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Is the concept of an odd imaginary number defined/well-defined/used in mathematics? I searched around but couldn't find anything. Thanks!

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    $\begingroup$ You might look around in the Gaussian integers. They are complex numbers $z=x+iy$, with $x,y \in \mathbb{Z}$. You can certainly look at the parity of $x$ and $y$. $\endgroup$ Mar 14, 2011 at 2:15
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    $\begingroup$ I tried to manually calculate e^ipi, and *ODDLY, I can't find 0. Why is that ? $\endgroup$
    – jokoon
    Mar 14, 2011 at 9:34
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    $\begingroup$ @gokoon, $$e^{i\pi}=cos(\pi) + isin(\pi)=1 + 0i=1$$, so maybe you want the zero to come in when you express that as $$e^{i\pi} - 1 = 0$$. $\endgroup$
    – JWL
    Oct 4, 2011 at 7:08
  • $\begingroup$ oh yes thanks. imaginary numbers are tricky... $\endgroup$
    – jokoon
    Oct 4, 2011 at 14:04

6 Answers 6

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"Odd" has several meanings in mathematics: you have odd integers (those which are not multiples of $2$); you have odd functions (those that satisfy $f(-x) = -f(x)$ for all $x$); and possibly others.

If you want to stick to the first meaning, then there's two things to keep in mind: even for just real numbers, "odd" in the sense of "not a multiple of $2$" doesn't really work very well, because every real number is a multiple of $2$: given $r\in\mathbb{R}$, $r = 2\left(\frac{r}{2}\right)$, and $\frac{r}{2}$ is a real number. The same is true for complex numbers: if $a+bi\in\mathbb{C}$, then $a+bi = 2\left(\frac{a}{2} + \frac{b}{2}i\right)$, so every complex number would be "a multiple of $2$", so no complex number would be odd. So this concept does not really do much for complex numbers as a whole.

On the other hand, you can restrict to those complex numbers which have integer real and complex part: $a+bi$ with $a,b\in\mathbb{Z}$ (instead of $a,b\in\mathbb{R}$, like in $\mathbb{C}$). These are called the Gaussian integers because they were first studied by Gauss.

For these numbers, you can talk about "multiples of $2$": a Gaussian integer $a+bi$ is a multiple of $2$ if and only if both $a$ and $b$ are even: because if $a+bi = 2(x+yi)$ with $a,b,x,y\in\mathbb{Z}$, then $a=2x$ is even and $b=2y$ is also even. So the "odd Gaussian integers" would be all Gaussian integers that are not multiples of $2$, namely the $a+bi$ that have either $a$ or $b$ odd. Note that $1$ would be an "odd Gaussian integer" (which looks good, because $1$ is an odd integer), but then again so would $1+2i$ (which may not look so good).

There is also a slight wrinkle: in the integers, if you add two integers of the same parity, you will always get something that is "even", and if you add two integers of different parity you will get something that is "odd." This does not happen with the above notion of "even" in the Gaussian integers. For instance, $1+2i$ is "odd", and so is $2+i$; if we add them, we get $3+3i$ which is also "odd." In fact, we have four different kinds of Gaussian integers: the "even" ones (both real and imaginary parts are even); the "even-odd" ones (real part even, imaginary part odd); the "odd-even" ones (real part odd, imaginary part even); and the "odd" ones (both real and imaginary part odd). It is only if you add two of the same kind that you will get an "even" Gaussian, and if you add two different kinds you will get an "odd" Gaussian. So this concept of "even" and "odd" does not seem to behave like it does in the integers. Added. What is worse, as Bill points out in comments, even this does not work well with multiplication, since for example the product of an "even-odd" by an "even-odd" gives an "odd-even", not an "even-odd".

We might, then, want to look at another possibility.

Another possibility is to notice that $2i = (1+i)^2$, and $i$ is invertible in the Gaussian integers. So instead of looking at the multiples of $2$, you can try looking at the multiples of $1+i$ (just like you don't define "odd" in terms of multiples of $4$ in the integers; I bring up $4$ because $4=2^2$). When is a Gaussian integer a multiple of $1+i$? $$(x+yi)(1+i) = (x-y) + (x+y)i.$$ Can we recognize such numbers? I claim they are precisely those Gaussian integers $a+bi$ with $a+b$ even.

