# Another way of proving sqrt 2 is a Real number

I'm aware of the common way to prove that sqrt 2 is a real number but I'm just wondering would it suffice to prove it is a real number by proving it is an irrational number. As we know that irrational numbers are Real numbers.

• Hint: irrational numbers are in a subset of all real numbers... Commented Apr 25, 2017 at 18:36
• What is this common way to prove that $\sqrt 2$ is a real number ?
– lhf
Commented Apr 25, 2017 at 18:46
• What is your definition of real number? To prove that $\sqrt{2}\in\mathbb{R}$ is the same as proving that for some $r\in\mathbb{R}^+$ we have $r^2=2$, by the definition of the square root function. Commented Apr 25, 2017 at 18:46
• Since $\mathbb{R}$ is complete and $x\mapsto x^2$ is continuous and increasing on $\mathbb{R}^+$, we may deduce that $\sqrt{2}\in\mathbb{R}$ from $1^2<2<2^2$, for instance. Commented Apr 25, 2017 at 18:48

Given each of these constructions, you would have a different proof that $\sqrt{2}$ is a real number. For example, given the Dedekind cut construction, it suffices to show $(A,B)$, such that \begin{align}A &= \left\{a \in \mathbb{Q}: a^2 < 2 \text{ or } a < 0 \right\} \\ B &= \left\{b \in \mathbb{Q} : b^2 > 2 \text{ and } b >0\right\} \end{align}, is a cut and that it is the square root of $2$. Under the Cauchy Sequence construction, we would typically work with a sequence such as \begin{align} A_0 &= 1 \\ A_{n+1} &= \frac{1}{2}(A_n + 2/A_n).\end{align} At the end of the day, proving a number is a real number, depends on how you construct your real numbers.