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 x → x × 0, and the erroneous step is to start with x×0 = y×0 and to conclude that therefore x=y.
| Table of contents |
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1.1 Proof that 1 equals −1
2 See also1.2 Proof that 1 is less than 0 1.3 Proof that 2 equals 1 1.4 Proof that a equals b 1.5 Proof that 0 equals 1 |
We start with
Examples
Proof that 1 equals −1
Then we convert these into fractions
Applying square roots on both sides gives
Which is equal to
But since (see imaginary number), we can substitute, obtaining
This proof is invalid since it applies the following principle for square roots wrongly:
Let us suppose that
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.
Let a and b be equal quantities. It follows that:
Proof that 1 is less than 0
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
Dividing by ln x gives
Q.E.DProof that 2 equals 1
| a | = | b |
| a2 | = | ab |
| a2 - b2 | = | ab - b2 |
| (a - b)(a + b) | = | b(a - b) |
| a + b | = | b |
| b + b | = | b |
| 2b | = | b |
| 2 | = | 1 |
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.
The catch is that since a-b=c, then a-b-c=0, and we have performed an illegal division by zero.
The following is a "proof" that 0 equals 1:
Proof that a equals b
Q.E.D(a - b - c) = b(a - b - c)
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.