In mathematics, certain kinds of err proof are frequently exhibited, and sometimes collected, as illustrations of a concept called mathematical fallacy. There is a distinction between a bare mistake and a mathematical fallacy in a proof, in that a error in a proofread leads to an invalid proofread while in the best-known examples of numerical fallacies there is some component of screen or magic trick in the presentation of the proof.
Reading: Mathematical fallacy
For case, the reason why robustness fails may be attributed to a division by zero that is hidden by algebraic note. There is a certain quality of the mathematical fallacy : as typically presented, it leads not entirely to an absurd result, but does so in a crafty or apt way. [ 1 ] Therefore, these fallacies, for pedagogical reasons, normally take the form of inauthentic proof of obvious contradictions. Although the proofread are flawed, the errors, normally by invention, are relatively insidious, or designed to show that certain steps are conditional, and are not applicable in the cases that are the exceptions to the rules. The traditional way of presenting a mathematical fallacy is to give an invalid mistreat of deduction mixed in with valid steps, so that the mean of fallacy is here slenderly different from the legitimate fallacy. The latter normally applies to a form of argument that does not comply with the valid inference rules of logic, whereas the baffling mathematical footprint is typically a right rule applied with a silent improper presumption. Beyond education, the resolving power of a fallacy can lead to deeper insights into a discipline ( for example, the insertion of Pasch ‘s axiom of Euclidean geometry, [ 2 ] the five colour theorem of graph theory ). Pseudaria, an ancient lost book of false proof, is attributed to Euclid. [ 3 ] mathematical fallacies exist in many branches of mathematics. In elementary algebra, typical examples may involve a footfall where division by zero is performed, where a root is falsely extracted or, more generally, where different values of a multiple valued serve are equated. long-familiar fallacies besides exist in elementary Euclidean geometry and calculus. [ 4 ] [ 5 ]
Howlers [edit ]
five hundred five hundred x 1 ten = five hundred five hundred 1 ten 2 = d ∖ vitamin d ∖ 1 ten 2 = − 1 x 2 { \displaystyle { \begin { array } { lambert } \ ; \ ; \ ; { \dfrac { d } { dx } } { \dfrac { 1 } { adam } } \\= { \dfrac { five hundred } { five hundred } } { \dfrac { 1 } { x^ { 2 } } } \\= { \dfrac { d\ ! \ ! \ ! \backslash } { d\ ! \ ! \ ! \backslash } } { \dfrac { 1 } { x^ { 2 } } } \\=- { \dfrac { 1 } { x^ { 2 } } } \end { array } } }
Anomalous cancellation in calculus anomalous cancellation in calculus Examples exist of mathematically correct results derived by incorrect lines of reasoning. Such an argument, however true the ending appears to be, is mathematically invalid and is normally known as a howler. The stick to is an model of a belly laugh involving anomalous cancellation : 16 64 = 16 / 6 / 4 = 1 4. { \displaystyle { \frac { 16 } { 64 } } = { \frac { 16\ ! \ ! \ ! / } { 6\ ! \ ! \ ! /4 } } = { \frac { 1 } { 4 } }. }
class by zero [edit ]
The division-by-zero fallacy has many variants. The stick to exemplar uses a disguise division by zero to “ prove ” that 2 = 1, but can be modified to prove that any number equals any early number .
- Let a and b be equal, nonzero quantities
- a = b { \displaystyle a=b }
- a = b { \displaystyle a=b }
- Multiply by a
- a 2 = a bacillus { \displaystyle a^ { 2 } =ab }
- a 2 = a bacillus { \displaystyle a^ { 2 } =ab }
- Subtract b2
- a 2 − b-complex vitamin 2 = a bacillus − b 2 { \displaystyle a^ { 2 } -b^ { 2 } =ab-b^ { 2 } }
- a 2 − b-complex vitamin 2 = a bacillus − b 2 { \displaystyle a^ { 2 } -b^ { 2 } =ab-b^ { 2 } }
- Factor both sides: the left factors as a difference of squares, the right is factored by extracting b from both terms
- ( a − barn ) ( a + bel ) = bel ( a − b-complex vitamin ) { \displaystyle ( a-b ) ( a+b ) =b ( a-b ) }
- ( a − barn ) ( a + bel ) = bel ( a − b-complex vitamin ) { \displaystyle ( a-b ) ( a+b ) =b ( a-b ) }
- Divide out (a − b)
- a + barn = b { \displaystyle a+b=b }
- a + barn = b { \displaystyle a+b=b }
- Observing that a = b
- b + b = b-complex vitamin { \displaystyle b+b=b }
- b + b = b-complex vitamin { \displaystyle b+b=b }
- Combine like terms on the left
- 2 boron = barn { \displaystyle 2b=b }
- 2 boron = barn { \displaystyle 2b=b }
- Divide by the non-zero b
- 2 = 1 { \displaystyle 2=1 }
- 2 = 1 { \displaystyle 2=1 }
The fallacy is in production line 5 : the progress from line 4 to line 5 involves division by a − b, which is zero since a = b. Since division by zero is undefined, the argument is invalid .
