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? .

Welcome to Mathematics Stack Exchange. Infinity is not a real number — J. W. Tanner, Jul 11 '19 at 16:31
$a/a=1$ and $a-a=0$ hold when $a$ is a real number. $\infty$ is not a real number. The first only when $a\neq 0.$ — Thomas Andrews, Jul 11 '19 at 16:32
Your first two expressions do not make sense because division and substraction are defined only for numbers (you can say real or complex numbers). Infinity is not a number. You might want to read more about limits, or watch some videos too. — evaristegd, Jul 11 '19 at 16:33

score 1 · Answer 1 · answered Jul 11 '19 at 16:44

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.

Mohammad Riazi-Kermani · Answer 2 · 2019-07-11T17:35:26.290

0

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.

edited Jul 11 '19 at 17:35

answered Jul 11 '19 at 16:44

Mohammad Riazi-Kermani

68,728

Addition is only (usually) defined for complex or real numbers. I think you should specify that the"addition" you're using there is not the same addition we use with numbers. – evaristegd Jul 11 '19 at 16:50
Good point. I edited my answer. – Mohammad Riazi-Kermani Jul 11 '19 at 17:36

score 0 · Answer 3 · answered Jul 11 '19 at 17:42

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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 $0=a-a$

answered Jul 11 '19 at 17:42

G Cab

35,272

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

3 Answers3