SOME FALLACIOUS MATH PROOFS

 Math has a lot of proofs. Some of these proofs prove statements that we all know cannot be true. When a fallacious statement is proven in math, then there is a violation in the process.

We should note well, some of these violations so that when they appear in a proof, we can easily point them out:

Ø  Dividing a number by zero: Dividing a number by zero is undefined and hence is a violation.

Ø  Any number multiplied by zero yields zero

Ø  If a² /b² = c² /d²  then a/b = ± c/d .This means that either the positive is true or the negative is true and so one should not always be quick to pick the positive as the answer.

Let us now proceed as more violations will be noticed in the process.

PROVE THAT 1 + 1 = 0

Step 1: a = 1

  Step 2:  b = 1

Step 3: a = b

Step 4: a² = b²

Step 5: a² -  b² =o this becomes (a – b)(a + b) = 0

Step 6: (a – b)(a + b) /(a – b) = 0/(a – b)

Step 7: 1(a + b) = 0

Step 8: a + b = 0

Step 9: 1 + 1 = 0 (this is because of step 1 and 2) and this end the proof that 1 + 1 = 0

Why is this a false  proof?

In step 6, (a – b)(a + b) was divided by (a – b). (a – b) is 0 because a = b. The division by (a – b) in step 6 is the same as dividing by zero which is undefined. In the same step 6, (a – b)(a + b) is 0 because (a – b) is 0. Any number multiplied by zero yields zero (Spencer, 1998).

PROVE THAT 2 = 1

Step 1: a² = ab( let a = b)

Step 2: a² – b² = ab – b² ( that is subtract b² from both sides)

Step 3: (a – b)(a + b)/(a – b) = b(a – b)/(a – b) ( that is dividing both sides by (a – b)

Note: ab – b² = b(a – b) (that is, b factorized out)

Step 4: (a + b) = b

Step 5: But a = b

Step 6: b + b = b

Step 7: 2b = b

Step 8: 2b/b = b/b

Step 9: 2 = 1

Which step makes this proof a fallacy?

We should note that a = b.

 In step  2, a² – b² = 0 , ab – b² = 0 and so 0 = 0 which means there is nothing to prove.

In step three ( a – b) = 0. (a – b)(a + b) = 0,

(a – b)(a + b)/(a – b) = 0/0 which is undefined.

In the same step, b(a –b)/(a – b) becomes b(0)/0 which is also undefined. Upon all these “violations”, the proof still continued and that makes it a fallacious proof.

 TO PROVE THAT 1 = 0

1.       Let x = 1

2.       Multiply both sides by x

          x² = x

3.       Subtract 1 from both sides

x² -1 = x – 1

4.       Divide both sides by x – 1

(x² – 1 )/(x – 1) = 1

5.       Simplify

(x – 1)(x + 1)/(x – 1) = 1 note that x² – 1 = (x – 1)(x + 1)

 x + 1 = 1

6.       Subtract 1 from both sides

X = 0   

Substitute the value of x from step 1

1 = 0

The fallacy here is subtle in step 2,

Multiplying both sides by x introduces an extraneous solution to the equation of x = 0. Then in step 4, there is division by x – 1 which is an illegal operation because x – 1 = 0 and you cannot divide by zero.

SEE THE FOLLOWING:

^1.                y = 100, z = 0

               x = (y + z)/2

              2x = y + z

              2x(y –z) = (y + z)(y –z)

              2xy – 2x² = y² – z²

              Z² = 2xz = y² – 2xy

             z² – 2xz + x² = y² – 2xy + x²

             (z – x)² = (y – x)²

              Z – x = y – x

              Z = y

             0 = 100

            Shaun; (December, 2008)

Then 1 + 1 = 3 becomes x + x = 3

2x = 3

2x/2 = 3/2

X = 1.5

Substituting into the expression we get 3 which imply that 1 + 1 = 3 as required.

Are the above two fallacious proofs? If they are, point out the wrong steps and explain why they are are wrong. You ideas will be appreciated.