analysis [edit ]
mathematical analysis as the mathematical learn of change and limits can lead to mathematical fallacies — if the properties of integrals and differentials are ignored. For case, a primitive use of consolidation by parts can be used to give a delusive proof that 0 = 1. [ 7 ] Letting u = 1/ log x and dv = dx / x, we may write :
- ∫ 1 x log x d x = 1 + ∫ 1 x log x five hundred x { \displaystyle \int { \frac { 1 } { x\, \log x } } \, dx=1+\int { \frac { 1 } { x\, \log ten } } \, dx }
after which the antiderivatives may be cancelled yielding 0 = 1. The problem is that antiderivatives are alone defined up to a changeless and shifting them by 1 or indeed any count is allowed. The error actually comes to fall when we introduce arbitrary integration limits a and b .
- ∫ a barn 1 adam log x five hundred x = 1 | a bacillus + ∫ a barn 1 x log x five hundred x = 0 + ∫ a b 1 ten log x five hundred x = ∫ a boron 1 ten log x five hundred x { \displaystyle \int _ { a } ^ { boron } { \frac { 1 } { x\, \log ten } } \, dx=1|_ { a } ^ { bacillus } +\int _ { a } ^ { b } { \frac { 1 } { x\, \log adam } } \, dx=0+\int _ { a } ^ { b } { \frac { 1 } { x\log ten } } \, dx=\int _ { a } ^ { b-complex vitamin } { \frac { 1 } { x\log ten } } \, dx }
Since the dispute between two values of a constant routine vanishes, the same definite integral appears on both sides of the equation .
Multivalued functions [edit ]
many functions do not have a unique inverse. For case, while squaring a number gives a unique rate, there are two possible straight roots of a cocksure issue. The square root is multivalued. One prize can be chosen by conventionality as the chief value ; in the case of the feather root the non-negative respect is the principal value, but there is no guarantee that the feather root given as the principal value of the square of a number will be equal to the original number ( e.g. the star square root of the squarely of −2 is 2 ). This remains true for nth roots .
positive and negative roots [edit ]
concern must be taken when taking the square root of both sides of an equality. Failing to do then results in a “ proof ” of [ 8 ] 5 = 4. validation :
- Start from
- − 20 = − 20 { \displaystyle -20=-20 }
- − 20 = − 20 { \displaystyle -20=-20 }
- Write this as
- 25 − 45 = 16 − 36 { \displaystyle 25-45=16-36 }
- 25 − 45 = 16 − 36 { \displaystyle 25-45=16-36 }
- Rewrite as
- 5 2 − 5 × 9 = 4 2 − 4 × 9 { \displaystyle 5^ { 2 } -5\times 9=4^ { 2 } -4\times 9 }
- 5 2 − 5 × 9 = 4 2 − 4 × 9 { \displaystyle 5^ { 2 } -5\times 9=4^ { 2 } -4\times 9 }
- Add
81
/
4
on both sides:
- 5 2 − 5 × 9 + 81 4 = 4 2 − 4 × 9 + 81 4 { \displaystyle 5^ { 2 } -5\times 9+ { \frac { 81 } { 4 } } =4^ { 2 } -4\times 9+ { \frac { 81 } { 4 } } }
- 5 2 − 5 × 9 + 81 4 = 4 2 − 4 × 9 + 81 4 { \displaystyle 5^ { 2 } -5\times 9+ { \frac { 81 } { 4 } } =4^ { 2 } -4\times 9+ { \frac { 81 } { 4 } } }
- These are perfect squares:
- ( 5 − 9 2 ) 2 = ( 4 − 9 2 ) 2 { \displaystyle \left ( 5- { \frac { 9 } { 2 } } \right ) ^ { 2 } =\left ( 4- { \frac { 9 } { 2 } } \right ) ^ { 2 } }
- ( 5 − 9 2 ) 2 = ( 4 − 9 2 ) 2 { \displaystyle \left ( 5- { \frac { 9 } { 2 } } \right ) ^ { 2 } =\left ( 4- { \frac { 9 } { 2 } } \right ) ^ { 2 } }
- Take the square root of both sides:
- 5 − 9 2 = 4 − 9 2 { \displaystyle 5- { \frac { 9 } { 2 } } =4- { \frac { 9 } { 2 } } }
- 5 − 9 2 = 4 − 9 2 { \displaystyle 5- { \frac { 9 } { 2 } } =4- { \frac { 9 } { 2 } } }
- Add
9
/
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2
on both sides:
- 5 = 4 { \displaystyle 5=4 }
- 5 = 4 { \displaystyle 5=4 }
- Q.E.D.
