Irrationality

Rational and irrational numbers are an interesting topic and yet not all the problems in the related field has been solved fully. I shall now introduce some basic properties of it.

1) Group as an algebraic structure have the following properties

For a group G there exists a binary operation, say * here.

This is a function $*:G\times G\to G$, mapping each ordered pair of elements to a unique element of G.

-Closure. i.e., if $a,b\in G$, then $a*b\in G$.

-Associativity: $(a*b)*c=a*(b*c)$

-Existence of identity: there exists $e\in G$ such that $a*e=e*a=a$ for every $a\in G$. For multiplication the identity is often written as 1; for addition it is 0.

-Existence of inverse element: for every $a\in G$, there exists $b\in G$ (it can be a) such that $a*b=b*a=e$. Within a group, $a*b=e$ also implies $b*a=e$. We say $a=b^{-1}$. For multiplication this is the reciprocal $1/b$, but for a general group the meaning depends on the operation.

Example: non-zero Rational number with multiplication is a group, identity 1.

2) Rational numbers can be written as $p/q$, where p is an integer and q is a positive integer. The expression is unique when $\gcd(|p|,q)=1$.

Conversely, irrational numbers cannot be expressed as a quotient of integers. This gives us the fundamental way to show that a number is irrational.

Example 2: Prove $\sqrt{2}$ is irrational.

Classic question. Assume $\sqrt{2}=\frac{p}{q},(p,q)=1$.

By squaring both sides and change of term, we have $2q^2=p^2$.

Then following logic finishes the prove by contradiction.

Since $p^2=2q^2$, p is even. Write $p=2k$; then $q^2=2k^2$, so q is also even. Therefore $\gcd(p,q)\geq2$, contradicting $\gcd(p,q)=1$.

Now we will introduce another method to show irrationality.

If $p/q$ and $r/s$ are distinct rational numbers with positive denominators, then $|ps-qr|\geq1$, so $\left|\frac{p}{q}-\frac{r}{s}\right|\geq\frac{1}{qs}$. Also, any finite sum of rational numbers is rational; hence an irrational number cannot be expressed as a finite sum of fractions of integers.

Corollary: A rational number has a terminating or eventually recurring decimal expansion.

Proof: In long division by a positive denominator q, there are only q possible remainders. Either the remainder becomes zero and the decimal terminates, or a remainder repeats and the subsequent digits recur.

For a prime $p\ne2,5$, Fermat's little theorem gives $10^{p-1}\equiv1\pmod p$. Thus $10^{p-1}-1=pk$ for a positive integer k.

Consequently $\frac1p=\frac{k}{10^{p-1}-1}=\sum_{j=1}^{\infty}k\,10^{-j(p-1)}$. Writing k as a block of p−1 digits, with leading zeros if necessary, gives a recurring decimal. For p=2 or p=5 the decimal terminates.

The smallest recurring period divides p−1. It need not increase monotonically as p increases.

However, for every fixed positive integer N, only finitely many primes can have period at most N: each must divide $\prod_{n=1}^{N}(10^n-1)$. Therefore the periods eventually exceed any fixed bound as the primes increase, excluding 2 and 5.

A real number is irrational exactly when its decimal expansion neither terminates nor eventually repeats. An infinite decimal expansion alone does not imply irrationality; for example, $1/3=0.333\ldots$ is rational.

Example 3: Prove e is irrational.

Consider $e=\sum_{k=0}^{\infty}\frac1{k!}$ and let $S_n=\sum_{k=0}^{n}\frac1{k!}$. Then $n!S_n$ is an integer.

Assume $e=a/b$, where a and b are integers and b is positive. Choose an integer $n\geq\max(b,2)$. Since b divides n!, $n!e$ is also an integer.

But $0<n!(e-S_n)=\sum_{j=1}^{\infty}\frac1{(n+1)(n+2)\cdots(n+j)}<\sum_{j=1}^{\infty}\frac1{(n+1)^j}=\frac1n<1$. The middle inequality is strict because the terms with $j\geq2$ are strictly smaller. This makes $n!e-n!S_n$ an integer strictly between 0 and 1, a contradiction. Hence e is irrational.