Indeed, if $a+bi$ is a multiple of $1+i$, then as above we have $a=x-y$ and $b=x+y$ for some integers $x$ and $y$, so $a+b = (x-y)+(x+y) = 2x$ is even. Conversely, suppose that $a+bi$ has $a+b$ even, $a+b = 2k$. Then $a-b$ is also even (since $a-b = (a+b)-2b$), so we can write $a-b = 2\ell$. Then \begin{align*} (k -\ell i)(1+i) &= (k+\ell) + (k-\ell)i\\ &= \left( \frac{a+b}{2} + \frac{a-b}{2}\right) + \left(\frac{a+b}{2} - \frac{a-b}{2}\right)i\\ &= a + bi, \end{align*} so $a+bi$ is a multiple of $1+i$. So if you define "odd" in terms of "multiple of $1+i$, then it corresponds precisely to whether or not $a\equiv b\pmod{2}$: if $a$ and $b$ have the same parity, then $a+bi$ is "even"; if $a$ and $b$ have different parity then $a+bi$ is "odd".

It also has the advantage of mirroring a bit better what happens with parity in the integers: if you add two "even" or two "odd" Gaussian integers (under this definition), then the sum is "even"; and if you add an "even" and an "odd" Gaussian integer you get an "odd" Gaussian integer. Also, if you multiply an "even" Gaussian by any Gaussian you get an "even" Gaussian: for if $a$ and $b$ have the same parity, then $$(a+bi) (x+yi) = (ax-by) + (ay+bx)i.$$ If both $a$ and $b$ are even, then so are $ax-by$ and $ay+bx$, so the result is even. If both $a$ and $b$ are odd, then either $x$ and $y$ have the same parity, in which case both $ax-by$ and $ay+bx$ are even; or else $x$ and $y$ have different parity, so that both $ax-by$ and $ay+bx$ are odd. Either way, the product is "even." Similarly, if you multiply two "odd" Gaussians, the result will be "odd."

So I think the latter concept is a bit more intuitive, but that may be just me.

Post data. There is in fact a lot of very interesting stuff in the background of the above; considering $1+i$ instead of $2$ in the Gaussian integers comes from Algebraic Number Theory.

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    $\begingroup$ If you accept Arturo's definition of odd as odd $a+b$ (which makes sense to me) the odd ones are one color of a checkerboard and the even ones are the other. $\endgroup$ Mar 14, 2011 at 2:52
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    $\begingroup$ The insightful and well-sequenced content of Mr. M's posts is pretty amazing. The quantity of answers just blows my mind! You are a true ambassador. (+1, obviously) $\endgroup$ Mar 14, 2011 at 3:01
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    $\begingroup$ Referring to your first line: '"Odd" has several meanings in mathematics'. There is an old riddle asking how you three boys can eat a total of 40 potatoes, with each one eating an odd number of potatoes. 38 is an odd number of potatoes to eat... $\endgroup$ Mar 14, 2011 at 4:27
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    $\begingroup$ @Ross: And of course, 2 is the oddest prime number... But both of these would be a meaning of "odd" in natural language, I think. (-: $\endgroup$ Mar 14, 2011 at 4:29
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    $\begingroup$ @Matt: Well, yes: there's all sorts of very good reasons for the definition, stemming from the fact that $(2)=(1+i)^2$, as discussed in Bill's post. The post data was meant to suggest that I didn't pull it out of my sleeve, but had good reasons for it. And since there are very good algebro-arithmetic reasons for it, it should not be surprising that it has a lot of very good algebro-arithmetic consequences. (-: $\endgroup$ Mar 14, 2011 at 21:07
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As I remarked in a post on casting out nines, we can apply parity arguments in any ring with $\:\!\Bbb Z/2\:\! $ as an image, e.g. all rationals with odd denominator, or the Gaussian integers $\,\mathbb Z[i],\,$ where the image $\ \mathbb Z[i]/(2,i\!-\!1) \cong \mathbb Z/2\ $ yields the parity definition that $\, a+b\,i\, $ is even iff $\, a\equiv b\pmod{\!2},\,$ i.e. if $\ a+b\,i\ $ maps to $\,0\,$ via the above isomorphism, which maps $\ 2\to 0,\ i\to 1$.

