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In mathematics, there are numerous "proofs" that show impossible results. These proofs, while seemingly valid to the casual observer, always contains an invalid step where a principle of mathematics is violated. These are normally regarded as mere curiosities, but can be used to show the importance of rigour in mathematics.

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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 both x and y are positive numbers. In the "proof" above, one is a negative number, thus making the whole proof 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

Suppose that

<math>a = b</math>

Multiplying by <math>a</math> gives

<math>a^2 = ab</math>

Subtracting <math>b^2</math> gives

<math>a^2 - b^2 = ab - b^2</math>

Factoring out <math>(a - b)</math>,

<math>(a + b)(a - b) = b(a - b)</math>

then cancelling the common factor <math>(a - b)</math> gives

<math>a + b = b</math>

But since <math>a = b</math>, we can substitute <math>a</math> for <math>b</math>

<math>2a = a</math>.

As <math>a</math> is arbitrary it can be cancelled from both sides, therefore,

<math>2 = 1</math>

Q.E.D.

The violation is found in the step where the common factor <math>(a - b)</math> is cancelled. This step is wrong because that factor is equal to zero, and in cancelling it, an implicit division by zero is made. This invalidates the succeeding steps and the proof is no longer valid. Before that last step, it can be said that we proved x*0=0.



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