# Imperfect Math

Mathematics has its limits. Extraneous roots and some interesting fraction properties reveal this.

## Extraneous Roots

Solve the following for x:

x - 3 = √30 - 2x

(x -3)2 = 30 - 2x

x2 - 6x + 9 = 30 - 2x

x2 - 6x + 9 + 2x =30 - 2x + 2x

x2 - 4x + 9 = 30

x2 - 4x + 9 - 30 = 30 - 30

x2 - 4x - 21 = 0

Now use the Quadratic Formula to solve for x:

x =
 -b ± √b2 - 4ac 2a

Quadratic Equations are written as such:

ax2 + bx + c = 0

Determine a, b, & c:

x2 - 4x - 21 = 0

(1)x2 + (-4)x + (-21) = 0

a = 1; b = -4; c = -21.

Now plugging into the Quadratic Formula, we get the following:

x =
 -(-4) ± √(-4)2 - 4(1)(-21) 2(1)

x =
 4 ± √16 - (-84) 2

x =
 4 ± √100 2

x =
 4 ± 10 2

x = 2 ± 5
x = 7 or x = -3

Now, let's check this in our original equation. Start with x = 7:

x - 3 = √30 - 2x

7 - 3 = √30 - 2(7)

4 = √30 - 14

4 = √16

4 = 4

It checks out!

Now, let's check this in our original equation. Start with x = -3:

x-3 = √30-2x

(-3)-3=√30-2(-3)

-6=√30+6

-6=√36

-6=6

BUT -6≠6

It DOES NOT check out!

This is our extraneous root.

## Unusual Fractions

(3)(1/3)=1 and (9)(1/9)=1

This is true, right?

If we convert to decimals, we get the following:

(3)(0.333333333333333333…)=1 and (9)(0.111111111111111111…)=1

Regardless of which one you solve, you get the following:

0.999999999999999… = 1

However, we know that they are not exactly the same. This seems to be the case for certain others as well (e.g. 11*1/11=1). However, they are essentially the same. For some more fun, visit http://mathforum.org/library/drmath/view/52365.html.

## Just for fun

### Does 1+1=1?

a = 1
b = 1
a = b
a2 = b2
a2 - b2 = 0
(a-b)(a+b) = 0
(a-b)(a+b)/(a-b) = 0/(a-b)
1(a+b) = 0
(a+b) = 0
1 + 1 = 0
2 = 0
1 = 0
1 + 1 = 1

The reason why this is not true and does not work is that a=b=1, then a-b=0. Yet, in the "solution," dividing by (a-b) is used, which is the same as dividing by 0, which is not allowed.

### Does 2 = 1?

a = b
a2 = ab
a2 - b2 = ab-b2
(a-b)(a+b) = b(a-b)
a+b = b
b+b = b
2b = b
2 = 1

The problem with this one is that if a=b, then a-b=0. Yet, the "solution" breaks the rules by dividing by a-b. You cannot divide by 0!

### Again: Does 2=1?

a = b
a2 = ab
a2 + a2 = a2 + ab
2a2 = a2 + ab
2a2 - 2ab = a2 + ab - 2ab
2a2 - 2ab = a2 - ab
∴ 2(a2 - ab) = 1(a2 - ab)
2 = 1

What's the fallacy? In the last step to get to 2 = 1, you have to divide both sides by (a2 - ab).
Since a = b, a2

- ab = a2 - a2 = 0. AND once again, you CANNOT divide by 0.

### Without dividing by 0: Does 2 = 1?

-2 = -2
4 - 6 = 1 - 3
4 - 6 + 9/4 = 1 - 3 + 9/4
(2 - 3/2)(2 - 3/2) = (1 - 3/2)(1 - 3/2)
(2 - 3/2)2 = (1 - 3/2)2
(2 - 3/2)2 = √(1 - 3/2)2
2 - 3/2 = 1 - 3/2
2 = 1

Here is the problem: miscalculation of square root. When you take the square root of the sides, you should calculate to the absolute value as follows:

|(2 - 3/2)| = |(1 - 3/2)|
|(2 - 3/2)| = |(1 - 3/2)|
|1/2| = |-1/2|
1/2 = 1/2