impossible statements

[Dan Zollman]

I've been using these proofs for a while, and after seeing a math and science forum, I decided to mention them.

Proof #1:
A + A - 2A = A + A - 2A (both members are identical)
A + (A - 2A) = (A + A) - 2A (associative property)
A + (-A) = (2A) - 2A (simplify contents of parentheses)
A - A = 2A - 2A
1(A - A) = 2A - 2A (factoring)
1(A - A) = 2(A - A) (factoring again)
1(A - A)/(A - A) = 2(A - A)/(A - A) (divide both sides by the quantity)
1 = 2

I have a supporting argument for this one. When someone finds the counterstatement, I'll continue.

Proof #2:
I found this while answering the question of 'what is 1 to the i power?'.
Since 1 = i^4, then
1^i = (i^4)^i
We also know that 1 to any power is 1, so we can add:
1 = (i^4)^i
and...
= i^4i = (i^i)^4
...take the 4th root of 1...
1^i = i^i
1=i, etc.

The simple answer to Proof #2 is that exponentiation is not defined for imaginary numbers (which has been confirmed to me by a few teachers and a math professor).
Now that leaves me with the questions: If exponentiation is not defined for imaginary numbers, then how can it be said that e^i pi = -1, and why does a TI graphing calculator say that i^i is .20something (while confirming that 1^i = 1)?
I haven't taken trig, calc, or stat yet in school, so part of my confusion may be because I haven't learned anything about that yet.

I know that these proofs are somewhat ignorant to the general purpose of the math systems involved. They aren't useful in the actualI'm just wondering what people would have to say about both of these proofs, and if they knew the answers to my questions.

[Dan Zollman]

I actually just realized I should have done a search on this topic before posting so that I don't post the same topic again, and I found this:
http://www.chiefdelphi.com/forums/sh...ad.php?t=24368

A variation of the first proof was in that thread, and even though the thread is a year and a half old, I shouldn't have mentioned it.

But I still want to know what people have to say about thes kind of mathematical proof. I think it's a very interesting subject.

[sanddrag]

I don't know about that one (I actually don't have time to read it at the moment) but what about this.

1/9 = .11111 continuing
2/9 = .222222 continuing
etc etc
8/9 = .88888 continuing

9/9 = ? .9999 continuing or 1 exactly? I know .9999 continuing is realy close to 1 but still it makes you wonder.

[Alan Anderson]

Originally Posted by sanddrag

9/9 = ? .9999 continuing or 1 exactly? I know .9999 continuing is realy close to 1 but still it makes you wonder.

".9999 continuing" is equal to 1.

[sanddrag]

Originally Posted by Alan Anderson

".9999 continuing" is equal to 1.

How do you figure? You are always going to have that really small (but existant) 0.00000...00001 left over. I see 0.999 continuing as approaching 1 but I am inclined to think it never actually gets there.

[Alan Anderson]

Originally Posted by Alan Anderson

".9999 continuing" is equal to 1.

Originally Posted by sanddrag

How do you figure? You are always going to have that really small (but existant) 0.00000...00001 left over. I see 0.999 continuing as approaching 1 but I am inclined to think it never actually gets there.

Your position is based on the false impression that the 9s eventually run out, leaving a bit left unaccounted for. They don't; that's what repeating decimals are all about.

Figuring it is easy:

       x = 0.9999...
     10x = 9.9999...
 10x - x = 9.9999... - 0.9999...
      9x = 9
       x = 1

See also http://en.wikipedia.org/wiki/Recurri..._of_0.99999... for other ways to demonstrate it.

[jdiwnab]

Originally Posted by Alan Anderson

Your position is based on the false impression that the 9s eventually run out, leaving a bit left unaccounted for. They don't; that's what repeating decimals are all about.

