How to Prove That √2 Is the Supremum of {x ∈ ℝ : x² < 2}

Let

$$S=\{x\in\mathbb{R}:x^2<2\}.$$

To prove that $\sqrt{2}$ is the supremum of $S$, you must show two things: $\sqrt{2}$ is an upper bound, and no smaller number is an upper bound. This second part is what establishes that the bound is the least one.

1. Show that √2 is an upper bound

If $x\in S$, then $x^2<2$. Equivalently, $|x|<\sqrt{2}$, so in particular $x<\sqrt{2}$. Therefore every element of $S$ is less than $\sqrt{2}$, and $\sqrt{2}$ is an upper bound.

In fact, the inequality also gives $x>-\sqrt{2}$. Thus $S=(-\sqrt{2},\sqrt{2})$. The endpoints are not included because their squares equal $2$, not a number strictly less than $2$.

2. Show that no smaller number is an upper bound

Take any real number $b<\sqrt{2}$. We will find an element of $S$ that is greater than $b$.

• If $b<0$, choose $x=0$. Then $x^2=0<2$, so $x\in S$, and $x=0>b$.
• If $0\le b<\sqrt{2}$, choose the midpoint between $b$ and $\sqrt{2}$:

$$x=\frac{b+\sqrt{2}}{2}.$$

Since $b<\sqrt{2}$, this midpoint satisfies $b<x<\sqrt{2}$. It is positive, so $x^2< (\sqrt{2})^2=2$. Hence $x\in S$ and $x>b$.

In either case, every number below $\sqrt{2}$ fails to be an upper bound for $S$. This proves that

$$\sup S=\sqrt{2}.$$

A concrete check

For example, take a proposed upper bound $b=1.4$. Since $0\le1.4<\sqrt{2}$, the midpoint construction gives

$$x=\frac{1.4+\sqrt{2}}{2}\approx1.4071.$$

This number is greater than $1.4$ but less than $\sqrt{2}$, so $x^2<2$. Thus $1.4$ is not an upper bound. The same construction works for every nonnegative $b<\sqrt{2}$, not just for this example.

Why the supremum is not a maximum

A maximum must be an element of the set. But $\sqrt{2}\notin S$, because $(\sqrt{2})^2=2$, and the defining condition is the strict inequality $x^2<2$. So $S$ has no maximum even though its supremum is $\sqrt{2}$. The supremum can be approached as closely as desired by elements of $S$, without belonging to the set.

Common mistakes

• Checking only that $\sqrt{2}$ is an upper bound. An upper bound need not be the least upper bound; you must also rule out every smaller upper bound.
• Claiming that the endpoint is in the set. The strict inequality excludes both $\sqrt{2}$ and $-\sqrt{2}$.
• Ignoring negative proposed bounds. The midpoint argument above is intended for $b\ge0$. For $b<0$, the simpler witness $x=0$ works.
• Treating $x^2<2$ as $x<\sqrt{2}$ without considering negative values. The full equivalent condition is $-\sqrt{2}<x<\sqrt{2}$. The upper-bound argument uses only its right-hand side, but the two-sided interval describes the whole set.