Real Numbers:Irrational

The number ${\sqrt {2}}$ is irrational.

In mathematics, the irrational numbers (from in- prefix assimilated to ir- (negative prefix, privative) + rational) are all the real numbers which are not rational numbers. That is, irrational numbers cannot be expressed as the ratio of two integers. When the ratio of lengths of two line segments is an irrational number, the line segments are also described as being incommensurable, meaning that they share no "measure" in common, that is, there is no length ("the measure"), no matter how short, that could be used to express the lengths of both of the two given segments as integer multiples of itself.

Among irrational numbers are the ratio $\pi$ of a circle's circumference to its diameter, Euler's number e, the golden ratio φ, and the square root of two. In fact, all square roots of natural numbers, other than of perfect squares, are irrational.

Like all real numbers, irrational numbers can be expressed in positional notation, notably as a decimal number. In the case of irrational numbers, the decimal expansion does not terminate, nor end with a repeating sequence. For example, the decimal representation of $\pi$ starts with 3.14159, but no finite number of digits can represent $\pi$ exactly, nor does it repeat. Conversely, a decimal expansion that terminates or repeats must be a rational number. These are provable properties of rational numbers and positional number systems, and are not used as definitions in mathematics.

Irrational numbers can also be expressed as non-terminating continued fractions and many other ways.

As a consequence of Cantor's proof that the real numbers are uncountable and the rationals countable, it follows that almost all real numbers are irrational.

Examples

Square roots

The square root of 2 was the first number proved irrational, and that article contains a number of proofs. The golden ratio is another famous quadratic irrational number. The square roots of all natural numbers which are not perfect squares are irrational and a proof may be found in quadratic irrationals.

Proof that ${\sqrt {2}}$ is irrational

One proof of the number's irrationality is the following proof by infinite descent. It is also a proof by contradiction, also known as an indirect proof, in that the proposition is proved by assuming that the opposite of the proposition is true and showing that this assumption is false, thereby implying that the proposition must be true.

1. Assume that ${\sqrt {2}}$ is a rational number, meaning that there exists a pair of integers whose ratio is exactly ${\sqrt {2}}$.
2. If the two integers have a common factor, it can be eliminated using the Euclidean algorithm.
3. Then ${\sqrt {2}}$ can be written as an irreducible fraction ${\frac {a}{b}}$ such that a and b are coprime integers (having no common factor) which additionally means that at least one of a or b must be odd .
4. It follows that ${\frac {a^{2}}{b^{2}}}=2$ and $a^{2}=2b^{2}$. ($({\frac {a}{b}})^{n}={\frac {a^{n}}{b^{n}}}$  )   ($a^{2}$ and $b^{2}$ are integers)
5. Therefore, $a^{2}$ is even because it is equal to $2b^{2}$. ($2b^{2}$ is necessarily even because it is 2 times another whole number and multiples of 2 are even.)
6. It follows that a must be even (as squares of odd integers are never even).
7. Because a is even, there exists an integer k that fulfills: a = 2k.
8. Substituting 2k from step 7 for a in the second equation of step 4: $2b^{2}$ = $(2k)^{2}$ is equivalent to $2b^{2}$ = $4k^{2}$, which is equivalent to $b^{2}$ = $2k^{2}$.
9. Because $2k^{2}$ is divisible by two and therefore even, and because $2k^{2}$ = $b^{2}$, it follows that $b^{2}$ is also even which means that b is even.
10. By steps 5 and 8 a and b are both even, which contradicts that ${\frac {a}{b}}$ is irreducible as stated in step 3.

Q.E.D.

Because there is a contradiction, the assumption (1) that ${\sqrt {2}}$ is a rational number must be false. This means that ${\sqrt {2}}$ is not a rational number. That is, ${\sqrt {2}}$ is irrational.