It is also similer to Limit of x as x approches 1. This isn't what is x when x is one, but you could treat it like that because x can be 1. you are so close to one that it might as well be one

[Michael Hill]

A + A - 2A = A + A - 2A (both members are identical)
A + (A - 2A) = (A + A) - 2A (associative property)
A + (-A) = (2A) - 2A (simplify contents of parentheses)
A - A = 2A - 2A
1(A - A) = 2A - 2A (factoring)
1(A - A) = 2(A - A) (factoring again)
1(A - A)/(A - A) = 2(A - A)/(A - A) (divide both sides by the quantity)
1 = 2

Error...can't divide by 0

[mechanicalbrain]

Originally Posted by Michael Hill

A + A - 2A = A + A - 2A (both members are identical)
A + (A - 2A) = (A + A) - 2A (associative property)
A + (-A) = (2A) - 2A (simplify contents of parentheses)
A - A = 2A - 2A
1(A - A) = 2A - 2A (factoring)
1(A - A) = 2(A - A) (factoring again)
1(A - A)/(A - A) = 2(A - A)/(A - A) (divide both sides by the quantity)
1 = 2

Error...can't divide by 0

How about 1(A-A)=2(A-A) can automattically be factored out because even if they didn't equal zero which they still are automatically factored out of the equation.

[Greg Marra]

Originally Posted by mechanicalbrain

How about 1(A-A)=2(A-A) can automattically be factored out because even if they didn't equal zero which they still are automatically factored out of the equation.

A - A = 2A - 2A

This step is already 0 = 0.

[mechanicalbrain]

OK here's a new one:
Theorem: n=n+1

Proof:
(n+1)^2 = n^2 + 2*n + 1

Bring 2n+1 to the left:
(n+1)^2 - (2n+1) = n^2

Substract n(2n+1) from both sides and factoring, we have:
(n+1)^2 - (n+1)(2n+1) = n^2 - n(2n+1)

Adding 1/4(2n+1)^2 to both sides yields:
(n+1)^2 - (n+1)(2n+1) + 1/4(2n+1)^2 = n^2 - n(2n+1) + 1/4(2n+1)^2

This may be written:
[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2

Taking the square roots of both sides:
(n+1) - 1/2(2n+1) = n - 1/2(2n+1)

Add 1/2(2n+1) to both sides:
n+1 = n

[Alan Anderson]

Originally Posted by mechanicalbrain

[ (n+1) - 1/2(2n+1) ]^2 = [ n - 1/2(2n+1) ]^2

Taking the square roots of both sides:

I hate it when square roots go bad.

Seriously, there are times when you have to remember that there is usually more than one root for a given expression. For example, the square roots of 9 are 3 and -3.

If you evaluate the expressions inside the brackets, you find that they can be reduced to 1/2 on the left side and -1/2 on the right. Naively taking the square root of their squares loses the minus sign, and breaks the equality. A proper result at this step requires an arbitrary choice of the negative root for one of the sides of the equation.

[Dan Zollman]

Here's my argument:

If we go back to the elementary school definition of division, we can say that in division, the answer is equal to the size of each part when something is divided into a certain number of parts. In other words, if we have the expressions 12/3 = x, x is 4 because when 12 is divided into 3 parts, there is four in each part. Also, x is the number of times that 3 goes into four.

In another definition, when we have the equations x/y = z, then z is the number that you can multipy by y to get x. In other words, if 12/3 = z, then z = 4 because 4 is the number you can multiply 3 by to get 12.

Having those two definitions to use, we can look at the problem of dividing by zero. When we ask, "What is 5/0?" we do not come up with an answer because there is no number that you can mulitiply by 0 to get 5. This is the main reason why the rules of algebra do not allow you to divide by zero.

In the 1 = 2 proof above, it is true that one step divides by zero. However, in that step, the numerator is also zero. Let's look at that situation:
0/0 = ?
What number can we multiply by zero to get zero? Any number!
And this is why I say that 0/0 = 0/0 can give you 1 = 2.

Even though this proof is logically valid, the idea that 0/0 is not practical in Algebra. This is why the rules of Algebra say that you cannot divide by zero, even when the numerator is zero.The fact alone that 1 does not equal 2 is enough reason that it will ultimately be said that this proof (and others like it) are useless.

I still think it's a fun idea to play with.

Now, what about the second proof?

[Alan Anderson]

Originally Posted by worldbringer

Now, what about the second proof?

It's another example of roots gone bad:

...take the 4th root of 1...

There are four 4th roots of 1:
1
-1
i
-i
Raise any of these to the 4th power and you get 1.

The main thing to remember from this is that taking roots is not something you can safely do to both sides of an equation unless you can figure out which root is required in order to satisfy the result.

[Greg Ross]

Proof by intimidation:

• Horses have an even number of legs.
• Behind they have two legs, and in front they have fore-legs.
• This makes six legs (which is certainly an odd number of legs for a horse.)
• The only number that is both even and odd is infinity.
• Therefore, horses have an infinite number of legs.

Q.E.D.