Invalid proof

From Exampleproblems

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.

Contents 1 Examples 1.1 Proof that 1 equals −1 1.2 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 1.6 Another proof that 2 = 1 2 Conclusion 3 See also 4 External links

Examples

Proof that 1 equals −1

• We start with

  − 1 = − 1

• Then we convert these into vulgar fractions

  

• Applying square roots on both sides gives

  

• Which is equal to

  

• If we now clear fractions by multiplying both sides by and then , we have

  

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

  1 = − 1

Q.E.D.

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

This principle is only correct when 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 first suppose that

  0 < x < 1

• 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

  lnx < 0

• Dividing by ln x gives

  1 < 0

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 as can clearly be seen in the previous line. 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 non-zero quantities

  a = b

• Multiply through by a

  a² = ab

• Subtract b²

  a² − b² = ab − b²

• Factor both sides

  (a − b)(a + b) = b(a − b)

• Divide out (a − b)

  a + b = b

• Observing that a = b

  b + b = b

• Combine like terms on the left

  2b = b

• Divide by the non-zero 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.

A variation:

• Let x and y be equal, non-zero quantities

  x = y

• Add x to both sides

  2x = x + y

• Take 2y from both sides

  2x − 2y = x − y

• Factorise

  2(x − y) = x − y

• Divide out (x − y)

  2 = 1

Q.E.D.

The fallacy here is the same as above in that by dividing by (x − y), you are dividing by zero and as such, this 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, substitute:

  a² − 2ab + b² = (a − b)c

• Write out the multiplication:

  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, a − b − c = 0, and as a result we have performed an illegal division by zero.

Proof that 0 equals 1

• Start with the addition of an infinite succession of zero's

  

• Then recognize that 0 = 1 − 1

  

• Applying the associative law of addition results in

  

• Of course − 1 + 1 = 0

  

• And the addition of an infinite string of zero's can be discarded leaving

  0 = 1

Q.E.D.

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 2 = 1

• By the common intuitive meaning of multiplication we can see that

  4 * 3 = 3 + 3 + 3 + 3

• It can also be seen that for a non-zero x

  x = 1 + 1 + ... + 1 (x terms)

• Now we multiply through by x

  x² = x + x + ... + x (x terms)

• Then we take the derivative with respect to x

  2x = 1 + 1 + ... + 1 (x terms)

• Now we see that the right hand side is x which gives us

  2x = x

• Finally, dividing by our non-zero x we have

  2 = 1

Q.E.D.

The error here is that in line two our definition of x assumed that x was an integer, non-integer real numbers are precluded by this definition. For that reason the function x² is not a continuous function and this thus not differentiable.

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!

On the other hand it is possible to construct useful mathematical systems where 1 is technically equal to 2. Mathematics in domains modulo 1 are one example. In such domains 0.5 + 0.5 = 0 = 1 = 2...

See also

• Paradox

External links

• Audience of Two's proof that 28 is the largest number
• More invalid proofs from AhaJokes.comes:demostración falsa