Infinity and solution of $x+2=x$

Question

I know that if we divide both sides by $x$ the equation becomes $1+\frac{2}{x}={1} \implies x=\frac{2}{0}$ which is undetermined.

According to wolfram alpha $\infty +2=\infty$ , logically speaking shouldn't we say that $x= \infty$?

Answer 1

Since $\infty$ is not a real number, it becomes rather difficult to do arithmetic with it. In fact, treating it as a real number means that we have to sacrifice some other property of arithmetic (for example, treating it as a real number leads to the statement that $2 = 0$, which we certainly don't want to have). Hence, we would simply say that no solution exists.

Answer 2

It depends on what kind of number $x$ is. If $x$ is a surreal number, ordinal number, or real number, then there is no solution. But if $x$ is a cardinal number, there are solutions such as $x = \aleph_k = \aleph_k + 2$ for any ordinal $k$. Although there are no solutions of $x = x + 2$ in ordinal numbers, there are solutions of $x = 2 + x$, for example $x = \omega = 2 + \omega$. $\aleph_k$ is an infinite cardinal number, and $\omega$ is an infinite ordinal number, and both are different types of infinity.