Q.Prove that $\sqrt3$ is an irrational number. (Hint: Follow the method used to prove $\sqrt2\notin Q$.)

The real numbers are built up in stages, each solving a problem the previous set could not: $N=\{1,2,3,\ldots\}$ (counting) $\subset W=\{0,1,2,\ldots\}$ (adds zero) $\subset Z=\{\ldots,-1,0,1,\ldots\}$ (adds negatives, for debts) $\subset Q=\{m/n:m,n\in Z,n\ne0\}$ (adds ratios, for even division) $\subset R$ (fills in every remaining point of the number line, including the irrationals).

A rational number's decimal expansion always terminates or eventually repeats; conversely, any terminating or repeating decimal is rational. Numbers whose decimals neither terminate nor repeat are irrational, denoted $Q'$, with $R=Q\cup Q'$ and $Q\cap Q'=\varnothing$.

Proving a number irrational (by contradiction). To show $\sqrt2\notin Q$: assume $\sqrt2=m/n$ in lowest terms; squaring gives $m^2=2n^2$, so $m$ is even; writing $m=2k$ gives $n^2=2k^2$, so $n$ is even too -- contradicting that $m,n$ share no common factor. The identical technique proves $\sqrt3$, $\sqrt5$, $\pi$, etc. irrational.

Density. Between any two distinct rationals there is always another rational (e.g. their average); yet the rationals still leave gaps on the number line -- the diagonal of a unit square, $\sqrt2$, is a point no rational reaches.

Combining irrationals. Sums, differences, and products of irrational numbers are NOT automatically irrational: $(1+\sqrt2)-\sqrt2=1\in Q$, and $\sqrt2\times\sqrt8=4\in Q$ -- the irrational parts can cancel.