x} \ \sqrt[3]{x-3} = 1\,$

Taking the cube again produces

$(1-x) (x-3) = 1\,$

$-x^2 + 4 x - 3 = 1\,$

$x^2 - 4 x + 4 = 0\,$

Which produces the solution x = 2. Substituting this value into the original equation, one obtains

$\sqrt[3] {1-2} + \sqrt[3] {2-3} = 1\,$

So therefore

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

$-2 = 1\,$

Q.E.D.

In the forward direction, the argument merely shows that no x exists satisfying the given equation. If you work backward from x=2, taking the cube root of both sides ignores the possible factors of $\frac{-1}{2} \pm \frac{\sqrt{3}}{2}i$ which are non-principal cube roots of one. An equation altered by raising both sides to a power is a consequence, but not necessarily equivalent to, the original equation, so it may produce more solutions. This is indeed the case in this example, where the solution x = 2 is arrived at while it is clear that this is not a solution to the original equation. Also, every number has 3 cube roots, 2 complex and one either real or complex. Also the substitution of the first equation into the second to get the third would be begging the question when working backwards.

3 = 0

^{[citation needed]}

Assume the following equation for a complex x :

$x^2+x+1=0\,$

Then:

$x^2=-x-1 \,$

Divide by x (assume x is not 0)

$x=-1 -1/x \,$

Substituting the last expression for x in the original equation we get:

$x^2+(x)+1=0\,$

$x^2 + (-1-1/x) +1 = 0 \,$

$x^2 + (-1/x) = 0 \,$

$x^2=1/x \,$

$x^3=1 \,$

$x=1 \,$

Substituting x=1 in the original equation yields:

$1^2+1+1 = 0 \,$

$3 = 0 \,$

Q.E.D.

The fallacy here is in assuming that $x 3 = 1$ implies $x = 1$ . There are in fact three cubed roots of unity. Two of these roots, which are complex, are the solutions of the original equation. The substitution has introduced the third one, which is real, as an extraneous solution. The equation after the substitution step is implied by the equation before the substitution, but not the other way around, which means that the substitution step could and did introduce new solutions.

Note that if we restrict our attention to the real numbers, so that $x 3 = 1$ implies $x = 1$ is in fact true, the proof will still fail. In fact, what it is saying is that if there is some real number solution to the equation, then 0=3. This is true, but it is a big "if". The quadratic formula erases any lingering fears that 0=3 by showing that no real number solution exists.

Complex numbers

1 = 3

^{[citation needed]}

From Euler's formula we see that

$e^{\pi i} = \cos(\pi) + i \sin(\pi) = -1 + 0i = -1 \,$

and

$e^{3\pi i} = \cos(3\pi) + i \sin(3\pi) = -1 + 0i = -1 \,$

so we have

$e^{\pi i} = e^{3\pi i} \,$

Taking logarithms gives

$\ln(e^{\pi i}) = \ln(e^{3\pi i}) \,$

and hence

$\pi i = {3\pi i} \,$

Dividing by πi gives

$1 = 3 \,$

QED.

The mistake is that the rule $ln(e x) = x$ is in general only valid for real x, not for complex x. The complex logarithm is actually multi-valued; and $ln( - 1) = (2 k + 1)π i$ for any integer k, so we see that $π i$ and $3π i$ are two among the infinite possible values for ln(-1).

x = y for any real x, y

^{[citation needed]}

Let x and y be any two numbers Then let $a = x - y \ , \ b = (x + y)^2 \ , \ c = 4 (x^2 - x y + y^2)\,$ Let $u = a \ , \ v = b - c\,$ Let's compute:

$(u + \sqrt {v})^3 = u^3 + 3 u^2 \sqrt{v} + 3 u v + v \sqrt{v}$

$(u - \sqrt {v})^3 = u^3 - 3 u^2 \sqrt{v} + 3 u v - v \sqrt{v}$

Replacing $u = a \ , \ v = b - c$ , we get:

$(a + \sqrt {b - c})^3 = a \ (a^2 + 3 b - 3 c) + (3 a^2 + b - c)\sqrt{b - c}$

$(a - \sqrt {b - c})^3 = a \ (a^2 + 3 b - 3 c) - (3 a^2 + b - c)\sqrt{b - c}$

Let's compute $3 a^2 + b - c\,$ Replacing $a = x - y \ , \ b = (x + y)^2 \ , \ c = 4 (x^2 - x y + y^2)\,$ :

$3 a^2 + b - c = 3 (x^2 - 2 x y + y^2) + x^2 + 2 x y + y^2 - 4 (x^2 - x y + y^2) = 0\,$

So:

$(a + \sqrt{b - c})^3 = (a - \sqrt {b - c})^3\,$

$a + \sqrt{b - c} = a - \sqrt {b - c}\,$

$\sqrt {b - c} = 0\,$

$b - c = 0\,$

$b = c\,$

Replacing $b = (x + y)^2 \ , \ c = 4 (x^2 - x y + y^2)\,$ :

$(x + y)^2 = 4 (x^2 - x y + y^2)\,$

$x^2 + 2 x y + y^2 = 4 (x^2 - x y + y^2)\,$

$-3 x^2 + 6 x y - 3 y^2 = 0\,$

$(x - y)^2 = 0\,$

$x - y = 0\,$

$x = y\,$

Q.E.D.

The mistake here is that from z³ = w³ one may not in general deduce z = w (unless z and w are both real, which they are not in our case).

