# Combinatorial Puzzle

Standard

This is a continuation of previous post:

How many distinct numbers can be formed by using four 2s and the four arithmetic operations $+,-,\times, \div$.

For example:
$1 = \frac{2+2}{2+2}=\frac{2}{2}\times\frac{2}{2}$
$2 = 2+\frac{2-2}{2}=\frac{2}{2}+\frac{2}{2}$
$3 = 2+2 - \frac{2}{2}$
$4 = 2+2+2-2 = (2\times 2) + (2-2)$
(note that some binary operations do not make sense without parenthesis)

I have no idea about how to approach this problem (since I am not very comfortable with combinatorics). So any help will be appreciated.

Edit[29 May 2017]: This problem has been solved in the comments below.

# Arithmetic Operations

Standard

There are only 4 binary operations which we call “arithmetic operations”. These are:

• Subtractions (-)
• Multiplication (×)
• Division (÷)

Reading this fact, an obvious question is:

Why only four out of the infinitely many possible binary operations are said to be arithmetical?

Before presenting my attempt to answer this question, I would like to remind you that these are the operations you were taught when you learnt about numbers i.e. arithmetic.

In high school when $\sqrt{2}$ is introduced, we are told that real numbers are of two types: “rational” and “irrational”. Then in college when $\sqrt{-1}$ is introduced, we should be told that complex numbers are of two types: “algebraic” and “transcendental“.

As I have commented before, there are various number systems. And for each number system we have some valid arithmetical operations leading to a valid algebraic structure. So, only these 4 operations are entitled to be arithmetic operations because only these operations lead to valid algebraic numbers when operated on algebraic numbers.

Now this leads to another obvious question:

Why so much concerned about algebraic numbers?

To answer this question, we will have to look into the motivation for construction of various number systems like integers, rational, irrationals, complex numbers… The construction of these number systems has been motivated by our need to be able to solve polynomials of various degree (linear, quadratic, cubic…). And the Fundamental Theorem of Algebra says:

Every polynomial with rational coefficients and of degree n in variable $x$ has n solutions in  complex number system.

But, here is a catch. The number of complex numbers which can’t satisfy any polynomial (called transcendental numbers) is much more than the number of complex numbers which can satisfy a polynomial equation (called algebraic numbers). And we wish to find solutions of a polynomial equation (ie.e algebraic numbers) in terms of sum, difference, product, division or $m^{th}$ root of rational numbers (since coefficients were rational numbers). Therefore, sum, difference, product and division are only 4 possible arithmetic operations.

My previous statement may lead to a doubt that:

Why taking $m^{th}$ root isn’t an arithmetic operation?

This is because it isn’t a binary operation to start with, since we have fixed $m$. Also, taking $m^{th}$ root is allowed because of the multiplication property.

CAUTION: The reverse of $m^{th}$ root is multiplying a number with itself m times and it is obviously allowed. But, this doesn’t make the binary operation of taking exponents, $\alpha^{\beta}$ where $\alpha$ and $\beta$ are algebraic numbers, an arithmetic operation. For example, $2^{\sqrt{2}}$ is transcendental (called Gelfond–Schneider constant or Hilbert number) even though 2 and $\sqrt{2}$ are algebraic.