Irrational numbers

Two kinds of decimals, two kinds of numbers. A rational number $\dfrac{p}{q}$ ($p, q$ integers, $q \neq 0$) always produces a decimal that either terminates (like $\dfrac14 = 0.25$) or is non-terminating but recurring, repeating some block of digits forever (like $\dfrac13 = 0.333\ldots$ or $\dfrac{22}{7} = 3.\overline{142857}$). An irrational number is the opposite of both: a real number whose decimal expansion is non-terminating and non-recurring — it never ends, and no block of digits ever repeats indefinitely. Familiar examples:

where: each right-hand side is a decimal that continues forever with no repeating pattern, which is precisely the test for irrationality.

Rationals and irrationals share the number line. Every point on the number line stands for exactly one real number, and every real number — rational or irrational — has exactly one point. Together the two sets fill the line with no gaps, forming the set of real numbers. This matters because it tells you irrational numbers are not some abstract leftover; $\sqrt2$ sits at a definite, locatable spot between $1.4$ and $1.5$.

Combining rationals with irrationals. If $a$ is rational (and nonzero where needed) and $b$ is irrational, then $a+b$, $a-b$, $ab$, and $\dfrac{a}{b}$ are all irrational. But combining two irrationals can land back on a rational number — for instance $(2+\sqrt3)+(2-\sqrt3) = 4$, and $\sqrt2 \times \sqrt2 = 2$. So "irrational plus irrational" is not automatically irrational; only "rational (nonzero) combined with irrational" is guaranteed irrational.

Locating irrationals geometrically. Since $\sqrt2$ is the length of the diagonal of a unit square ($1^2+1^2=2$), a right triangle with both legs of length $1$ has a hypotenuse of exactly $\sqrt2$. Swinging that hypotenuse onto the number line with a compass marks $\sqrt2$'s exact position — no measuring tape, no approximation.

Why "non-terminating and non-recurring" is the right test. It helps to see why every rational number must terminate or recur in the first place. Long division of $p$ by $q$ can only ever produce remainders from the list $0, 1, 2, \ldots, q-1$ — finitely many possibilities. If a remainder $0$ ever turns up, the division stops and the decimal terminates. If it never turns up, some remainder must eventually repeat (there are only $q-1$ nonzero remainders to choose from), and once a remainder repeats, the digits that follow it repeat too, forever. An irrational number's decimal never falls into either trap: no remainder cycle exists because there is no fraction $\dfrac{p}{q}$ generating it at all. This is the deeper reason the decimal test works, not just a rule to memorise.

Classify each number as rational or irrational, with a reason: (a) $0.10110111011110\ldots$, (b) $\sqrt9$, (c) $2+\sqrt5$, (d) $\dfrac{22}{7}$.

Step 1. Check (a). The decimal $0.10110111011110\ldots$ is built with a growing pattern (one more $1$ each block), but the digits themselves never settle into a fixed repeating block — it is non-terminating and non-recurring. So it is irrational.

Step 2. Check (b). $\sqrt9 = 3 = \dfrac31$, a ratio of integers, so it is rational. This confirms that the square root of a perfect square is always rational, not automatically irrational.

Step 3. Check (c). Here $2$ is rational and $\sqrt5$ is irrational; since a nonzero rational combined with an irrational (by addition here) is always irrational, $2+\sqrt5$ is irrational — its decimal, approximately $4.2360679\ldots$, never terminates or repeats.

Step 4. Check (d). $\dfrac{22}{7} = 3.142857142857\ldots$ is non-terminating, but the block "$142857$" recurs forever, so it is a repeating decimal and hence rational. Note this is only an approximation to $\pi$; $\pi$ itself is irrational.

Answer: (a) irrational, (b) rational, (c) irrational, (d) rational.

Competitive stretch

Locate $\sqrt3$ on the number line, starting from the point $P$ representing $\sqrt2$ found in the derivation above.

Step 1. At $P$ (which is at distance $\sqrt2$ from $O$), draw a segment $PQ$ of length $1$ unit, perpendicular to the number line.

Step 2. Join $O$ to $Q$. Since $OP = \sqrt2$ and $PQ = 1$ with a right angle at $P$, the Pythagorean relation gives $OQ^2 = OP^2 + PQ^2 = (\sqrt2)^2 + 1^2 = 2+1 = 3$, so $OQ = \sqrt3$.

Step 3. With centre $O$ and radius $OQ = \sqrt3$, draw an arc cutting the number line at a point $R$. Then $R$ represents $\sqrt3$ exactly. Answer: repeating the same right-triangle idea one level up, starting from the already-located $\sqrt2$, pins down $\sqrt3$ on the number line.

• Thinking $\dfrac{22}{7}$ is $\pi$ — it is a rational approximation; the true value of $\pi$ is irrational and never equals $\dfrac{22}{7}$ exactly.
• Judging irrationality by how "complicated" a decimal looks. The only real test is whether the digits eventually repeat (rational) or never settle into a repeating block (irrational), regardless of how many digits are shown.
• Assuming every square root is irrational. Only non-perfect squares like $\sqrt2, \sqrt3, \sqrt5$ are irrational; $\sqrt4, \sqrt9, \sqrt{16}$ equal $2, 3, 4$ and are rational.
• Concluding two irrationals must combine to an irrational. Counterexamples: $\sqrt2 \times \sqrt2 = 2$ and $(3+\sqrt2)+(3-\sqrt2)=6$, both rational.

Ask the tutor: Want to see why the compass-and-arc trick works for $\sqrt5$ or $\sqrt7$ too, or practise classifying more decimals as rational or irrational?