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Every non negative real number has a unique non negative root, called the principle square root. Does this mean that any combination of roots added will also produce a unique value?

For example, $$\sqrt{2}+\sqrt{3} = 1.414213562373095\ldots + 1.732050807568877\ldots = 3.146264369941972\ldots$$ Is this resulting value as unique as each of the two principle square roots that were added?

Is there a way to prove this uniqueness mathematically or logically without producing a huge table of all possible combinations to check against each other?

Blue
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Perpetual Fool
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    $\sqrt2+\sqrt 3$ is the sum of the two aforementioned numbers, therefore it is that specific real number. –  Jul 19 '20 at 19:03
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    No: $\sqrt 1+\sqrt{16}=\sqrt 4+\sqrt 9$ – Andrei Jul 19 '20 at 19:05
  • @Andrei Either you've misunderstood the question or I have. – J.G. Jul 19 '20 at 19:06
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    If $a$ is uniquely-defined, and if $b$ is uniquely-defined, then $a+b$ is uniquely-defined. – Blue Jul 19 '20 at 19:06
  • It’s really hard to know what you’re asking here. – Lubin Jul 19 '20 at 19:10
  • Perhaps you are thinking about minimal polynomials or something along those lines? $\sqrt 2+\sqrt 3$ has the minimal polynomial $x^4-10x^2+1$ and that quartic has two positive roots. – lulu Jul 19 '20 at 19:11
  • Writing $\sqrt{2}+\sqrt{3}$ expresses an unambiguous sum of unambiguous values. However, writing $x+y$, and all I know is that $x^2=2$ and $y^2=3$, expresses an ambiguous sum that has no inherent value ... although I can say that there are only four values attainable by the expression (namely, $\sqrt2+\sqrt3$, $\sqrt2-\sqrt3$, $-\sqrt2+\sqrt3$, and $-\sqrt2-\sqrt3$). – Blue Jul 19 '20 at 19:23

2 Answers2

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$\sqrt{2}$ and $\sqrt{3}$ are linearly independent over $\mathbb{Q}$. See here

David Lui
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Since you are assuming the existence and uniqueness of roots of non-negative real numbers (in fact, this is not hard to prove rigorously), Yes. Because $\sqrt2+\sqrt3$ is the principle square root of $5+\sqrt{24}.$

Bumblebee
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