Irrational Numbers

Theorem. $\sqrt2$ is not a rational number.

Proof (by contradiction). Suppose $\sqrt2=\dfrac mn$ for positive integers $m,n$ sharing no common factor greater than $1$. Squaring, $m^2=2n^2$, so $m^2$ is even, and hence $m$ itself is even (an odd number squared is odd). Write $m=2k$; then $4k^2=2n^2$, so $n^2=2k^2$, making $n$ even too. But then $m$ and $n$ are both even, contradicting that they share no common factor greater than $1$. Hence $\sqrt2$ cannot be rational. $\blacksquare$

There are points on the number line, like the one at distance $\sqrt2$ from the origin, that no rational number reaches. The real numbers that are not rational are called irrational numbers, and their set is written $Q'$. Every real number is rational or irrational but never both:

Since every terminating/periodic decimal is rational, an irrational number's decimal expansion is exactly the kind that neither terminates nor repeats. The real numbers $R$ can be pictured as filling every point of the number line, with $x<y$ meaning $x$ lies to the left of $y$. …