The fallacy is in the second to final line, where the square etymon of both sides is taken : a 2 = b 2 only implies a = b if a and b have the lapp sign, which is not the encase here. In this case, it implies that a = – b, so the equation should read
- 5 − 9 2 = − ( 4 − 9 2 ) { \displaystyle 5- { \frac { 9 } { 2 } } =-\left ( 4- { \frac { 9 } { 2 } } \right ) }
which, by adding 9/2 on both sides, correctly reduces to 5 = 5. Another case illustrating the danger of taking the feather beginning of both sides of an equation involves the watch fundamental identity [ 9 ]
- cosine 2 x = 1 − sine 2 x { \displaystyle \cos ^ { 2 } x=1-\sin ^ { 2 } ten }
which holds as a consequence of the Pythagorean theorem. then, by taking a square root ,
- carbon monoxide x = 1 − sin 2 x { \displaystyle \cos x= { \sqrt { 1-\sin ^ { 2 } adam } } }
Evaluating this when x = π, we get that
- − 1 = 1 − 0 { \displaystyle -1= { \sqrt { 1-0 } } }
or
- − 1 = 1 { \displaystyle -1=1 }
which is faulty. The error in each of these examples basically lies in the fact that any equation of the form
- ten 2 = a 2 { \displaystyle x^ { 2 } =a^ { 2 } }
where a ≠ 0 { \displaystyle a\neq 0 } 
- x = ± a { \displaystyle x=\pm a }
and it is essential to check which of these solutions is relevant to the trouble at hand. [ 10 ] In the above fallacy, the squarely root that allowed the second equation to be deduced from the first is valid only when cobalt x is positivist. In particular, when x is set to π, the second equality is rendered invalid .
Square roots of negative numbers [edit ]
Invalid proof utilizing powers and roots are much of the take after kind :
- 1 = 1 = ( − 1 ) ( − 1 ) = − 1 − 1 = one ⋅ iodine = − 1. { \displaystyle 1= { \sqrt { 1 } } = { \sqrt { ( -1 ) ( -1 ) } } = { \sqrt { -1 } } { \sqrt { -1 } } =i\cdot i=-1. }
The fallacy is that the rule x y = x y { \displaystyle { \sqrt { xy } } = { \sqrt { adam } } { \sqrt { y } } } 
y
{\displaystyle y}
- i = − 1 = ( − 1 ) 2 4 = ( ( − 1 ) 2 ) 1 4 = 1 1 4 = 1 { \displaystyle i= { \sqrt { -1 } } =\left ( -1\right ) ^ { \frac { 2 } { 4 } } =\left ( \left ( -1\right ) ^ { 2 } \right ) ^ { \frac { 1 } { 4 } } =1^ { \frac { 1 } { 4 } } =1 }
The error here lies in the third base equality, as the rule a barn cytosine = ( a boron ) cytosine { \displaystyle a^ { bc } = ( a^ { b } ) ^ { c } } 
complex exponents [edit ]
When a issue is raised to a complex might, the result is not uniquely defined ( see failure of world power and logarithm identities ). If this property is not recognized, then errors such as the keep up can result :
- e 2 π one = 1 ( e 2 π one ) i = 1 iodine e − 2 π = 1 { \displaystyle { \begin { aligned } e^ { 2\pi one } & =1\\\left ( e^ { 2\pi i } \right ) ^ { one } & =1^ { iodine } \\e^ { -2\pi } & =1\\\end { aligned } } }
The mistake here is that the rule of multiplying exponents as when going to the one-third channel does not apply unmodified with complex exponents, even if when putting both sides to the baron i only the star rate is chosen. When treated as multivalued functions, both sides produce the same bent of values, being { e 2π n | n ∈ ℤ } .
