# Why $2\sqrt{x} + \sqrt{3}$ can’t be simplified any further?

In How do you simplify the addition of square root, it has been pointed out that it cannot be simplified any further.

As obvious as it seems, the following questions seemed need to be addressed:

1. What is meant by a simpler form? As matter of fact, what determines what is simpler with respect to something else?

2. Suppose we bypass 1, and we all agree that we can decide a simpler form between given any two forms. How do we justify that something is in its simplest form (i.e. it has reached the simplest form and no further simplification can be performed.)?

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"As matter of fact what determines what is simpler with respect to something else?" - Even for polynomials, it's a bit hard to settle on a particular form as "simplest". $x^{100}-1$ is simpler in expanded form, while $(x+1)^{100}$ is simpler in factored form... – J. M. Dec 11 '11 at 10:57
@J.M. : Your comment opens the prequestion "Is simplicity of equations an orederable property?, if yes then we can ask what is the ordering based on? – Arjang Dec 17 '11 at 13:37

The simplest way any polynomial expression can be written is

$$\sum_i{a_i \cdot x^i} .$$

Example: there is no simpler way to write $3x^4+2x^2-7x+2$ where $a_4=3, a_2=2, a_1=-7, a_0=2$.

Your expression can also be written as $2 \cdot x^{1/2} + 3 \cdot x^0$ where $a_{1/2}=2,a_0=3$. This expression can't be simplified, even if your $i$ isn't an integer.

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You've seen my comment above? – J. M. Dec 11 '11 at 11:42
Oops, no, didn't see it. – Oltarus Dec 11 '11 at 14:05

Well, there are at least some cases where one can tell no further simplification is necessary. E.g., the simplest form for a quotient of two integers is that in which the numerator and denominator have no common factor greater than 1. Euclid's algorithm will get you there, and tell you when you have arrived.

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One approach to simplification is asking whether $2\sqrt{x} + \sqrt{3}$ lives in a simpler field than $\mathbb Q(\sqrt{x},\sqrt{3})$, which has degree 4 over $\mathbb Q(x)$. Any simpler field than that must be a quadratic extension of $\mathbb Q(x)$. So to prove that $2\sqrt{x} + \sqrt{3}$ cannot be simplified any further in this context, one just has to prove that it is not quadratic over $\mathbb Q(x)$. This shouldn't be too hard to do.

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