General roots

The proof above for the square root of two can be generalized using the fundamental theorem of arithmetic. This asserts that every integer has a unique factorization into primes. Using it we can show that if a rational number is not an integer then no integral power of it can be an integer, as in lowest terms there must be a prime in the denominator that does not divide into the numerator whatever power each is raised to. Therefore, if an integer is not an exact kth power of another integer, then that first integer's kth root is irrational.

Logarithms

Perhaps the numbers most easy to prove irrational are certain logarithms. Here is a proof by contradiction that log₂ 3 is irrational (log₂ 3 ≈ 1.58 > 0).

Assume log₂ 3 is rational. For some positive integers m and n, we have

$\log _{2}3={\frac {m}{n}}.$

It follows that

$2^{m/n}=3$

$(2^{m/n})^{n}=3^{n}$

$2^{m}=3^{n}.$

However, the number 2 raised to any positive integer power must be even (because it is divisible by 2) and the number 3 raised to any positive integer power must be odd (since none of its prime factors will be 2). Clearly, an integer cannot be both odd and even at the same time: we have a contradiction. The only assumption we made was that log₂ 3 is rational (and so expressible as a quotient of integers m/n with n ≠ 0). The contradiction means that this assumption must be false, i.e. log₂ 3 is irrational, and can never be expressed as a quotient of integers m/n with n ≠ 0.

Cases such as log₁₀ 2 can be treated similarly.

Types

Transcendental/algebraic

Almost all irrational numbers are transcendental and all real transcendental numbers are irrational (there are also complex transcendental numbers): the article on transcendental numbers lists several examples. So e ^r and π ^r are irrational for all nonzero rational r, and, e.g., e^π is irrational, too.

Irrational numbers can also be found within the countable set of real algebraic numbers (essentially defined as the real roots of polynomials with integer coefficients), i.e., as real solutions of polynomial equations

$p(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x+a_{0}=0\;,$

where the coefficients $a_{i}$ are integers and $a_{n}\neq 0$. Any rational root of this polynomial equation must be of the form r /s, where r is a divisor of a₀ and s is a divisor of a_n. If a real root $x_{0}$ of a polynomial $p$ is not among these finitely many possibilities, it must be an irrational algebraic number. An exemplary proof for the existence of such algebraic irrationals is by showing that x₀  = (2^1/2 + 1)^1/3 is an irrational root of a polynomial with integer coefficients: it satisfies (x³ − 1)² = 2 and hence x⁶ − 2x³ − 1 = 0, and this latter polynomial has no rational roots (the only candidates to check are ±1, and x₀, being greater than 1, is neither of these), so x₀ is an irrational algebraic number.

Because the algebraic numbers form a subfield of the real numbers, many irrational real numbers can be constructed by combining transcendental and algebraic numbers. For example, 3$\pi$ + 2, $\pi$ + ${\sqrt {2}}$ and $e{\sqrt {3}}$ are irrational (and even transcendental).

Decimal expansions

The decimal expansion of an irrational number never repeats or terminates (the latter being equivalent to repeating zeroes), unlike any rational number. The same is true for binary, octal or hexadecimal expansions, and in general for expansions in every positional notation with natural bases.

To show this, suppose we divide integers n by m (where m is nonzero). When long division is applied to the division of n by m, only m remainders are possible. If 0 appears as a remainder, the decimal expansion terminates. If 0 never occurs, then the algorithm can run at most m − 1 steps without using any remainder more than once. After that, a remainder must recur, and then the decimal expansion repeats.

Conversely, suppose we are faced with a repeating decimal, we can prove that it is a fraction of two integers. For example, consider:

$A=0.7\,162\,162\,162\,\ldots$

Here the repetend is 162 and the length of the repetend is 3. First, we multiply by an appropriate power of 10 to move the decimal point to the right so that it is just in front of a repetend. In this example we would multiply by 10 to obtain:

$10A=7.162\,162\,162\,\ldots$

Now we multiply this equation by 10^r where r is the length of the repetend. This has the effect of moving the decimal point to be in front of the "next" repetend. In our example, multiply by 10³:

$10,000A=7\,162.162\,162\,\ldots$

The result of the two multiplications gives two different expressions with exactly the same "decimal portion", that is, the tail end of 10,000A matches the tail end of 10A exactly. Here, both 10,000A and 10A have .162 162 162 ... after the decimal point.

