Does this expression represent the largest real number?

Question

I'm not very good at this, so hopefully I'm not making a silly mistake here...

Assuming that $\infty$ is larger than any real number, we can then assume that:

$\dfrac{1}{\infty}$ is the smallest possible positive real number.

It then stands to reason that anything less than $\infty$ is a real number. Therefore, if we take the smallest possible quantity from $\infty$, we end up with:

$\infty-\dfrac{1}{\infty}$

Does that expression represent the largest real number? If not, what am I doing wrong?

Answer 1

You are looking at the extended real number system. In particular, we adjoin $+ \infty$ and $- \infty$ to $\mathbb{R}$. So in the context of the extended real numbers, $+\infty- \frac{1}{+ \infty} = + \infty$.

Answer 2

Since $\infty$ is not a real number, you cannot assume that $\dfrac{1}{\infty}$ is a meaningful statement. It is not a real number.

You might want to investigate non-standard, hyperreal and surreal numbers and infinitesimals.

Answer 3

George, the symbol you have written $\infty$ is not a real number. It is a concept which we use (when we are dealing with numbers) to indicate something is unbounded.

The real numbers are unbounded so there is no largest or smallest real number.

Answer 4

Another way to think of infinity is as follows. One way to construct the real numbers from the rationals is via Cauchy sequences, i.e. rational numbers "converging" (in effect) to the real number. For example, what is the meaning of $\pi$? Given a list of digits $3.14159\ldots$, we may think of $\pi$ as a sequence of rational converging to $\pi$, for example $$3, 3.1, 3.14, 3.141, 3.1415, 3.14159, \ldots$$

We can do the same with infinity! Let's an infinite number as a sequence diverging to infinity. One example is $$1,2,3,4,5,6,\ldots$$ We can then define all the usual arithmetic operations as usual, for example $$1,2,3,4,5,6,\ldots + 1,2,3,4,5,6,\ldots = 2,4,6,8,10,12,\ldots$$ There is now a very concrete meaning of "the limit of a function $f$ at $1,2,3,4,5,6$" - that is just the limit of the sequence $f(1),f(2),f(3),\ldots$. If a function converges to some value at $\infty$, then it will converge to the same value under all divergent sequence. Compare this to the definition of limit at a point $x$ - we should get the same limit whatever the sequence converging to $x$ is.

Things become more complicated when we try to understand when $a < b$ for two infinite numbers. If $a,b$ are finite and represented by some sequences $a_i,b_i$ converging to them, then if $a<b$ we know that eventually $a_i < b_i$. However, that is not the case with infinite sequences - consider for example $1,2,3,4,5,6,\ldots$ vs. $0,4,0,8,0,12,\ldots$. This can be fixed using some "decision rule" implemented using an ultrafilter, which I'm not going to explain here.

Having done this, we have a full fledged number system, and for each infinite "number" $\alpha$ we can form a corresponding infinitesimal $1/\alpha$ which is positive but smaller than any "real" number (i.e. smaller than any constant sequence $x,x,\ldots$ where $x$ is real).

If you find this interesting, look up nonstandard analysis.

Answer 5

You could use the same reasoning to say that since $1$ is a real number then $\infty - 1$ has to be a real number. So you see that if you subtract anything from $\infty$ it does NOT become a real number. Actually $\lim_{x \to \infty} x-1 = \infty$.

To visualize this: Given that you have infinite money and spend 1 dollar. How much money have you left, yes its still infinite. This is a bit hard to understand when starting to deal with infinity, you can for example split your money up into two parts and both parts will remain infinite.

You see that infinity cannot be made finite by subtracting some finite number.