Invalid proof

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In mathematics, there are multiple mathematical proofs of obvious contradictions. Although the demos are wrong, the errors are subtle, and most of the time, intentional. These fallacies are normally considered mere curiosities, but they can be used to illustrate the importance of rigor in this area.

Most of these demos rely on variants of the same bug. The error consists in using a function f that is not bijective, to observe that f(x) = f(y) for certain x and y, concluding (wrongly) that therefore x = y. Division by zero is a special case: the function f is x → x × 0, and the wrong step is to start with x × 0 = y × 0 and thus conclude that x = y.

Examples

Proof that 1 is equal to −1

Suppose we are working on the set of Complex Numbers and start with the following:

1=1 is equal to the fact that the elements are reflective

Now, we convert them into fractions

{displaystyle {frac {1}{-1}}={frac {-1}{1}}}

Applying the square root on both sides we obtain

{displaystyle {sqrt {frac {1}{-1}}}={sqrt {frac {-1}{1}}}}

which is equivalent to

{displaystyle {frac {sqrt {1}}{sqrt {-1}}}={frac {sqrt {-1}}{sqrt {1}}}}

But since ${displaystyle i={sqrt {-1}}}$ (see imaginary number), we can replace it, obtaining

{displaystyle {frac {1}{i}}={frac {i}{1}}}

Rearranging the equation to eliminate the fractions, we get

{displaystyle 1^{2}=i^{2} }

And since $i^2 = -1$ we have as a result

{displaystyle 1=-1 }

Q.E.D.

This proof is a mathematical fallacy. Although it is correct that:

${displaystyle {sqrt {frac {1}{-1}}}={sqrt {frac {-1}{1}}}}$

The following step is invalid:

${displaystyle {frac {sqrt {1}}{sqrt {-1}}}={frac {sqrt {-1}}{sqrt {1}}}}$

Since:

${displaystyle {sqrt {frac {1}{-1}}}neq {frac {sqrt {1}}{sqrt {-1}}}}$

Proof that 1 is less than 0

Suppose that

<math alttext="{displaystyle xx.1{displaystyle x vis1} <img alt="{displaystyle x

Now we take the logarithm of both sides of the inequality. We can do this whenever x > 0, because the logarithms increase monotonically. If we take into account that the logarithm of 1 is 0, we will obtain

<math alttext="{displaystyle ln xln x.0{displaystyle ln x vis0} <img alt="{displaystyle ln x

Divide by ln x returns

<math alttext="{displaystyle 11.0{displaystyle 1 tax0} <img alt="{displaystyle 1

Q.E.D.

The error is in the last step, the division. This procedure is incorrect because being x < 1 (by original assumption) the result of ln x should be a negative number. If we divide both members by a negative term, it is necessary to invert the inequality symbol. For this reason, the correct result is:

1 > 0.

(See the correct proof in "Mathematical Proof)".

Proof that 2 is equal to 1

Let a and b be two equal quantities. It follows that:

a	=	b
a2	=	ab
a2 - b2	=	ab - b2
(a - b)(a + b)	=	b(a - b)
a + b	=	b
b	=	b
2b	=	b
2	=	1

Q.E.D.

The fallacy is found from line 4 to 5: where, being a=b, in the same term a² - b² cancel out giving zero in the same term and since division by zero does not is defined, the proof is invalid.

The other fallacy is that it would also prove that a = 0, since if: a + b = b => a = b - b => to = 0

Another proof that 1 equals 2

By definition of multiplication, you have to, x 0,
${displaystyle x=1+1+...+1}$ (x terms)
Multiplying both sides by x,
${displaystyle x^{2}=x+x+...+x}$ (x terms)
Deriving with respect x,
${displaystyle 2x=1+1+...+1}$ (x terms)
Now, returning to the front line, you see that the right side of that equality is xand therefore,
${displaystyle 2x=x}$
Splitting both sides by x (which is possible, since it is a number does not mean that x 0), has
${displaystyle 2=1}$

Q.E.D.

The bug: in the first line of the supposed proof, x was assumed to be an integer; such an expression does not make sense for non-integer numbers. On the other hand, to derive functions it is necessary a continuous domain like the reals, not the integers; for each integer x you have a correct equation, but to derive both sides you need an equation of functions, not integers, and the function x + x +... + x "with x terms" does not make sense in general (how can you have x terms?), so it is not differentiable.

Another way of looking at the error is that you are differentiating two different functions with different derivatives but that intersect at a point. In this sense, it is confirmed that F(x) = G(x) but it is wrongly assumed that F'(x) = G'(x).

Proof that 4 equals 2

4 = 4

Subtract from both sides of the equation

4 - 4 = 4 - 4

On one side we factor using the "sum times its difference" and on the other side it is factored by 2

(2 - 2) (2 + 2) = 2 (2 - 2)

We cancel the equal terms on each side of the equation (2 - 2)

(2 + 2) = 2

We are left as a result

4 = 2

Q.E.D.

The fallacy is found in the passage from line 3 to 4, since it implies a division by (2 - 2), which is zero. Since division by zero is not defined, the proof is invalid.

Proof that a is equal to b

We start with

a - b = c

We square both sides

a2 - 2ab + b2 = c2

Since (a - b)(c) = c² = ac - bc, we can rewrite it as

a2 - 2ab + b2 = ac - bc

If we rearrange it, we get

a2 - ab - ac = ab - b2 - bc

We factor both members

a(a - b - c) = b(a - b - c)

We divide both members by (a - b -c)

a~~(a-b-c)~~ =~~(a-b-c)~~

In the end

a = b

Q.E.D.

The fallacy is that if a - b = c, then a - b - c = 0, so we have performed division by zero, which invalidates the proof.

Proof that 0 is equal to 1

The following is a n#34;demo#34; that 0 is equal to 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
	=	1

Q.E.D.

The error is that the associative law cannot be freely applied to infinite sums unless they are absolutely convergent. The latter was, according to Guido Ubaldus, the demonstration that God exists, since he had been "created" something out of nothing Outside of this, if the associative law could be applied, then this would hold for all numbers, that is, all numbers would be equal to 0, and, by transitivity of equality, all numbers would be the same number.

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