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Is $0.\sqrt9$ valid number? Are such numbers allowed?

First, I thought the value of the above number can be 0.3 but then it occurred how I would interpret this number: $0.65\sqrt2$ or $0.65\sqrt229$

Are such numbers valid?

user33786
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    It depends. Notation can always be used freely, but must be carefully explained, well-defined, and can't certainly be ambiguous. –  Jan 04 '19 at 18:39
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    I would think that $0.\sqrt{9}$ is not great notation... if you are using decimal representation, shoulnd't all digits be a selection of $0-9$? – Eleven-Eleven Jan 04 '19 at 18:40

2 Answers2

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They're valid or invalid depending on the definition you use. What do you take these expressions to mean? If you divide each digit (treating surds as honorary digits) by successively larger powers of $10$, then sum, you'll have a definition that generalises the usual no-surds-allowed one. I sincerely doubt it'll catch on, though; you're getting expressions of the form $a+b\sqrt{c}$, so you might as well just write that.

J.G.
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I asked myself a similar question a while ago about a supposed function $f(x) = 0.x$. It turns out you can express this function in terms of a more widely accepted notation as

$$f(x) = 10^{-\lfloor \log_{10}(x)+1 \rfloor}x.$$

Using this definition, then indeed it turns out

$$f(\sqrt{9}) = 0.3.$$

$65 \sqrt 2$ is commonly interpreted as $65 \cdot \sqrt 2 = 91.9239...$ as opposed to the concatenation of the two, $65 || \sqrt 2 = 651.41421...$. In either case that you mean, we have $$f(65 \cdot \sqrt 2) = \frac{13}{10\sqrt 2}, \ \ \ f(65||\sqrt2) = \frac{650+\sqrt 2}{1000}$$

and so on.

mallan
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