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I: irrational or rational? 
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#1
Jul408, 11:39 PM

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I've been thinkng about this one for a while. Is i rational or irrational. i is an imaginary number, so logically, it would be irrational. But [tex]\frac{1}{i}[/tex] = i so it has a fractional equivilant. But then, it doesn't have a real number decimal equivilant...
So, what is it? Is i rational or irrational? 


#2
Jul408, 11:44 PM

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Just invoke the definition of rational numbers. x is a rational number if it can be expressed as a fraction of integers. Can you write "i" as a ratio of 2 integers?



#3
Jul508, 12:04 AM

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#4
Jul508, 12:15 AM

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I: irrational or rational?
Well yeah, I suppose I inadvertently conveyed the impression that imaginary numbers were either irrational or rational.



#5
Jul508, 02:57 AM

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You could extend the concept by considering complex numbers whose components are both rational. This might be useful in proving things, as it's dense in C.



#6
Jul508, 09:48 AM

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Is it correct to say that the property of being rational or irrational applies only to the individual parts of a complex number? In other words, if the complex number c=(a,b), it makes sense to ask if a or b are rational, but not if c is rational? After all, in the complex number c=(a,b) , a could be rational, and b could be irrational.



#7
Jul508, 09:51 AM

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If you're considering algebraic number theory, then you're usually interested in some base ring, and the word "rational" means it's a quotient of things in that base ring, whereas "irrational" means its not.
e.g. if we're working in Z, then both [itex]i / 2[/itex] and [itex]i \sqrt{2}[/itex] are irrational (over Z). However, if our ring of interest is Z[i], the Gaussian integers, then [itex]i / 2[/itex] would be rational, but [itex]i \sqrt{2}[/itex] would be irrational. (over Z[i]) And if we were working in [itex]\mathbf{Z}[i \sqrt{2}][/itex], then [itex]i \sqrt{2}[/itex] would be rational. 


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Jul508, 10:44 AM

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#9
Jul508, 04:30 PM

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#10
Jul508, 06:20 PM

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I’ve seen a dialog like this before, where a post expecting an answer requiring skill level A to understand is followed by a discourse at an advanced skill level B.
My understanding of the HallsOfIvy response is that under the “basic” concept of “rational” and “”irrational”, these terms can be applied to the a and b parts of a complex number c = (a, b), but not to c itself, because the terms apply only to real numbers. A picky point, but pertinent to the original post. 


#11
Jul508, 08:21 PM

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In general, we this syntax denotes 'adjoining' a new element  Z[i] denotes the ring you get by taking the elements of Z, the additional element i of the complex numbers, and everything else you can produce through algebraic manipulation. (+, , and *, since we're thinking of rings) So Z[i] is the set of all polynomial expressions with 'variable' i and integer coefficients. i² simplifies, of course, meaning Z[i] is the set of all numbers of the form a+bi with a,b being integers  and that is precisely the ring of Gaussian integers. As far as I know, this mainly becomes a useful notion only in algebraic number theory and algebraic geometry. In ANT, because you often study an individual number field, or the relationship between different number fields, and so it is useful to distinguish between numbers in your number field and complex numbers not in your number field. In AG, you often face the same situation, but with a finite field. For example, you might be studying curve defined by a polynomial with coefficients in a field, but be interested in its 'points' with coordinates possibly in an extension field. 


#12
Jul508, 09:07 PM

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#13
Jul608, 06:18 AM

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No, Hurkyl's definition: "Numbers of the for m+ ni with m and n integers" is correct.



#14
Jul608, 06:57 AM

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Well, what is "[itex]\mathbb{Z}[i]/i^2+1[/itex]" supposed to mean? It looks like Luke is adjoining an indeterminate variable, denoted by i, and then modding out by the the ideal generated by the polynomial i^2+1. If this is the case, then [itex]\mathbb{Z}[i]/(i^2+1)[/itex] is also the Gaussian integers. Maybe using 'i' is a bit out of place here, because if we take it to mean [itex]\sqrt{1}[/itex], then modding out by (i^2+1) accomplishes nothing. What I'm saying is that [itex]\mathbb{Z}[i][/itex] is really shorthand for [itex]\mathbb{Z}[x]/(x^2+1)[/itex].



#15
Jul608, 08:29 AM

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Morphism is right  but I just wanted to elaborate a bit more. Adjoining can be used in (at least) two different ways:
(1) To adjoin an 'existing' variable, such as in the construction of the Gaussian integers Z[i] (2) To adjoin a 'free' variable, such as in the construction of the polynomial ring Z[x] The technical execution is the same, at least from one POV; the only difference is for the 'free' variable we do not impose any relations aside from those implied by the algebraic structure (e.g. x+x=2x), whereas we let the existing variable retain those relations it already satisfies in its parent structure (e.g. i²=1). One common use of (1) is when you are working within some ambient structure. For example, it's appropriate if you view your number field as a subrings of C rather than as abstract finite field extensions of Q. Another example is if you are, for some reason, interested in the subring of even polynomials: the subring Z[x²] of Z[x]. 


#16
Jul608, 03:59 PM

P: 355

Ah ok. I just hadn't seen that notation before.



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