May I Assign $\infty$ as a Value to a Variable?

Question

Now first something that I already know; \begin{eqnarray} ∞/ ∞ = undetermined ( ≠1 ) \\ ∞- ∞ = undetermined (≠0)\\ \end{eqnarray}

So basically one reason for this is that the $∞$ I assume is not as same as the $∞$ someone else will assume as $ ∞$ is a very large number with no definite value.....but what if I assign the $ ∞$ to a certain variable....that way the infinity is always same.

For eg:

What if I assign $ a=∞$;

Now infinity is always the same if I use $a $ instead of directly using $∞$......so my question is are the same laws mentioned above applicable here.....or can i solve it like solving any other equation; \begin{eqnarray} a/a = 1 \\ a-a = 0\\ \end{eqnarray} Or are these still undetermined? .

Answer 1

Strange things can happen when considering the infinite.

Consider the infinite set $\lbrace 1, 2, 3, \ldots \rbrace$. If I multiply all the elements of this set by $2$, I get $\lbrace 2, 4, 6, \ldots \rbrace$. Have I added or taken anything away? No. Therefore, both of these sets must contain exactly the same number of elements even though it appears that every element in the second set is in the first with infinitely many taken away.

On top of all this, we recognize that not all "infinities" are the same.

There are infinitely many infinities each of different cardinalities (measures of the number of elements in the underlying set).

So, when we assign $a$ a value, it must be a finite one.

Answer 2

Infinity is not a real number but you can do some algebra with infinity.

For example we define $$ \lambda +\infty =\infty$$for any real $\lambda$

$$ \infty +\infty=\infty$$ $$\lambda \times \infty =\infty $$ for positive $\lambda$

$$\infty \times \infty =\infty$$

But there are some undefined expressions such as $$\infty -\infty$$ or $$\frac {\infty}{\infty}$$ which do not follow the usual algebraic rules.

There is no way around it and we have to deal with them case by case.

Answer 3

As the other answers explained $\infty$ is not a number. Already Galileo had to admit that.

However, it is true that $1=a/a$ is valid for all $0 \ne a$, including the limit for $a \to \pm \infty$.
And same for $0=a-a$