
We start with
Then we convert these into fractions
Applying square roots on both sides gives
Which is equal to
But since <math>i = \sqrt{1}</math> (see imaginary number), we can substitute, obtaining
By rearranging the equation to remove the fractions, we get
And since <math>i^2 = 1</math>, we therefore have
This proof is invalid since it applies the following principle for square roots wrongly:
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.
Let us suppose that
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
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.
Suppose that
Multiplying by <math>a</math> gives
Subtracting <math>b^2</math> gives
Factoring out <math>(a  b)</math>,
then cancelling the common factor <math>(a  b)</math> gives
But since <math>a = b</math>, we can substitute <math>a</math> for <math>b</math>
As <math>a</math> is arbitrary it can be cancelled from both sides, therefore,
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 This invalidates the succeeding steps and the proof is no longer valid.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.
Search Encyclopedia

Featured Article
