Is it just for aesthetic purposes, or is there a deeper reason why we write $2\sqrt{3}$ and not $\sqrt{3}2$?
Why not write $\sqrt{3}2$?
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40The format $\sqrt{3}2$ is easily confused with $\sqrt{32}$. Indeed, when I saw the subject, my initial instinct was to correct it to $\sqrt{32}$. – 2012-10-12
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5I suspect it was also common for early typesetters to skip the overline, and just typeset $\sqrt{3}$ as $\sqrt{}3$, which would then be clearly ambiguous. – 2012-10-12
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0We write $2\sqrt{3}$ to just to simplify the number, while $\sqrt{3}2$ will make confusion – 2012-10-12
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3@ThomasAndrews : Writing √3 is not an instance of typesetters skipping the overline; rather it is a case where no overline is called for. The overline in $\sqrt{3x}$ indicates that the whole $3x$ is within the radical, rather than just the $3$. – 2012-10-12
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2The proper name for the 'overline' is 'vinculum' – 2013-04-12
4 Answers
The format $\sqrt{3}2$ is easliy confused with $\sqrt{32}$.
I also suspect that many early typesetters would skip the overline, so that $\sqrt{3}$ would be typeset as $\sqrt{\vphantom{3}}3$. In that case, $2\sqrt{\vphantom{3}}3$ is unambiguous but $\sqrt{\vphantom{3}}32$ highly ambiguous.
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7A related question might be, why do we usually write "2x" rather than "$x2$." Perhaps trying to avoid ambiguity with $x^2$, but that seems unlikely. – 2012-10-12
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6I suspect that we write '$2x$' because the '2' there is secretly acting as an _operator_, and operator composition is usually written as a left action; $2x$ is actually $2(x)$, where $2()$ is the 'multiply by $2$' operator applied to the unknown $x$. (This is related, obviously, to adam's answer) – 2012-10-12
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0Yeah, I was about to post that point. In general, we put operators to the left. – 2012-10-12
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0I suspect the fact that we can read the former aloud as "...two ecks" while the latter has to be read as "...ecks times two" in order to be unambiguous with ".... multiplied by two" is a factor as well. – 2012-10-12
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1Writing $\sqrt{}\,3$ is not an instance of typesetters skipping the overline; rather it is a case where no overline is called for. The overline in $\sqrt{3x}$ indicates that the whole $3x$ is within the radical, rather than just the $3$. – 2012-10-12
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0@ThomasAndrews there's also $x_2$ to confuse it with especially when you pronounce $x_2$ as "ecks two" – 2012-10-13
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0To be even more explicit, we sometime end the overline on the square root by a small "leg". – 2012-10-14
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0@StevenStadnicki What do you mean by "operator" here? – 2012-10-14
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0@DougSpoonwood http://en.wikipedia.org/wiki/Operator_%28mathematics%29 - loosely, a function; in this case, a function from mathematical terms to mathematical terms that represents the 'multiply this term by 2' operation. – 2012-10-14
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0@StevenStadnicki If that holds for 2x, then you have an infinity of functions behind ordinary multiplication. You have the multiply x by 1 operation, the multiply x by 3 operation, you have the multiply 1 by x operation, the multiply 4 by x operation, and so on. Isn't multiplication just one operation instead of an infinity of operations? Or is multiplication both, and thus 1 equals infinity? – 2012-10-14
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0@DougSpoonwood In fact, it's often useful to consider a multiplication operator for each number (and even an addition operator for each number, e.g. the 'add 3 to my argument' operator). Multiplication can be treated as a single function/operator with multiple operands or a plethora of operators each with a single operand, depending on context, and both are completely consistent; look up the concept of _currying_ functions. – 2012-10-15
One possibility - would you rather think of the number as "two of the thing known as $\sqrt3$," or as "$\sqrt3$ many of the number two?"
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1Does it matter? – 2013-04-12
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0I for one would rather count naturally, but I would agree that beyond that it does not matter, since either way involves something radical. – 2013-04-12
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0It's interesting because with $i$ for example, I think people tend to write $2i$, but $i \sqrt{2}$ instead of $\sqrt{2}i$. In this case do you consider $i$ as a 'counting number' or $\sqrt{2}$? – 2013-04-12
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0counting = natural in my example. But if I had to choose one as the "counting number" between the two, I would choose $i$, and I always write it first, at least for example with unitary numbers such as $e^{i\sqrt{2}}$. – 2013-04-13
Certainly one can find old books in which $\sqrt{x}$ was set as $\sqrt{\vphantom{x}}x$, and just as $32$ does not mean $3\cdot2$, so also $\sqrt{\vphantom{32}}32$ would not mean $\sqrt{3}\cdot 2$, but rather $\sqrt{32}$. An overline was once used where round brackets are used today, so that, where we now write $(a+b)^2$, people would write $\overline{a+b}^2$. Probably that's how the overline in $\sqrt{a+b}$ originated. Today, an incessant battle that will never end tries to call students' attention to the fact that $\sqrt{5}z$ is not the same as $\sqrt{5z}$ and $\sqrt{b^2-4ac}$ is not the same as $\sqrt{b^2-4}ac$, the latter being what one sees written by students.
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2The proper name for the 'overline' is 'vinculum' – 2013-04-12
It's simply a matter of clarity. If you write $\sqrt 3 2$ meaning $2 \times \sqrt 3$ rather than $\sqrt{32}$, it would be clearer to write $(\sqrt 3) 2$ or $\sqrt 3 \times 2$, but then you have to say: oh, what the heck, just go with $2 \sqrt 3$.
Another thing to consider is that neglecting to properly extend overlines is a tell-tale sign of a TeX novice. As you are already aware, to get $\sqrt{32}$ you need to write \sqrt{32} in your source.