Mathematical fallacy

Care must be taken when taking the square root of both sides of an equality. Failing to do so results in a "proof" of[8] 5 = 4.

Proof:

Start from

${\displaystyle -20=-20}$

Write this as

${\displaystyle 25-45=16-36}$

Rewrite as

${\displaystyle 5^{2}-5\times 9=4^{2}-4\times 9}$

Add 81/4 on both sides:

${\displaystyle 5^{2}-5\times 9+{\frac {81}{4}}=4^{2}-4\times 9+{\frac {81}{4}}}$

These are perfect squares:

${\displaystyle \left(5-{\frac {9}{2}}\right)^{2}=\left(4-{\frac {9}{2}}\right)^{2}}$

Take the square root of both sides:

${\displaystyle 5-{\frac {9}{2}}=4-{\frac {9}{2}}}$

Add 9/2 on both sides:

${\displaystyle 5=4}$

Q.E.D.

The fallacy is in the second to last line, where the square root of both sides is taken: a² = b² only implies a = b if a and b have the same sign, which is not the case here. In this case, it implies that a = –b, so the equation should read

${\displaystyle 5-{\frac {9}{2}}=-\left(4-{\frac {9}{2}}\right)}$

which, by adding 9/2 on both sides, correctly reduces to 5 = 5.

Another example illustrating the danger of taking the square root of both sides of an equation involves the following fundamental identity[9]

${\displaystyle \cos ^{2}x=1-\sin ^{2}x}$

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

${\displaystyle \cos x={\sqrt {1-\sin ^{2}x}}}$

Evaluating this when x = π , we get that

${\displaystyle -1={\sqrt {1-0}}}$

or

${\displaystyle -1=1}$

which is incorrect.

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

${\displaystyle x^{2}=a^{2}}$

where ${\displaystyle a\neq 0}$, has two solutions:

${\displaystyle x=\pm a}$

and it is essential to check which of these solutions is relevant to the problem at hand.[10] 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.

Invalid proofs utilizing powers and roots are often of the following kind:

${\displaystyle 1={\sqrt {1}}={\sqrt {(-1)(-1)}}={\sqrt {-1}}{\sqrt {-1}}=i\cdot i=-1.}$

The fallacy is that the rule ${\displaystyle {\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}}$ is generally valid only if at least one of ${\displaystyle x}$ and ${\displaystyle y}$ is non-negative (when dealing with real numbers), which is not the case here.[11]

Alternatively, imaginary roots are obfuscated in the following:

${\displaystyle i={\sqrt {-1}}=\left(-1\right)^{\frac {2}{4}}=\left(\left(-1\right)^{2}\right)^{\frac {1}{4}}=1^{\frac {1}{4}}=1}$

The error here lies in the third equality, as the rule ${\displaystyle a^{bc}=(a^{b})^{c}}$ only holds for positive real a and real b, c.

When a number is raised to a complex power, the result is not uniquely defined (see Exponentiation § Failure of power and logarithm identities). If this property is not recognized, then errors such as the following can result:

${\displaystyle {\begin{aligned}e^{2\pi i}&=1\\\left(e^{2\pi i}\right)^{i}&=1^{i}\\e^{-2\pi }&=1\\\end{aligned}}}$

The error here is that the rule of multiplying exponents as when going to the third line does not apply unmodified with complex exponents, even if when putting both sides to the power i only the principal value is chosen. When treated as multivalued functions, both sides produce the same set of values, being {e^2πn | n ∈ ℤ}.

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. This fallacy was known to Lewis Carroll and may have been discovered by him. It was published in 1899.[12][13]

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

1. Draw a line bisecting ∠A.
2. Draw the perpendicular bisector of segment BC, which bisects BC at a point D.
3. Let these two lines meet at a point O.
4. Draw line OR perpendicular to AB, line OQ perpendicular to AC.
5. Draw lines OB and OC.
6. By AAS, △RAO ≅ △QAO (∠ORA = ∠OQA = 90°; ∠RAO = ∠QAO; AO = AO (common side)).
7. By RHS,[note 2] △ROB ≅ △QOC (∠BRO = ∠CQO = 90°; BO = OC (hypotenuse); RO = OQ (leg)).
8. 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 triangle. In fact, O always lies on 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 drawn 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 thus the lengths are not necessarily the same.