Inequalities

1 < 0

^{[citation needed]}

Let us first suppose that

$\displaystyle{0 < x < 1}$

Now we will take the logarithm of both sides. As long as x > 0, we can do this because logarithms are monotonically increasing. Observing that the logarithm of 1 is 0, we get

$\displaystyle{\ln (x) < 0}$

Dividing by ln (x) gives

$\displaystyle{1 < 0}$

Q.E.D.

The violation is found in the last step, the division. This step is invalid because ln(x) is negative for 0 < x < 1. While multiplication or division by a positive number preserves the inequality, multiplication or division by a negative number reverses the inequality, resulting in the correct expression 1 > 0.

Examples in geometry

Any angle is zero

^{[citation needed]}

Construct a rectangle ABCD. Now identify a point E such that CD = CE and the angle DCE is a non-zero angle. Take the perpendicular bisector of AD, crossing at F, and the perpendicular bisector of AE, crossing at G. Label where the two perpendicular bisectors intersect as H and join this point to A, B, C, D, and E.

Now, AH=DH because FH is a perpendicular bisector; similarly BH = CH. AH=EH because GH is a perpendicular bisector, so DH = EH. And by construction BA = CD = CE. So the triangles ABH, DCH and ECH are congruent, and so the angles ABH, DCH and ECH are equal.

But if the angles DCH and ECH are equal then the angle DCE must be zero.

Q.E.D.

The error in the proof comes in the diagram and the final point. An accurate diagram would show that the triangle ECH is a reflection of the triangle DCH in the line CH rather than being on the same side, and so while the angles DCH and ECH are equal in magnitude, there is no justification for subtracting one from the other; to find the angle DCE you need to subtract the angles DCH and ECH from the angle of a full circle (2π or 360°).

Fallacy of the isosceles triangle

The fallacy of the isosceles triangle, from (Maxwell 1959, Chapter II, § 1), purports to show that every triangle is isosceles, meaning that two sides of the triangle are congruent.

Given a triangle △ABC, proof that AB = AC:

Draw a line bisecting ∠A
Call the midpoint of line segment BC, D
Draw the perpendicular bisector of segment BC, which contains D
If these two lines are parallel, AB = AC; otherwise they intersect at point O
Draw line OR perpendicular to AB, line OQ perpendicular to AC
Draw lines OB and OC
By AAS, △RAO ≅ △QAO (AO = AO; ∠OAQ ≅ ∠OAR since AO bisects ∠A; ∠ARO ≅ ∠AQO are both right angles)
By HL, △ODB ≅ △ODC (∠ODB,∠ODC are right angles; OD = OD; BD = CD because OD bisects BC)
By SAS, △ROB ≅ △QOC (RO = QO since △RAO ≅ △QAO; BO = CO since △ODB ≅ △ODC; ∠ROB ≅ ∠QOC since they are vertical angles)
Thus, AR ≅ AQ, RB ≅ QC, and AB = AR + RB = AQ + QC = AC

Q.E.D.

As a corollary, one can show that all triangles are equilateral, by showing that AB = BC and AC = BC in the same way.

All but the last step of the proof is indeed correct (those three triangles are indeed congruent). The error in the proof is the assumption in the diagram that the point O is inside the triangle. In fact, whenever AB ≠ AC, O lies outside the triangle. Furthermore, it can be further shown that, if AB is shorter than AC, then E will lie outside of AB, while F will lie within AC (and vice versa). (Any diagram drawn with sufficiently accurate instruments will verify the above two facts.) Because of this, AB is actually AR - RB, whereas AC is still AQ + QC; and thus the lengths are not necessarily the same.

Notes

^ Maxwell 1959, p. 9
^ Maxwell 1959
^ Heath & Helberg 1908, Chapter II, §I
^ Maxwell 1959, p. 7
^ Maxwell 1959
^ Maxwell 1959, Chapter VI, §I.2
^ Maxwell 1959, Chapter VI, §I.1
^ Maxwell 1959, Chapter VI, §II
^ In general, the expression $\sqrt[n]{1}$ evaluates to n complex numbers, called the nth roots of unity.
^ Harro Heuser: Lehrbuch der Analysis - Teil 1, 6th edition, Teubner 1989, ISBN 978-3835101319, page 51 (German).

References

Barbeau, Edward J. (2000), Mathematical fallacies, flaws, and flimflam, MAA Spectrum, Mathematical Association of America, MR 1725831, ISBN 978-0-88385-529-4 .
Bunch, Bryan (1997), Mathematical fallacies and paradoxes, New York: Dover Publications, MR 1461270, ISBN 978-0-486-29664-7 .
Heath, Sir Thomas Little; Heiberg, Johan Ludvig (1908), The thirteen books of Euclid's Elements, Volume 1, The University Press .
Maxwell, E. A. (1959), Fallacies in mathematics, Cambridge University Press, MR 0099907 .

External links

Invalid proofs at Cut-the-knot (including literature references)
More invalid proofs from AhaJokes.com
More invalid proofs also on this page

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Mathematical fallacy

Contents

Simple examples

Power and root

Division by zero

2 = 1

All numbers are equal to 1

Multivalued functions

2π = 0

Calculus

0 = 1

1 = 0

Infinite series

0 = 1

The sum of all positive integers is negative

The sum of all positive integral powers of 2 is -1

Extraneous solutions

−2 = 1

3 = 0

Complex numbers

1 = 3

x = y for any real x, y

Inequalities

1 < 0

Examples in geometry

Any angle is zero

Fallacy of the isosceles triangle

See also

Notes

References

External links

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