Numbers That Are Not Rational
The rational numbers include every integer and every ratio of integers with a nonzero denominator. They nevertheless do not account for all the familiar locations on a number line. For example, the diagonal of a square with side length \(1\) has length \(\sqrt{2}\), a number that cannot be represented as a ratio of integers. Numbers of this kind are called irrational.
In this tutorial, we use the familiar number line as an ambient setting: rational numbers sit on it, and some of its other numbers are irrational. The formal construction and properties of the real number system come next. Our main task here is to learn what it means for a number to be irrational and how to prove that particular numbers have this property.
This definition makes irrationality a statement about whether a fraction representation exists. To prove that a number is irrational, we can suppose that such a representation exists and derive a contradiction. For square roots, the key is often to use the assumed fraction in an equation, then show that its numerator and denominator must share a factor. A carefully chosen representation with the least possible positive denominator makes that contradiction precise.
A Least-Denominator Argument
Every rational number has at least one representation \(a/b\) with integers \(a,b\) and \(b>0\): if the original denominator is negative, change the signs of both entries. A rational number may have many such representations. When needed, we can choose one with the smallest positive denominator. This choice is justified by the well-ordering property of the positive integers: every nonempty set of positive integers has a least member.
Proof. Suppose, for a contradiction, that \(\sqrt{2}\) is rational. Among all representations \(\sqrt{2}=a/b\) with integers \(a,b\) and \(b>0\), choose one with the least possible denominator \(b\). Squaring the equality gives
The right side is even, so \(a^2\) is even. An odd integer has the form \(2k+1\), and its square is
which is odd. Therefore \(a\) cannot be odd; it must be even. Write \(a=2c\) for an integer \(c\). Substituting this into the equation gives \(4c^2=2b^2\), so \(b^2=2c^2\). The same parity argument shows that \(b\) is even. Write \(b=2d\), where \(d\) is a positive integer. The original fraction then satisfies \(a/b=2c/(2d)=c/d\). Thus \(\sqrt{2}=c/d\), but \(d=b/2\) is a positive integer strictly smaller than \(b\). This contradicts the choice of \(b\) as the least positive denominator. The supposition was false, so \(\sqrt{2}\) is irrational. \(\square\)
The contradiction does not come merely from finding an even numerator or an even denominator. It comes from finding both: the fraction can then be rewritten with a smaller positive denominator while representing the same number. The least-denominator choice rules out that possibility.
Worked Example: Proving That the Square Root of 2 Is Not a Fraction
The proof can be tracked through the equations. If \(\sqrt{2}=a/b\) with \(b>0\), then \(a^2=2b^2\). For instance, an odd candidate numerator would have the form \(a=2k+1\), and then
which is odd. But \(2b^2\) is even, so the equation forces \(a\) to be even. Setting \(a=2c\) gives \(4c^2=2b^2\), hence \(b^2=2c^2\), which forces \(b\) to be even as well. If the chosen representation has \(a=2c\) and \(b=2d\), then \(a/b=c/d\) and \(0<d<b\). Thus any proposed representation leads to another with a smaller positive denominator. There can be no representation to begin with.
Using Divisibility to Prove Irrationality
The same strategy works when a prime number other than \(2\) appears in a square equation. For \(\sqrt{3}\), the relevant fact is that if \(3\) divides the square of an integer, then \(3\) divides the integer itself. To see this, every integer has remainder \(0\), \(1\), or \(2\) upon division by \(3\). The squares of these remainders are \(0\), \(1\), and \(4\); modulo \(3\), these are \(0\), \(1\), and \(1\). A square is therefore divisible by \(3\) only when its base is divisible by \(3\).
Worked Example: Proving That the Square Root of 3 Is Irrational
Suppose, for a contradiction, that \(\sqrt{3}=a/b\) for integers \(a,b\) with \(b>0\), choosing a representation with the least positive denominator \(b\). Squaring gives \(a^2=3b^2\), so \(3\) divides \(a^2\). By the remainder argument above, \(3\) divides \(a\); write \(a=3c\). Substitution yields \(9c^2=3b^2\), and dividing both sides by \(3\) gives \(b^2=3c^2\). Thus \(3\) divides \(b^2\), so \(3\) divides \(b\) as well. Write \(b=3d\), with \(d\) a positive integer. Then \(a/b=3c/(3d)=c/d\), and \(0<d<b\). This is a representation of \(\sqrt{3}\) with a smaller positive denominator, contradicting the choice of \(b\). Therefore \(\sqrt{3}\) is irrational.
The proof uses divisibility rather than parity, but its structure is the same as the proof for \(\sqrt{2}\): assume a fraction exists, derive that both entries are divisible by the same integer, and reduce the denominator.
