Invalid proof

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In mathematics, there are a variety of spurious proofs of obvious contradictions. Although the proofs are flawed, the errors are comparatively subtle, usually by design. These fallacies are normally regarded as mere curiosities, but can be used to show the importance of rigor in mathematics. Most of these proofs depend on some variation of the same error. The error is to take a function ''f'' that is not one-to-one, to observe that ''f(x)'' = ''f(y)'' for some ''x'' and ''y'', and to (erroneously) conclude that therefore ''x'' = ''y''. Division by zero is a special case of this; the function ''f'' is ''x'' → ''x'' × 0, and the erroneous step is to start with ''x''×0 = ''y''×0 and to conclude that therefore ''x''=''y''. == Examples == === Proof that 1 equals −1 === We start with :<math>-1 = -1\ </math> Then we convert these into fractions :<math>\frac{1}{-1} = \frac{-1}{1}</math> Applying square roots on both sides gives :<math>\sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}</math> Which is equal to :<math>\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}</math> But since <math>i = \sqrt{-1}</math> (see imaginary number), we can substitute, obtaining :<math>\frac{1}{i} = \frac{i}{1}</math> By rearranging the equation to remove the fractions, we get :<math>1^2 = i^2\ </math> And since <math>i^2 = -1</math>, we therefore have :<math>1 = -1\ </math> ''Q.E.D.'' This proof is invalid since it applies the following principle for square roots wrongly: :<math>\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}</math> This principle is only correct when the product of ''x'' and ''y'' is a positive number. In the "proof" above, this is not the case. Thus the proof is invalid. === Proof that 1 is less than 0 === Let us suppose that :<math>x < 1</math> Now we will take the logarithm on 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 :<math>\ln x < 0</math> Dividing by ln ''x'' gives :<math>1 < 0</math> ''Q.E.D.'' The violation is found in the last step, the division. This step is wrong because the number we are dividing by is negative, which in turn is because the argument to the logarithm is less than 1, our original assumption. A multiplication with or division by a negative number flips the inequality sign; in other words, we should obtain 1 > 0, which is indeed correct. === Proof that 2 equals 1 === Let ''a'' and ''b'' be equal quantities. It follows that: :<math>a=b</math> :<math>a^2 = ab</math> :<math>a^2 - b^2 = ab - b^2</math> :<math>(a - b)(a + b) = b(a - b)</math> :<math>a + b = b</math> :<math>b + b = b</math> :<math>2b = b</math> :<math>2 = 1</math> ''Q.E.D.'' 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. === Proof that a equals b === *we start with: *:a - b = c *now, square both sides: *:a² - 2ab + b² = c² *since (a - b)(c) = c² = ac - bc, we can rewrite: *:a² - 2ab + b² = ac - bc *rearranging all, we get: *:a² - ab - ac = ab - b² - bc *factorize both members: *:a(a - b - c) = b(a - b - c) *cancel the common factor: *:a = b ''Q.E.D.'' The catch is that since a-b=c, then a-b-c=0, and we have performed an illegal division by zero. === Proof that 0 equals 1 === The following is a "proof" that 0 equals 1:

0	=	0 + 0 + 0 + ...
	=	(1 − 1) + (1 − 1) + (1 − 1) + ...
	=	1 + (−1 + 1) + (−1 + 1) + (−1 + 1) + ...	(associative law)
	=	1 + 0 + 0 + 0 + ...
	=	1

''Q.E.D.'' The error here is that the associative law cannot be applied freely to infinite sums unless they are absolutely convergent. In fact, it is possible to show that in any field, 0 is not equal to 1. == See also == * Paradoxes:demostración falsa