Existence of numbers such as $\pi^{-1}$

Question

For my non-mathematics students (this particular class are computing), I would define $\displaystyle \frac{1}{n}$ for $n\in\mathbb{N}$ as the solution of the equation

$$nx=1,$$

and then

$$\frac{m}{n}=m\cdot \frac{1}{n}.$$

I like this because it defines fractions in terms of one, natural numbers and multiplication and it accounts for fraction arithmetic.

Sometimes I see the more practical student doesn't really go for my approach and says, e.g. I know what $\displaystyle \frac13$ is...

That is kind of off the point but while I accept his point, I would then ask myself what is, e.g. $\displaystyle \frac{1}{\pi}$ but, via the real number axioms, the solution of

$$\pi \cdot x=1.$$

My questions are,

1. To the 'constructionist' mathematicians... does $\displaystyle \frac{1}{\pi}$ exist?
2. How do you calculate it to an arbitrary number of digits?
3. Do you 'accept' $\pi\in\mathbb{R}$ but not $\displaystyle \frac{1}{\pi}$?

Feel free to substitute $\pi$ for your favourite irrational number (if you believe in them!).

I myself am perfectly happy with the number $\displaystyle \frac{1}{\pi}$ from the axioms of the real numbers but was interested in what other people thought.

Answer 1

Imagine we have a sequence of numbers: $$\{3,3.1,3.14,3.141,3.1415\dots\}$$ (In fact, I could use any increasing sequence that approaches $\pi$.)

$\pi$ could be defined as the smallest number $x$ that is greater than all of those numbers. That is, I could define $\pi$ to be the smallest number that is greater than $3$, $3.1$, and $3.14$, and $3.1415$, etc.

(Obviously, if we restrict ourselves to the rationals, such a number does not exist. But there is a property of the real numbers, saying that any increasing and bounded sequence has an supremum—that is, a smallest possible number that is greater than every term in the sequence. This sequence is obviously increasing. It's bounded because each term is less than, say, $4$.)

Now, look at this sequence of numbers: $$\left\{\frac13,\frac1{3.1},\frac1{3.14},\frac1{3.141},\frac1{3.1415}\dots\right\}$$ Each of those numbers "exist," because those are rational numbers (and we know how to take the reciprocal of a rational). Now, let me define $\frac1\pi$ to be the largest number that is smaller than each of those numbers.

(This is called the infimum. In the real number system, any decreasing and bounded-below sequence has one. This sequence is obviously decreasing. It's bounded-below because each term is greater than, say, $\frac14$.)

$\frac1\pi\approx.0.3183098862$.

Answer 2

The real numbers $\mathbb R$ are an example of a mathematical field. In other words, they satisfy a number of axioms (essentially, the basic laws of middle school mathematics (distributive property, commutative property of multiplication, etc...)). One of these field axioms is the following:

For each nonzero $x\in\mathbb R$, there exists a number ${1\over x}\in\mathbb R$ such that $x\cdot {1\over x} = 1$.

This is essentially your question: the real numbers are defined as a certain field, and since $\pi$ is a real number it is guaranteed to come with a real number $1\over\pi$.

Answer 3

If you accept the relationship between algebra and geometry, you might argue like this:

Every line $f(x) = a x$ with $a \ne 0$ has exactly one intersection with the line $g(x) = 1$.