For other square roots, a related divisibility argument can often be used. If a prime \(p\) divides \(a^2\), then \(p\) divides \(a\). When an assumed fraction for \(\sqrt{p}\) gives \(a^2=pb^2\), this fact forces \(p\) to divide \(a\), and substituting \(a=pc\) then forces \(p\) to divide \(b\). The least-denominator argument rules out the resulting reduction. The examples here illustrate the method for \(p=2\) and \(p=3\); they do not require us to assume a general theorem about every possible square root.
Worked Example: Why \(4+\sqrt{2}\) Is Irrational
Suppose instead that \(4+\sqrt{2}\) were rational. Since \(4\) is rational, subtracting \(4\) would show that \(\sqrt{2}=(4+\sqrt{2})-4\) is rational: the difference of two rational numbers is rational by rational arithmetic. This contradicts the theorem that \(\sqrt{2}\) is irrational. Therefore \(4+\sqrt{2}\) is irrational.
This argument proves the claim without trying to write \(4+\sqrt{2}\) as a fraction directly. It reverses a familiar operation: if adding a rational number had turned an irrational number into a rational one, subtracting the same rational number would recover an irrational number as a rational difference, which is impossible.
How Rational Arithmetic Preserves Irrationality
The preceding example illustrates a broader principle. Adding a rational number cannot turn an irrational number into a rational one. Nor can multiplying an irrational number by a nonzero rational number. Both claims follow by supposing the result is rational and undoing the rational operation.
Proof. First suppose, contrary to the first claim, that \(x+r\) is rational. Since \(r\) is rational, the difference \((x+r)-r\) is rational by rational addition and the existence of rational opposites. But \((x+r)-r=x\), contradicting the assumption that \(x\) is irrational. Hence \(x+r\) is irrational.
Now suppose \(r\ne0\) and, contrary to the second claim, that \(rx\) is rational. A nonzero rational number has a rational reciprocal, so \(1/r\) is rational. The product of rational numbers is rational; consequently \((1/r)(rx)\) is rational. Since \((1/r)(rx)=x\), this again contradicts the irrationality of \(x\). Therefore \(rx\) is irrational whenever \(r\ne0\). \(\square\)
The nonzero condition in the multiplication claim is essential. If \(r=0\), then \(rx=0\), which is rational for every \(x\). The proof also explains why the result is useful: once one irrational number is known, rational translations and nonzero rational rescalings produce more irrational numbers.
Worked Example: A Nonzero Rational Multiple of \(\sqrt{3}\)
Because \(\sqrt{3}\) is irrational and \(2/5\) is a nonzero rational number, the theorem implies that \((2/5)\sqrt{3}\) is irrational. To check why the nonzero condition matters, suppose this product were rational. Multiplying it by the rational number \(5/2\) would give a rational number, but
That would make \(\sqrt{3}\) rational, contradicting the worked proof above. In contrast, \(0\cdot\sqrt{3}=0\) is rational; multiplication by zero cannot be reversed.
What Irrationality Does—and Does Not—Mean
An irrational number is not a fraction of integers, but this does not mean it is an approximation or that it is impossible to locate. For example, the positive square root of \(2\) lies between \(1\) and \(2\), since \(1^2<2<2^2\). The irrationality proof says that no fraction equals that number exactly; it does not prevent rational numbers from being used as approximations.
A common shortcut says that an irrational number has a decimal expansion that “does not end.” The underlying idea is that a rational number has a decimal expansion that either terminates or eventually repeats, but decimal-expansion conventions require care. For instance, \(1.000\ldots\) and \(0.999\ldots\) denote the same number. For proving irrationality at this stage, a fraction contradiction is more direct: it specifies exactly what must be ruled out and avoids relying on decimal notation.
The central method is to translate a proposed fraction into an integer equation. Parity or divisibility then reveals a shared factor, and choosing a representation with a least positive denominator turns that factor into a contradiction. Once a first irrational number has been established, rational arithmetic can extend the collection of known examples without repeating the original argument.
Check Your Understanding
Use the definition of irrationality, fraction representations, and rational arithmetic to answer the following questions.
- What must be shown to prove that a number is irrational?
- In the least-denominator proof for \(\sqrt{2}\), why does \(a^2=2b^2\) imply that \(a\) is even?
- In the proof for \(\sqrt{3}\), what are the possible remainders of an integer square upon division by \(3\)?
- Explain why the theorem about rational multiples requires the rational multiplier to be nonzero.
- Use the theorem on rational shifts to decide whether \(7-\sqrt{2}\) is irrational.