Does infinite equal infinite? I have a question.
Let $x$ be infinite.
$$2x=\infty\times2, \quad 2x=\infty$$
So actually, does $2x=x$?
 A: Start with a set, let's take $\mathbb Z$. Its elements are integers and on it there is a multiplication. That means that every pair of integers is connected with a integer wich we call the product. E.g $(2,5)$  has product $10$ . You can add 'infinity' (whatever it is) to this set and denote the result by $\mathbb{Z}\cup\left\{ \infty\right\}$. If you want to extend the multiplication then it must be 'decided' what the product is for pairs like $(2,\infty)$ and $(\infty,\infty)$. These decisions/conventions must be taken in such a way that the rules of multiplication (e.g. $x\times y=y\times x$) remain valid as much as possible. Quite a job! Your intuition says that for $(2,\infty)$ it is a good thing to choose $\infty$ as product. That confirms to me that your intuition is to be respected. And remember: intuition is very important in mathematics!
A: In every commonly used number system that has a number called $\infty$, it is indeed true that $2 \infty = \infty$ -- e.g. the Riemann sphere, the extended real line, or the cardinal numbers.
However, there are various number systems -- e.g. the hyperreal numbers or the ordinal numbers -- that have infinite numbers that do not satisfy this property. Note that we usually never use the symbol $\infty$ when referring to an infinite number in these number systems.
(the principal exception I know of is the extended hyperreal line, which has many infinite numbers obeying the 'usual' laws of arithmetic, and a pair of additional numbers we call $+\infty$ and $-\infty$ that have the largest magnitude of all infinite numbers, and do not obey the 'usual' laws of arithmetic)
The answer to your question of whether $2x = x$ when $x$ is infinite, thus, depends very much on what number system you're using.
Examples include: if you're studying calculus of real variables, you're probably using the extended real line; if you're quantifying the number of elements in a collection, you're probably using the cardinal numbers.
A: As drhab stated in his answer, your intuition tells you that $\infty \times 2$ should be $\infty$. But that intuition depends on what you understood $\infty$ to mean. A very 'layman' definition could go something like "a quantity with larger magnitude than any finite number", where "finite" = "has a smaller magnitude than some positive integer". Clearly then $\infty \times 2$ also has larger magnitude than any finite number, and so according to this definition it is also $\infty$. But this definition also shows us why, given that $2x=x$ and that $x$ is non-zero but may be $\infty$, we cannot divide both sides by $x$. It is akin to asking, if John runs twice as fast as Jack and both run off away from me, can I divide John's final position by Jack's final position, which are both further away from me than I can ever go, and get $2=1$? (Of course neither John nor Jack themselves can reach their "final position", but the process by which they 'approach' it explains the situation quite well.)
A: This depends on what definition for "infinite" you compare.
For instance, if we generalize the Laplace transform to divergent integrals, we will see that
$$\int_0^\infty 1 dx=\int_0^\infty\frac1{x^2}\ne\int_0^\infty x \,dx$$
All three integrals are divergent and infinite and have the regularized value zero, but two of them are equal but not equal to the third one.
In other cases of divergent integrals or series, the regularized value and/or growth rate (germ at infinity) or behavior at a singularity can differ as well or the differences can compensate for each other as in the example above.
So no, at least in terms of divergent integrals and series, the infinities may differ.
A: Yes. If you work with cardinals, you can show that if two cardinals are finite (i.e. integers), then the product rule is as usual ; if you have two positive cardinals that are infinite, say $\kappa, \lambda$, then 
$$
\kappa + \lambda = \kappa \lambda = \max \{ \kappa, \lambda\}.
$$
In particular, if $\kappa$ is infinite, then $\kappa + \kappa = \max \{\kappa,\kappa\} = \kappa$. 
See http://en.wikipedia.org/wiki/Cardinal_number#Cardinal_addition to understand them a bit better.
Hope that helps,
A: It depends on how you use the term "infinite". If you speak in terms of cardinal numbers (for counting of objects), then yes, they're the same infinity. This is because any countable set containing an infinite number of objects can be counted in a way to have exactly the same number of objects. For instance, the set of all integers is clearly twice as big as the set of all even integers... and yet, if you just multiply the set of all integers by 2, you get the set of all even integers, thus showing that there's just as many even integers as integers.
If, on the other hand, you're using it as a number, you cannot evaluate it, as infinity is not a number.
On a third hand, if you think in terms of limits, then they are not equal. That is,
$$
\lim_{x\to\infty} \frac{x}{2x} = \frac12
$$
and thus the limits are not the same.
A: From the question itself, arises the fact that you are looking at it at an intuitive way. The answer to your intuitive question "is 2x = x?" then is YES. But notice that x is an "infinite number", and so, saying that two times infinite is infinite is not a big deal.
Anyway, as you pointed out, in maths there are indeed frames in which 2x = x...
From my point of view, the infinite question has not been solved. We still don't know WHAT infinite actually is, although we have described many of its properties pretty well. Maybe, if we start researching outside the boundaries of the stablished axioms we may get to some conclutions...
A: You can't do this, as I have learned.  Take all numbers between 0,1 and compare it to all numbers between 0,2.  We assume there are double the amount of numbers between 0,2 compared to 0,1, but its not that simple.
Imagine all numbers between 0,1 that only go out one decimal place.$$0.0,0.1,0.2,0.3,\cdots1.0$$Now double them.$$0.0,0.2,0.4,0.6,\cdots2.0$$We can clearly see this is only half of the possible one digit values between 0,2.
However, let us add the next digit.
Then we can see that I can indeed reach all of your values between 0 and 2, but you will argue that now I am missing some of the values in the 2nd decimal place.
So we go out another decimal, and another, in fact, we go infinitely.
There will always be a number $n$ that when doubled, equals a number $m$, where $n$ is between 0,1 and $m$ is between 0,2.
However, there is never a number $m$ where there is not some $n$ I can double to reach your $m$.
Therefore, I have argued that there is, indeed, the same amount of numbers between $0,1$ and $0,2$.  Measuring infinity like this is called the Cardinality of a set.
However, I will say that there is both an equivalent amount of numbers between 0,1 and 0,2 and double the amount of numbers between them, at the same time.
A: There is a solution to $2x=x$. It's $x=0$. That being said, where does infinity come from? If it's the result of adding one to itself forever, $1+1+1+...$, then if we say $f = 1+1+...$, it seems intuitive that if we define $g$ as $2+2+...$, then if we divide $g$ by $f$, then the result will be 2. To be more precise, if $f(x) = x$ and $g(x)=2x$, then it seems obvious that for all values of x, even $∞$, that $g(x)/f(x) = 2$, and $f(x)/g(x) = 1/2$. In calculus, with limits one might say that $1/2$ and $2$ are the limits of f(x)/g(x) and g(x)/f(x), respectively, as $x → ∞$ or 'as x increases without bound'. Some say this means there are different sizes of infinities, or that some infinities or functions grow at different rates. If you think a system which handles infinity well has certain properties, then find one which has those properties. If you think systems should say infinity doesn't exist, and questions concerning it are nonsense, impossible, or undefined, then find one. Many types of systems exist (however much the adherents of a given system may deny it). 
If we say that 'infinite' is the property that a number is greater than any finite number, then since doubling a number does not decrease its value, if the function $i(x)$ takes x and returns $1$ if x is infinite, and $0$ otherwise, then for any number $n$ such that $i(n) = 1$, $i(2n)$ and $i(n+1)$ is also $1$.
