# Invalid proof

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 xx × 0, and the erroneous step is to start with x × 0 = y × 0 and to conclude that therefore x = y.

 Contents

## Examples

### Proof that 1 equals −1

[itex]-1 = -1\ [itex]

Then we convert these into vulgar fractions

[itex]\frac{1}{-1} = \frac{-1}{1}[itex]

Applying square roots on both sides gives

[itex]\sqrt{\frac{1}{-1}} = \sqrt{\frac{-1}{1}}[itex]

Which is equal to

[itex]\frac{\sqrt{1}}{\sqrt{-1}} = \frac{\sqrt{-1}}{\sqrt{1}}[itex]

If we now clear fractions by multiplying both sides by [itex]\sqrt{-1}[itex] and then [itex]\sqrt{1}[itex], we have

[itex]\sqrt{1}\sqrt{1} = \sqrt{-1}\sqrt{-1}[itex]

But any number's square root squared gives the original number, so

[itex]1 = -1\ [itex]

This proof is invalid since it applies the following principle for square roots wrongly:

[itex]\sqrt{\frac{x}{y}} = \frac{\sqrt{x}}{\sqrt{y}}[itex]

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

[itex]x < 1[itex]

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

[itex]\ln x < 0[itex]

Dividing by ln x gives

[itex]1 < 0[itex]

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:

1. [itex]a = b[itex]
2. [itex]a^2 = ab[itex]
3. [itex]a^2 - b^2 = ab - b^2[itex]
4. [itex](a - b)(a + b) = b(a - b)[itex]
5. [itex]a + b = b[itex]
6. [itex]b + b = b[itex]
7. [itex]2b = b[itex]
8. [itex]2 = 1[itex]

The fallacy is in line 5: the progression from line 4 to line 5 involves division by ab, which is zero since a equals b. Since division by zero is undefined, the argument is invalid.

### Proof that a equals b

[itex]a - b = c[itex]
• now, square both sides:
[itex]a^2 - 2ab + b^2 = c^2[itex]
• since [itex]a - b = c[itex], substitute:
[itex]a^2 - 2ab + b^2 = (a - b)c[itex]
• write out the multiplication:
[itex]a^2 - 2ab + b^2 = ac - bc[itex]
• rearranging all, we get:
[itex]a^2 - ab - ac = ab - b^2 - bc[itex]
• factorize both members:
[itex]a(a - b - c) = b(a - b - c)[itex]
• cancel the common factor:
[itex]a = b[itex]

The catch is that since ab = c, abc = 0, and as a result we have performed an illegal division by zero.

### Proof that 0 equals 1

The following is a "proof" that 0 equals 1:

 [itex]0[itex] [itex]=[itex] [itex]0 + 0 + 0 + \ldots[itex] [itex]=[itex] [itex](1 - 1) + (1 - 1) + (1 - 1) + \ldots[itex] [itex]=[itex] [itex]1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + \ldots[itex] (associative law) [itex]=[itex] [itex]1 + 0 + 0 + 0 + \ldots[itex] [itex]=[itex] [itex]1[itex]

The error here is that the associative law cannot be applied freely to an infinite sum unless the sum would converge without any parentheses.  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 it is unclear in what sense these expressions can be considered equal.

### Another proof that any number equals zero

Obviously, 4 * 3 = 3 + 3 + 3 + 3

In general, X * Y = Y + Y + ... + Y (X terms, for any number X and Y)

Taking the derivative with respect to X, we get: Y = 0 + 0 + ... + 0 (X terms)

In other words Y = 0, for any number Y.

The error here is that the second line only makes sense if X is a natural number. However, if it is, then the second line is not a continuous function and therefore its derivative cannot be taken.

## Conclusion

These arguments do constitute valid proofs, but not of the claimed assertions.  For example, there is no a priori reason why division by zero should be defined (it's not a field axiom, for example, though 1 ≠ 0, from which 2 ≠ 1 follows, is an axiom), and the "proof" that 2 = 1 is, in fact, simply a demonstration that division by zero cannot be defined in general.  A proof that division by zero could be defined would demonstrate a contradiction and show that the axiomatic system we are working under is logically inconsistent!

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