Real Numbers: Irrational Numbers and Decimal Expansions

Proving a number is irrational

The standard technique is proof by contradiction: assume the number is rational (equal to some fraction $\frac{a}{b}$ of integers), then show that assumption forces something impossible.

Prove that $3 + 2\sqrt{5}$ is irrational.

Solution: Suppose $3+2\sqrt{5}$ were rational, equal to $\frac{a}{b}$ for integers $a,b$ (with $b \ne 0$). Then:

The right side is a ratio of integers, so it's rational, but $\sqrt{5}$ is known to be irrational. That contradiction means the original assumption was false, so $3+2\sqrt{5}$ is irrational.

Spotting a terminating decimal

A fraction in lowest terms has a terminatingdecimal expansion exactly when its denominator's prime factorization contains only 2s and 5s. Otherwise the decimal is non-terminating and repeating.

Without doing the division, decide whether $\dfrac{15}{1600}$ and $\dfrac{29}{343}$ terminate.

Solution: $1600 = 2^6 \times 5^2$, only 2s and 5s, so $\dfrac{15}{1600}$ terminates (it equals $0.009375$). But $343 = 7^3$, which has a prime factor other than 2 or 5, so $\dfrac{29}{343}$ is non-terminating and repeating.