# Which is more simplified: $a\sqrt{b}$ or $\sqrt{c}$?

Which is considered more simplified (if it matters)?

$a \sqrt{b}$ or $\sqrt{c}$

For example: $2 \sqrt{3}$ or $\sqrt{12}$

• The word "simplified" does not have a rigorous meaning in mathematics, just an informal meaning. So there's no real answer to your question. – Gregory Grant Aug 25 '15 at 18:10
• I think that $2\sqrt3$ would generally be called more simplified. – Akiva Weinberger Aug 25 '15 at 18:16
• I've seen "simplified" to mean "the radicand is square-free". – The Chaz 2.0 Aug 25 '15 at 18:20
• I want to add, that Mathematica also says, that $2 \sqrt{3}$ is a simpler expression for $\sqrt{12}$, I also found this old discussion with some nice ideas as well – user190080 Aug 25 '15 at 19:57

Typically, one would say that $\sqrt{a^2 b}$ is not fully simplified, while $a\sqrt{b}$ is.

As Gregory Grant points out in his comment, generally "simplified" is not defined formally (rather by example and an often informal description), and there are varying degrees of simplification. It's also highly context-dependent: Certain forms are preferable to others, depending on what you'd like to do with your expression (not to mention, simplifying rational expressions isn't the same sort of beast as simplifying radical expressions, etc).

I suspect there are a few reasons we've come to prefer e.g. $2 \sqrt{3}$ over $\sqrt{12}$:

• A historical reason (and I suspect rationalizing denominators stems originally from this reason). Back when calculators were powered by food and not batteries, it probably made more sense to use as few square roots as possible. Since somebody had to compute both $\sqrt{3}$ and $\sqrt{12}$ if both were to be used numerically, it's helpful to know that $\sqrt{12} = 2\sqrt{3}$. This is because knowing $\sqrt{3}$ enables one to calculate $\sqrt{12}$ more efficiently (multiplying by $2$) than with some numerical method of calculating square roots.

• An algebraic reason. We know that radicals with the same radicand and root index can be combined algebraically; they're "like terms." Thus, when faced with something like $5\sqrt{3} + \sqrt{12}$, nothing can be done until one realizes that $\sqrt{12} = 2\sqrt{3}$ so that

$$5\sqrt{3} + \sqrt{12} = 5\sqrt{3} + 2\sqrt{3} = 7\sqrt{3},$$ or in a similar vein,

$$\frac{2+\sqrt{12}}{2} = \frac{2\big(1 + \sqrt{3}\big)}{2} = 1 + \sqrt{3},$$ to give some examples.

• The last example can be done another way: $\frac{2 + \sqrt{12}}{2} = \frac{2}{2} + \frac{\sqrt{12}}{2} = 1 + \frac{\sqrt{12}}{\sqrt{4}} = 1 + \sqrt{3}$, so simplifying as $2\sqrt{3}$ doesn't make things easier. – lisyarus Aug 25 '15 at 18:38

It highly depends on the reasons of the simplification.

In some circumstances, $2 \sqrt{3}$ will be better, since the radical is square free, which is sometimes useful. For example, in calculations by hand that involve lots of radicals and rationals (pjs36 answer gives some more reasons to use $2\sqrt{3}$).

In some other cases, $\sqrt{12}$ will be better, since it contains less operations: it is just a square root of a constant, while $2 \sqrt{3}$ is a constant multiplied by a square root of a constant. Technically, we can say that the abstract syntax tree of the expression has lesser depth. This is more convenient if this expression is just a subexpression of some highly complex expression, where the number of operations involved can affect readability and understanding. I usually prefer this, while keeping in mind that $\sqrt{12}=2\sqrt{3}$.

I would say that none of these is more simplified than the other.

As a general rule, simpler means having fewer symbols. So $\sqrt6$ is simpler than $\sqrt2\sqrt3$. In your case, the number of symbols is the same; so another rule kicks in: small numbers are simpler than big ones. So $2\sqrt3$ is simpler than $\sqrt{12}$. However, it's debatable whether $2^7$ is simpler than $128$, because decimal notation is considered more basic and familiar than taking powers. Ultimately, the question of which is simpler may come down to a matter of taste.