Paradox Extended and Clarification

JoËoL — Fri, 18 May 2012 14:32:19 +0000

A problem came up in the comments of the previous post ("A Funny Paradox") which I think could use some further explanation and will also help to explain the source and logic (which is not crazy logic if you can see it) behind the paradox in the linked post above. The problem is this:

Prove whether or not the following expression exists. If it does exist, find its value:

$\sqrt{7-\sqrt{7-\sqrt{7-\sqrt{\cdots}}}}.$

Assuming that an answer exists, it is easy to find a solution. We can let

$x=\sqrt{7-\sqrt{7-\sqrt{7-\sqrt{\cdots}}}}.$

Then by the nature of infinity, each part beginning with the square root is the exact same as the whole, therefore the equation becomes

$x=\sqrt{7-x}.$

So backwards-similar to the paradox in the previous post, we have made a finite equation out of an infinite expression, and this of course is easily solved as a quadratic, taking only the positive solution.

The problem comes in trying to show that it exists. I personally do not think it can be done, at least not in a real sense. My professor tried to prove that it does exist by starting with $\sqrt{7}$ then $\sqrt{7-\sqrt{7}}$ , and so on, showing that this converges as you repeat the process to infinity. And then my professor (a wonderful number theorist) concluded that this means the infinite expression exists. And then to solve it, it can be set up as before.

But my problem is that the seed of $\sqrt{7}$ is always there in the proof that it exists, while in fact there is no such seed, as it repeats to infinity. So in a counter-example, I can create a similar problem:

Prove whether or not the following expression has a solution, and if so solve it:

$1(1(1(1(\cdots)))).$

Then, to prove that it exists, start with

$1$ , then

$1(1)$ , and so on. This obviously converges as we repeat the process. Then to solve it, say

$x=1(1(1(1(\cdots)))).$

Then because of the nature of infinity, each part beginning with

$1$ is equal to the whole, thus

$x=1(x).$

Then

$x$ is any real number. Then the solution to the converging expression

$1(1(1(1(\cdots))))$ is

$x=1(1(1(1(\cdots))))=\text{ any real number }.$

And this is obviously nonsensical as we know very well that no matter how many times you multiply

$1$ by

$1$ you still have

$1$ . Then either our approach is wrong or

$1(1(1(1(\cdots))))=1=\text{ any real number }.$

Since the latter makes little mathematical sense, I would conclude that our approach is wrong. What approach can one take to solve the original problem? I have a feeling that for any approach, one could find a counterexample where the process does not hold. What do you think? Is it a poorly formulated problem, or is there in fact a solution?

A Funny Paradox

JoËoL — Thu, 19 Apr 2012 04:29:32 +0000

I have just discovered the funniest paradox. Well I think so at least.

Take

$a=3a$

$\Rightarrow a=3\underbrace{(3a)}_{=a}$

$a=3(3\underbrace{(3a)}_{=a})$

$a=3(3(3(3a)))$

$\vdots$

$a=\prod_{n=1}^{\infty}{3}$

According to our first instincts, $a$ should equal $0$ . So, either

$0=\prod_{n=1}^{\infty}{3}$

or the initial equation has two solutions

$a=0 \ \text{or} \ a=\prod_{n=1}^{\infty}{3}$

Lol.

*Note: For this example, we have used 3, but really any number would do the same thing (except anything between -1 and 1 whose infinite product would just converge to 0).

And we could extend this paradox further to say

$a=1a=1(1a)=1(1(1a))=\dots =\prod_{n=1}^{\infty}{1}=1$

$a=1$ . But basic math tells us that if

$a=1a$ then

$a$ can be anything, since anything multiplied by

$1$ is just itself. So

$a=1=\text{any real number}$