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 has been 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 as violations:

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 (
spencer 1998)

Ø
If a

^{2 }/b^{2 }= c^{2 }/d^{2 }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: a

^{2}= b^{2}
Step 5: a

^{2}- b^{2}=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: a

^{2}= ab( let a = b)
Step 2: a

^{2 }– b^{2}= ab – b^{2 }( that is subtract b^{2 }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 – b

^{2}= 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, a

^{2}– b^{2}= 0 , ab – b^{2}= 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

x

^{2}= x
3.
Subtract 1 from both sides

x

^{2}-1 = x – 1
4.
Divide both sides by x – 1

(x

^{2}– 1 )/(x – 1) = 1
5.
Simplify

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

^{2}– 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 – 2x

^{2 }= y^{2 }– z^{2}
Z

^{2 }= 2xz = y^{2}– 2xy
z

^{2}– 2xz + x^{2}= y^{2 }– 2xy + x^{2}
(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.

## No comments:

## Post a Comment