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Just as $\Im(z)$ and $\Re(z)$ denote the imaginary and real parts of $z$, respectively, do there exist symbols for the rational and irrational parts of a real number?

beep-boop
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  • What would you mean by "rational part of a number"? –  Jun 12 '14 at 20:08
  • e.g. $\rm{rationalpart}(3+\sqrt{2})=3$, and $\rm{irrationalpart}(3+\sqrt{2})=\sqrt{2}$. – beep-boop Jun 12 '14 at 20:09
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    this is not well defined because then Rationalpart$(3+\sqrt{2}) = -123$ and Irrationalpart$(3+\sqrt{2})=126+\sqrt{2}$ – Surb Jun 12 '14 at 20:11
  • Why not just define a symbol, say $Ra(x)$ and $Ir(x)$ to be the rational/irrational part respectively and just notify the reader about the notation? – cirpis Jun 12 '14 at 20:11
  • Why not have $\operatorname{rationalpart}(3+\sqrt 2) = 4$, $\operatorname{irrationalpart}(3+\sqrt 2) = \sqrt 2 - 1$? This seems not to be well-defined. –  Jun 12 '14 at 20:12
  • Why isn't $\operatorname{rationalpart}(3+\sqrt2)$ equal to $\frac{27}{19}$ and $\operatorname{irrationalpart}(3+\sqrt2) = \frac{30}{19}+\sqrt 2$? – MJD Jun 12 '14 at 20:12
  • I don't follow all these comments about it not being well-defined. – beep-boop Jun 12 '14 at 20:12
  • @alexqwx, $3+\sqrt{2}$ can be written as $4+(\sqrt{2}-1)$. Note that $\sqrt{2}-1$ is irrational. So we could say that the "rational" part is $4$ and the "irrational" part is $\sqrt{2}-1$. – Spencer Jun 12 '14 at 20:14
  • Notice that $3+\sqrt2 \approx 4.4142\ldots$. But you somehow want $\operatorname{rationalpart}(4.4142\ldots) = 3$ instead of $4$. Why? – MJD Jun 12 '14 at 20:14
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    The point is that there is NO UNIQUE pair of parts that can be defined from a given real. Just take your example and add 1 to the rational part and subtract 1 from the irrational part and you have a different pair which add up to the same real. – Ned Jun 12 '14 at 20:16
  • Furthermore, even if you constraint say $a\leq Rationalpart(x)\leq b,$ you will always infinitely many decomposition into a sum of a rational + an irrational. – Surb Jun 12 '14 at 20:17
  • @alexqwx Are you considering only numbers of the form $a+\sqrt{b}$ where $a,b\in\mathbb{Q}$ or all irrational numbers? (see transcendental numbers) – user5402 Jun 12 '14 at 20:47

4 Answers4

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You are talking in the realm of e.g. quadratic rings like $\mathbb{Q}(\sqrt{d})$. Often $d$ is negative (Gaussian integers, for instance), and (even when it isn't) you might as well use the notations $\Re(z)$ and $\Im(z)$. But make double sure your audience knows what you are talking about.

Note that if you want to talk about cubic or higher rings, you get more "basis vectors," and you'd need to extend the notation somehow.

vonbrand
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I had a similar issue and came across this question. Here is my own solution and I hope it still helps.

The real part and imaginary part always make sense because $\mathbb C$ is a degree 2 field extension of $\mathbb R$. Viewing $\mathbb C$ as a vector space over $\mathbb R$ we have a canonical basis $\{1,i\}$. Real part is the coefficient of $1$ while imaginary part is the coefficient of $i$.

Thus, for a field extension $K$ of $\mathbb Q$ of finite degree, we can make the notion of "rational part" meaningful by fixing a basis $B=\{1,e_1,e_2,\dots\}$, and define the coefficient of $1$ to be the "rational part". As such definition is sensitive to the basis $B$, I recommend the notation to be something like $\mathfrak Q_B(x)$.

For example in $\mathbb Q [\sqrt 2]$, we can take the basis $B_1=\{1,\sqrt 2\}$ or $B_2=\{1,126+\sqrt2\}$. With respect to different bases, $\mathfrak Q_{B_1}(3+\sqrt 2)=3$, $\mathfrak Q_{B_2}(3+\sqrt 2)=-123$. But as long as we are using the same basis, comparing the rational part $\mathfrak Q_B(x)$ always makes sense.

Tian Lan
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I think the closest thing you can get is the floor function. Where the "rational" or integer part of $x$ would be the largest integer less than or equal to $x$. But this doesn't really guarantee that what is left over will be irrational.

Edit

Thinking about it a bit more I think the following is at least well defined for real values of $x$,

$$ RationalPart(x) = \begin{cases} x \qquad x \in \mathbb{Q} \\ \lfloor x \rfloor \qquad x\notin \mathbb{Q} \end{cases} $$

Of course this doesn't have any of the nice properties like linearity that $Im$ and $Re$ have.

Spencer
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  • What do you mean by linear with regards to $\Im$ and $\Re$? – beep-boop Jun 12 '14 at 20:24
  • The $Im$ and $Re$ functions are linear if we consider the vector space of complex numbers over the field of real numbers. $Im(a z_1 + b z_2) = aIm(z_1)+bIm(z_2)$, where $a,b\in \mathbb{R}$. – Spencer Jun 12 '14 at 20:25
  • This contradicts OP's specific example that $\operatorname{rationalpart}(3+\sqrt2) = 3$; your suggestion makes it 4 instead. – MJD Jun 12 '14 at 21:03
  • @MJD, Thank you for your comment. I'm aware of the difference and I didn't mean to imply that my answer is consistent with that example. The point of this answer was to show that there is a sensible way to speak of something like the "rational part" of a number although it is not the same as OP's original idea of the concept. Since the actual question doesn't mention the example I think there is some room for interpretation here. – Spencer Jun 12 '14 at 21:19
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Not that I know of. Rational and irrational numbers are all elements of the same set complex numbers.

ronald
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