Generally it's easy to show that if $\,2\nmid f(x)\not\in\mathbb Z\,$ then the number of ways to define parity in the ring $\ \mathbb Z[w] \cong \mathbb Z[x]/f(x)\ $ is given by the number of roots of $\, f(x)\, $ modulo $2.\, $ For suppose that there exists a homomorphism $\ h\, :\, \mathbb Z[w]\to \mathbb Z/2.\,$ Then $\,w\,$ must map to a root of $\,f(x)\,$ in $\ \mathbb Z/2.\,$ Thus if $\, f(0)\equiv 0\pmod{\!2}\, $ then $\, \mathbb Z[w]/(2,w) \cong \mathbb Z[x]/(2,x,f(x)) \cong \mathbb Z/2\ $ by $\, x\mid f(x)\pmod{\!2},\,$ and if $\, f(1)\equiv 0\pmod{\!2}\ $ then $\, \mathbb Z[w]/(2,w\!-\!1) \cong \mathbb Z/2\ $ by $\, x\!-\!1\mid f(x)\pmod{\!2}. $

Let's consider some simple examples. Note $\ x^2+1\ $ has the unique root $\ x\equiv 1\pmod{\!2},\,$ so the Gaussian integers $\ \mathbb Z[i]\cong \mathbb Z[x]/(x^2+1)\ $ have a unique definition of parity - with $\,i\,$ being odd. Since $\ x^2+x+1\ $ has no roots modulo $\, 2,\, $ there is no way to define parity for the Eisenstein integers $\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2+x+1).\, $ Indeed since $\ w^3 = 1\ $ we infer that $\, w \equiv 1\pmod{\!2}\, $ contra $\ w^2+w+1 = 0.\, $ Otoh $\ \mathbb Z[w] \cong \mathbb Z[x]/(x^2+x+2)\ $ has $2$ parity structures since both $\,0\,$ and $\,1\,$ are roots of $\ x^2 + x + 2\ $ modul0 $\,2,\,$ so we can define $\,w\,$ to be either even or odd. A simple application is in this post where we use parity in the ring $\,\mathbb Z[\sqrt{5}]\,$ to prove that the integer $\,(9+4\sqrt{5})^n + (9-4\sqrt{5})^n\,$ is even.

Any parity argument that's truly ring-theoretic will generalize to any ring $\,R\,$ with parity, e.g.

Parity Root Test $\ $ Suppose that $\,f(x)\,$ is a polynomial with coefficients in a ring $\,R\,$ with parity. Then $\,f(x)\,$ has no roots in $\,R\,$ if $\,f(x)\,$ has constant coefficient and coefficient sum both being odd. Further, $\,f(x)\,$ has no roots in the fraction field of $\,R\,$ if $\,R\,$ is a domain and the leading coefficient of $\,f(x)\,$ is odd, and $\,0\,$ is the only element of $\,R\,$ that is divisible by arbitrarily large powers of $\,2.$

Proof $\ $ The test simply verifies $\ f(0)\equiv f(1)\equiv 1\pmod{\!2}\ $ i.e. $\, f(x)\,$ has no roots $\bmod 2.\, $ Therefore $\,f(x)\,$ has no roots in $\,R.\,$ For the fractional case, by the hypothesis, we may cancel powers from a fraction $\,a/b\,$ until $\,a,b\,$ are not both even. If $\,b\,$ is odd then $\,a/b\equiv a\pmod{\!2}\,$ would be a root, so $\,b\,$ is even and $\,a\,$ is odd. Then $\,0 = b^n\,f(a/b) = f_n a^n + b\,(\cdots) \equiv 1\pmod{\!2},\,$ a contradiction, since the leading coefficient $\,f_n\,$ and $\,a\,$ are both odd and $\,b\,$ is even.

As a corollary, we deduce that no ring $\,R\,$ with parity contains a primitive cube root of one. Indeed, if so, $\,R\,$ contains a root of $\ f(x) = x^2+x+1\,,\ $ contra $\,f(x)\,$ has no roots in $\,R\,$ by the parity root test. Hence, for example, the Gaussian integers do not contain a primitive cube root of one.

Analogous ideas apply to generalize the above to testing unsolvability by finite case analysis in any finite image ring $S$, e.g. modular root test - which generalizes the above from $S = \Bbb Z_2$ to $\,\Bbb Z_n$.

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    $\begingroup$ I like the negative examples with no (simple) version of parity. Maybe it is good to mention the algebraic words: if 2 is inert, you need 4 kinds (not just even and odd), if 2 ramifies you get a single natural definition of parity, and if 2 splits you get two contradictory but natural definitions of parity. $\endgroup$ Mar 14, 2011 at 3:32
  • $\begingroup$ (btw you have a few Z's that aren't mathbb'd) $\endgroup$ Mar 14, 2011 at 3:32
  • $\begingroup$ @Jack: I'm always short on $\rm\:\mathbb Z$'s ... I tried not to get too technical since it seems such posts don't get read. $\endgroup$ Mar 14, 2011 at 3:40
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    $\begingroup$ @Bill Dubuque You write in a very terse style, so your posts take some time to go through, for me at least, but I usually find time to read through them and they are always very rewarding. Your banging away at the Rational Root Test is not for 'naught...;) $\endgroup$
    – Uticensis
    Mar 14, 2011 at 3:44
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    $\begingroup$ @Jack: Your summary would work for quadratic fields; for arbitrary fields, you get a single definition of parity if $2$ ramifies completely; if it is inert you need $2^n$ kinds (where $n$ is the degree of $f(x)$); if it splits completely you get $n$ different incompatible definitions; and somewhere in between for the other cases. $\endgroup$ Mar 14, 2011 at 14:56
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Sure. If you define an even number as any number $k$ expressible as $k=2z$ where $z$ is a Gaussian integer, then an odd number would be any number $k$ for which no such $z$ can be found. This would be any number whose real or imaginary part (or both) is odd.