Therefore, when we subtract the 10A equation from the 10,000A equation, the tail end of 10A cancels out the tail end of 10,000A leaving us with:

$9990A=7155.$

Then

$A={\frac {7155}{9990}}={\frac {53}{74}}$

is a ratio of integers and therefore a rational number.

Irrational powers

Dov Jarden gave a simple non-constructive proof that there exist two irrational numbers a and b, such that a^b is rational:

Consider ${\sqrt {2}}^{\sqrt {2}}$; if this is rational, then take a = b = ${\sqrt {2}}$. Otherwise, take a to be the irrational number ${\sqrt {2}}^{\sqrt {2}}$ and b = ${\sqrt {2}}$. Then $a^{b}$ = $({\sqrt {2}}^{\sqrt {2}})^{\sqrt {2}}$ = ${\sqrt {2}}^{{\sqrt {2}}\cdot {\sqrt {2}}}$ = ${\sqrt {2}}^{2}$ = 2, which is rational.

Although the above argument does not decide between the two cases, the Gelfond–Schneider theorem shows that ${\sqrt {2}}^{\sqrt {2}}$ is transcendental, hence irrational. This theorem states that if a and b are both algebraic numbers, and a is not equal to 0 or 1, and b is not a rational number, then any value of a^b is a transcendental number (there can be more than one value if complex number exponentiation is used).

An example that provides a simple constructive proof is

$\left({\sqrt {2}}\right)^{\log _{\sqrt {2}}3}=3.$

The base of the left side is irrational and the right side is rational, so one must prove that the exponent on the left side, $\log _{\sqrt {2}}3$, is irrational. This is so because, by the formula relating logarithms with different bases,

$\log _{\sqrt {2}}3={\frac {\log _{2}3}{\log _{2}{\sqrt {2}}}}={\frac {\log _{2}3}{1/2}}=2\log _{2}3$

which we can assume, for the sake of establishing a contradiction, equals a ratio m/n of positive integers. Then $\log _{2}3=m/2n$ hence $2^{\log _{2}3}=2^{m/2n}$ hence $3=2^{m/2n}$ hence $3^{2n}=2^{m}$, which is a contradictory pair of prime factorizations and hence violates the fundamental theorem of arithmetic (unique prime factorization).

A stronger result is the following: Every rational number in the interval $((1/e)^{1/e},\infty )$ can be written either as a^a for some irrational number a or as nⁿ for some natural number n. Similarly, every positive rational number can be written either as $a^{a^{a}}$ for some irrational number a or as $n^{n^{n}}$ for some natural number n.

Set of all irrationals

Since the reals form an uncountable set, of which the rationals are a countable subset, the complementary set of irrationals is uncountable.

Under the usual (Euclidean) distance function d(x, y) = |x − y|, the real numbers are a metric space and hence also a topological space. Restricting the Euclidean distance function gives the irrationals the structure of a metric space. Since the subspace of irrationals is not closed, the induced metric is not complete. However, being a G-delta set—i.e., a countable intersection of open subsets—in a complete metric space, the space of irrationals is completely metrizable: that is, there is a metric on the irrationals inducing the same topology as the restriction of the Euclidean metric, but with respect to which the irrationals are complete. One can see this without knowing the aforementioned fact about G-delta sets: the continued fraction expansion of an irrational number defines a homeomorphism from the space of irrationals to the space of all sequences of positive integers, which is easily seen to be completely metrizable.

Furthermore, the set of all irrationals is a disconnected metrizable space. In fact, the irrationals equipped with the subspace topology have a basis of clopen sets so the space is zero-dimensional.

Licensing

Content obtained and/or adapted from:

• Irrational number, Wikipedia under a CC BY-SA license
• Square root of 2, Wikipedia under a CC BY-SA license