geometry [edit ]
many numerical fallacies in geometry arise from using an linear equality involving oriented quantities ( such as adding vectors along a given line or adding orient angles in the plane ) to a valid identity, but which fixes only the absolute prize of ( one of ) these quantities. This measure is then incorporated into the equation with the wrong predilection, so as to produce an absurd decision. This ill-timed orientation is normally suggested implicitly by supplying an imprecise diagram of the position, where relative positions of points or lines are chosen in a way that is actually impossible under the hypotheses of the argumentation, but non-obviously so. In general, such a fallacy is easy to expose by drawing a precise movie of the site, in which some relative positions will be different from those in the leave diagram. In ordain to avoid such fallacies, a decline geometric controversy using addition or subtraction of distances or angles should constantly prove that quantities are being incorporated with their right orientation course .
fallacy of the isosceles triangle [edit ]
The fallacy of the isosceles triangle, from ( Maxwell 1959, Chapter II, § 1 ), purports to show that every triangulum is isosceles, meaning that two sides of the triangle are congruous. This fallacy has been attributed to Lewis Carroll. [ 12 ] Given a triangle △ABC, prove that AB = AC :
- Draw a line bisecting ∠A.
- Draw the perpendicular bisector of segment BC, which bisects BC at a point D.
- Let these two lines meet at a point O.
- Draw line OR perpendicular to AB, line OQ perpendicular to AC.
- Draw lines OB and OC.
- By AAS, △RAO ≅ △QAO (∠ORA = ∠OQA = 90°; ∠RAO = ∠QAO; AO = AO (common side)).
- By RHS,[note 2] △ROB ≅ △QOC (∠BRO = ∠CQO = 90°; BO = OC (hypotenuse); RO = OQ (leg)).
- 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. The error in the proof is the assumption in the diagram that the point O is inside the triangulum. In fact, O always lies at the circumcircle of the △ABC ( except for isosceles and equilateral triangles where AO and OD coincide ). Furthermore, it can be shown that, if AB is longer than AC, then R will lie within AB, while Q will lie outside of AC, and vice versa ( in fact, any diagram draw with sufficiently accurate instruments will verify the above two facts ). Because of this, AB is still AR + RB, but AC is actually AQ − QC ; and therefore the lengths are not necessarily the same .
proof by generalization [edit ]
There exist several fallacious proofs by induction in which one of the components, basis case or inductive footfall, is wrong. intuitively, proof by induction ferment by arguing that if a statement is true in one case, it is true in the adjacent subject, and hence by repeatedly applying this, it can be shown to be truthful for all cases. The following “ proof ” shows that all horses are the like tinge. [ 13 ] [ note 3 ]
- Let us say that any group of N horses is all of the same colour.
- If we remove a horse from the group, we have a group of N − 1 horses of the same colour. If we add another horse, we have another group of N horses. By our previous assumption, all the horses are of the same colour in this new group, since it is a group of N horses.
- Thus we have constructed two groups of N horses all of the same colour, with N − 1 horses in common. Since these two groups have some horses in common, the two groups must be of the same colour as each other.
- Therefore, combining all the horses used, we have a group of N + 1 horses of the same colour.
- Thus if any N horses are all the same colour, any N + 1 horses are the same colour.
- This is clearly true for N = 1 (i.e. one horse is a group where all the horses are the same colour). Thus, by induction, N horses are the same colour for any positive integer N. i.e. all horses are the same colour.
The fallacy in this proof arises in line 3. For N = 1, the two groups of horses have N − 1 = 0 horses in common, and frankincense are not necessarily the lapp discolor as each other, so the group of N + 1 = 2 horses is not inevitably all of the lapp color. The deduction “ every N horses are of the same semblance, then N + 1 horses are of the like color ” works for any N > 1, but fails to be true when N = 1. The footing casing is chastise, but the generalization gradation has a fundamental flaw .
See besides [edit ]
Notes [edit ]
- ^ 19 95 = 19 / 9 / 5 = 1 5 26 65 = 26 / 6 / 5 = 2 5 49 98 = 49 / 9 / 8 = 4 8 = 1 2 { \displaystyle { \begin { aligned } { \frac { 19 } { 95 } } = { \frac { 19\ ! \ ! \ ! / } { 9\ ! \ ! \ ! /5 } } & = { \frac { 1 } { 5 } } \\ { \frac { 26 } { 65 } } = { \frac { 26\ ! \ ! \ ! / } { 6\ ! \ ! \ ! /5 } } & = { \frac { 2 } { 5 } } \\ { \frac { 49 } { 98 } } = { \frac { 49\ ! \ ! \ ! / } { 9\ ! \ ! \ ! /8 } } & = { \frac { 4 } { 8 } } = { \frac { 1 } { 2 } } \end { aligned } } }
The lapp fallacy besides applies to the follow :
Read more: Willem Dafoe
- ^ Hypotenuse–leg congruity
- ^n girls have the same colour eyes.George Pólya ‘s original “proof” was that anygirls have the same colour eyes.