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The concepts of "even" or "odd" come from the ideal structure of the ring of integers (I am not certain if this will mean anything to you). In the integers, numbers are classified as "even" if their image in $\mathbb{Z}/(2)$ is $0$ and "odd" if it is 1. The more useful concept of "even" or "odd" in the Gaussian integers would be to classify Gaussian integers by their image in $(\mathbb{Z}+\mathbb{Z}i)/(2)$. This would lead to four possibilities: numbers with both parts even, numbers with even real part and odd imaginary part, numbers with odd real part and even imaginary part, and numbers with both parts odd.

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I thought it might be interesting to consider when the norm $\bar{z} = a^2+b^2$ of a gaussian integer $z = a+bi$ $a,b \in \mathbb{Z}$ is even or odd. In particular, this agrees with Artuo Magidin's definition as it being even if $a+b$ is, since squaring doesn't change the parity of an integer.

Since the norm is multiplicative, we immediately recover that odd times odd is even, etc. Furthermore, since the norm of $z+x$, where x is another gaussian integer, $(a+c) + (b+d)i$, is $a^2 + c^2 + 2ac + b^2 + d^2 + 2bd = \bar{z} + \bar{x} + 2(ac+bd)$, and so also respects parity additively.

However, now we have a homomorphism to the integers. If we quotient the gaussian integers by its kernel (namely, 1 and i), the resulting ring is, of course, (isomorphic to) the integers.

So if we take the parity of the norm as the definition of parity, then in some sense this matches up with the parity of the original integers because when we go back to the integers, the parity is the same.

I hope this makes sense to someone besides me. As pointed out in the comments, this is incorrect.

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  • $\begingroup$ wait a minute, is that right? Z[i]/i should be isomorphic to Z, yes? $\endgroup$ Mar 14, 2011 at 18:20
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    $\begingroup$ You have to be a bit more careful in the third paragraph. The norm is a multiplicative semigroup homomorphism, but not a ring homomorphism: $$N((a+bi)+(c+di)) = N(a+bi) + N(c+di) \Longleftrightarrow ac=bd=0.$$ The kernel of the semigroup homomorphism (whose image is just the integers that can be written as the sum of two squares) consists of the pairs $(a+bi,c+di)$ with $a^2+b^2 = c^2+d^2$; this is more than merely the ideal generated by $1$ and $i$ (and the preimage of $1$ includes $-1$ and $-i$ as well). The quotient is the multiplicative semigroup of integers that are sums of two squares. $\endgroup$ Mar 14, 2011 at 18:25
  • $\begingroup$ oh yes, I knew there was something off. Thank you for pointing out my mistake. $\endgroup$ Mar 14, 2011 at 18:31
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Let $R$ be a unital, commutative ring. Let $I$ be any ideal such that $|R/I|=2$. Call an element $r \in R$ even if $r \in I$ and odd if $r \not\in I$. This implies that $r$ is even iff $r \equiv 0 \pmod I$ and $r$ is odd iff $r \equiv 1 \pmod I$. The operations of addition and multiplication behave as expected with respect to parity.

Examples:

$\mathbb{Z}$ is a PID whose ideals have the form $(p)$ for $p$ a prime. Thus, the only ideal of index 2 is $(2)$. This is the classical notion of even/odd.

$\mathbb{Z}[i]$ is a PID. One can show that the index of the ideal $(a) \subseteq \mathbb{Z}[i]$ is simply $|a|^{2}$. Thus, the only ideals of index 2 are those generated by $1+i$, $1-i$, $-1+i$, and $-1-i$.

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  • $\begingroup$ Since the link in the answer seems to be dead, here is link to the same site in Wayback Machine. $\endgroup$ Dec 8, 2016 at 3:38
  • $\begingroup$ This is the same parity definition explained in my answer (and its links) two years prior. Please don't post duplicate answers. $\endgroup$ Nov 22, 2021 at 0:07

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