The denominators of partial sums growing without bound would not, by itself, prove irrationality. The positive remainder bound above is what completes this proof.

4) Back to out main concern, we would like to show some properties of irrational numbers.

- Irrational numbers are not closed under addition, subtraction, multiplication or division. For example, $\sqrt2+(-\sqrt2)=0$, $\sqrt2-\sqrt2=0$, $\sqrt2\cdot\sqrt2=2$ and $\sqrt2/\sqrt2=1$.

Example 4: Prove the existence of infinitely many pairs (x,y) where x,y are irrational and x+y, xy are rational.

Proof: For each rational r, let $x=r+\sqrt2$ and $y=r-\sqrt2$. Both are irrational, while $x+y=2r$ and $xy=r^2-2$ are rational. Distinct values of r give distinct pairs, so there are infinitely many such pairs.

This does not mean that every fixed irrational x admits infinitely many y. If $x+y=s$ and $xy=t$ are rational, then $x^2-sx+t=0$, so x must be a quadratic irrational and y is its other root, or algebraic conjugate. Conversely, the conjugate of any real quadratic irrational satisfies these conditions.

For fixed irrational x there is at most one such y: if y and z both work, then $y-z=(x+y)-(x+z)$ and $x(y-z)=xy-xz$ are rational. If $y-z\ne0$, their quotient would make x rational, a contradiction. Thus y=z.

5) Show that $\cos \frac{\pi}{90}$ is irrational.

This is a bit tricky. We use the multiple-angle polynomial that also follows from De Moivre's theorem.

Define $P_0(t)=1$, $P_1(t)=t$ and $P_{n+1}(t)=2tP_n(t)-P_{n-1}(t)$. The cosine addition formula gives $\cos(n\theta)=P_n(\cos\theta)$, and each polynomial has integer coefficients. Thus rational $\cos\theta$ would imply rational $\cos(n\theta)$.

Take $\theta=\pi/90$ and n=15. Then $\cos(15\theta)=\cos(\pi/6)=\sqrt3/2$ is irrational, so $\cos(\pi/90)$ must be irrational. Here $\sqrt3$ is irrational by the same prime-divisibility argument used for $\sqrt2$. Using $\cos(\pi/3)$ instead would not work, since it equals $1/2$.

Exercise:

1) Show that (Z,+) is a group.

2) Show that $\sqrt{10}$ is irrational.

3) Show that $\sqrt2$ cannot be written as a finite sum of fractions of integers.

4) Show that irrational numbers can be written in continued fraction of infinite length.

5) Show that 37 is a factor of 999 considering 1/37.

6) Verify that $x=r+\sqrt2$ and $y=r-\sqrt2$, for rational r, give infinitely many distinct irrational pairs with rational sum and product. Explain why this does not give infinitely many partners for a fixed x.

7) Discuss the countability of irrational numbers.

8) Discuss the countability of irrational pairs (x,y) such that x+y and xy are rational.

9) If x,y are irrational while $x^2+y,x+y^2,xy$ is rational. Find all possible x+y with proof.

10) Show that $\tan \frac{\pi}{180}$ is irrational.

11) Let x be measured in radians, with $0\leq x\leq2\pi$ and $\cos x\ne0$. If $\tan x$ is rational, prove that x is irrational unless x=0. (This is an advanced exercise.)

12) Show that $\pi$ is irrational.

13) If x,y are rational numbers with x>y, show that there exist rational numbers a,b and an irrational number c such that x>a>c>b>y. Extend this to any two distinct real endpoints, and hence prove that both the rational and irrational numbers are dense in the real numbers.

Extra information: For certain numbers, sufficiently strong bounds on rational approximations can prove transcendence. Such a proof must rule out algebraic numbers of every degree, rather than only rational numbers. You can visit the page talking about Liouville's numbers. Then finish this question:

CGMO 2007 day 2 Q3

If a,b,c are integers that $|a|,|b|,|c|\leq 10$ and define $f(x)=x^3+ax^2+bx+c$ satisfying $\left | f(2+\sqrt{3}) \right |$ is smaller than 0.0001, prove that $\left | f(2+\sqrt{3}) \right |=0$.