MISSION · Number sense
Rational and Irrational Numbers
Why some decimals stop, some repeat forever, and some never settle down
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The idea in one sentence
A rational number is one you can write as a fraction of two whole numbers; an irrational number is a real number that no fraction can ever reach.
The dividing line
\[\text{rational} = \frac{a}{b}, \quad a, b \text{ integers}, \ b \neq 0\]That definition sweeps in more than it first appears. Every whole number is rational (7 = 7/1). Every terminating decimal is rational (0.25 = 1/4). Every repeating decimal is rational too, which is the part that surprises people.
Irrational numbers are what’s left over: \(\sqrt{2}\), \(\sqrt{7}\), \(\pi\), \(e\). Put the two groups together and you have the real numbers — every point on the number line.
Decimals give it away
Divide \(a\) by \(b\) by long division and there are only \(b\) possible remainders. So either a remainder of 0 shows up and the decimal stops, or a remainder repeats and the whole block of digits from then on repeats forever. There is no third outcome.
\[\frac{3}{8} = 0.375 \qquad \frac{3}{11} = 0.272727\ldots\]So the test works both ways: a decimal that terminates or repeats is rational, and a decimal that goes on forever without ever repeating is irrational.
That’s what \(\pi = 3.14159265\ldots\) is doing. Not a long pattern — no pattern.
Turning a repeating decimal into a fraction
Let \(x = 0.4444\ldots\) Multiply by 10 to shift one repeat along:
\[10x = 4.4444\ldots\]Subtract the first line from the second. The infinite tails are identical, so they cancel exactly:
\[9x = 4 \quad \Rightarrow \quad x = \frac{4}{9}\]For a two-digit repeat, multiply by 100 instead: \(0.\overline{27}\) gives \(99x = 27\), so \(x = 3/11\). The method is why every repeating decimal is rational — it always works.
Which roots are irrational
\(\sqrt{9} = 3\) is perfectly rational. A square root is irrational exactly when the number under it is not a perfect square, so \(\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}\) are irrational while \(\sqrt{4}, \sqrt{25}, \sqrt{1.44}\) are not.
The proof for \(\sqrt{2}\) is short and worth seeing. Suppose it equalled \(a/b\) in lowest terms. Squaring gives \(a^2 = 2b^2\), so \(a^2\) is even, so \(a\) is even, so \(a^2\) is a multiple of 4, so \(b^2\) is even, so \(b\) is even too — and a fraction in lowest terms cannot have both parts even. The assumption breaks itself.
Arithmetic mixes them in odd ways
Rational plus rational is always rational. But:
- rational + irrational is always irrational (\(1 + \sqrt{2}\))
- irrational × irrational can be either (\(\sqrt{2} \times \sqrt{2} = 2\))
- irrational + irrational can be either (\(\sqrt{2} + (-\sqrt{2}) = 0\))
So “irrational” is not a property that survives arithmetic the way “even” does.
Common mix-ups
- Calling every root irrational. Check whether it’s a perfect square first.
- Thinking irrational means “goes on forever”. 1/3 goes on forever and is rational. The test is whether it repeats.
- Treating 22/7 as π. It’s a rational approximation, and it isn’t even a very good one past the third decimal.
- Believing your calculator. A screen showing 1.414213562 has stopped, but the number hasn’t.
Check yourself
Write 0.4444… as a fraction, then decide which of √7, √9, 0.25 and 3/11 is irrational.
Ten times it minus itself gives 9x = 4, so x = 4/9. And √7 is the irrational one: 9 is a perfect square, 0.25 terminates, and 3/11 repeats.
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