বাস্তব সংখ্যা

This is a loose introduction to what the real numbers are. We denote this set by $\mathbb R$.

Irrationality of $\sqrt 2$

First, let's check that the rationals have "gaps". Intuitively, we know there exists a positive "number" $x$ such that $x^2=2$. But we can prove that it is impossible to express $x$ as a rational number $\tfrac mn$.

Indeed, suppose $m^2 = 2n^2$ for positive integers $m$ and $n$.

We know that prime factorizations are unique, so this is impossible. Moreover, with trivial modifications we can also prove that $\sqrt 3$, $\sqrt 5$, and so on are irrational.

Dedekind Cuts

Now we know that our rationals have "holes" in them. So we need some way of extending the rationals to the set $\mathbb R$. However, we need a way to do so formally in terms of only $\mathbb Q$.

Here's one way to do so. Imagine we have the real number line, of which the rationals (which we understand) are a dense subset. For each point on the line, we can make a cut -- some numbers fall to the left of the cut and some numbers fall to the right. The idea is to associate each real number with a position we can place the knife. This gives us a way to use $\mathbb Q$ to define $\mathbb R$.

Take a moment to think about what kind of definition this might lead to, then click Continue.

To formalize this, we use a Dedekind cut.

A Dedekind cut $A | B$ consists of two sets $A$ and $B$ which satisfies the following conditions:

(i) $A$ and $B$ are nonempty sets whose union is $\mathbb Q$.

(ii) $A$ has no largest element.

(iii) Whenever $a \in A$ and $b \in B$, $a < b$. In particular, $A$ and $B$ are disjoint.

Then we define a real number to be a Dedekind cut; hence $\mathbb R$ is the set of Dedekind cuts.

Now let's see if we can define the square root of $2$! One definition we might try is $$A = \{ x \mid x \in \mathbb Q, x^2 < 2 \}$$ and $$B = \{ x \mid x \in \mathbb Q, x^2 \geq 2 \}.$$

Fortunately, this isn't hard to fix. Just add the negative numbers to $A$ and remove them from $B$.

Now we have a way to define $\sqrt 2$ using only rational numbers!

Here's another question to think about.

Exactly -- it's precisely the rational numbers.

Bounds and Inequalities in $\mathbb R$

It takes a lot more work to show that you can add, subtract, etc. two real numbers, and this explanation is already getting long. However, using the definition above we can derive a very important property of $\mathbb R$.

First, we need to define what it means for $x < y$ in $\mathbb R$.

This gives us one of the most important properties of $\mathbb R$, the least upper bound property.

Let $S$ be a subset of $\mathbb R$. Let $\mathcal B$ be the set of real numbers $M$ with the property that for any $s \in S$ we have $s \le M$. Then if $\mathcal B$ is nonempty, it has a smallest element.

The real numbers in $\mathcal B$ are called upper bounds. Essentially this says if $S$ is bounded above, then there is a least such bound. When this least upper bound exists, it is called the supremum for the set. Note that the supremum for $S$ need not belong to it!

The answer remains the same when $S = [0,3]$. In both cases, the least upper bound is $3$.

Finally, a brief sketch of the proof of our theorem -- basically, we take all the cuts in $\mathcal B$ and consider the union of their left halves. Set this equal to $X$, and let $Y = \mathbb Q \setminus X$. One can then verify that $b = X | Y$ is a cut in $\mathcal B$, and moreover any cuts less than $b$ are not bounds.