## Little Problem with Complex Mathematics on Calculators

General discussion about calculators, Swiss Micros or otherwise
Thomas Okken
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### Re: Little Problem with Complex Mathematics on Calculators

H2X wrote:
Wed Feb 20, 2019 6:51 am
My intuition demands not that infinity be a number or have limits, but it insists that there is structure. It would be surprised if there is a finite "number" of such structures.
Allow me to repeat myself:

https://en.m.wikipedia.org/wiki/Aleph_number

There are infinitely many different infinite cardinalities. The cardinality of the integers, pairs of integers, rational numbers, etc., is aleph 0 in that structure, while the cardinality of the reals is aleph 1. I've personally never felt the urge to explore deeper into that rabbit-hole.

H2X
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### Re: Little Problem with Complex Mathematics on Calculators

Thomas Okken wrote:
Wed Feb 20, 2019 7:17 am
H2X wrote:
Wed Feb 20, 2019 6:51 am
My intuition demands not that infinity be a number or have limits, but it insists that there is structure. It would be surprised if there is a finite "number" of such structures.
Allow me to repeat myself:

https://en.m.wikipedia.org/wiki/Aleph_number

There are infinitely many different infinite cardinalities. The cardinality of the integers, pairs of integers, rational numbers, etc., is aleph 0 in that structure, while the cardinality of the reals is aleph 1. I've personally never felt the urge to explore deeper into that rabbit-hole.
Thanks for repeating yourself, Thomas. I should have read it the first time!

Personally, I'd love to see the singularity. As long as it doesn't require jumping into a black hole...
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J-F Garnier
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### Re: Little Problem with Complex Mathematics on Calculators

Walter wrote:
Sat Feb 16, 2019 10:16 am
My basic question was: why is i ∞ = 0 + i ∞ but it looks that (i ∞)^2 ≠ (0 + i ∞)^2 ?

Reasoning for the second equation:
(i ∞)^2 = -∞ while (0 + i ∞)^2 = 0 - ∞ + 2 i 0 ∞ (this expansion of the squared parenthesis doesn't assume anything IMO).
For making both sides equal, 0 × ∞ must be zero.
OTOH, WolframAlpha claims 0 × ∞ being undefined, and the 34S returns 0 × ∞ being 'not numeric'.
At the same time, WolframAlpha claims (0 + i ∞)^2 = -∞ while the 34S claims (0 + i ∞)^2 = -∞ + i × NaN; and I don't see that NaN = 0.
There must be a fundamental error in this reasoning but I can't find it.
If I remember a course I watched some time ago, the root cause is that when you add the Infinity quantity to the Real number set, you can't apply all the usual arithemtic rules to Infinity.
A trivial example, from 2+Inf = 3+Inf , you can't deduce 2 = 3. This is similar to the more familiar kind of expressions 2*0 = 3*0 from which we don't deduce 2=3 since both sides are 0.
So I think (but not 100% sure) that you can't deduce the 0*Inf value from (i ∞)^2 = (0 + i ∞)^2 since both sides are Infinity.

J-F

ijabbott
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### Re: Little Problem with Complex Mathematics on Calculators

rprosperi wrote:
Tue Feb 19, 2019 7:49 pm
ijabbott wrote:
Tue Feb 19, 2019 6:00 pm
It seems intuitive that there should be more rational numbers than integers, but this is where intuition is wrong. The set of integers and the set of rational numbers have the same size in the same way that the set of even integers has the same size as the set of integers. However, the set of real numbers is larger (and is the same size as the set of complex numbers).
I could not begin to disagree, but could you please explain that, as neither of these things you note as having the same size feel intuitive to me, and that Reals and Complex are larger (but equal) is just as unclear. When you have time is fine, this is only a curiosity for me, and if I knew how to just google this, I'd do that.
I see Thomas Okken has already done a marvellous job, but I think the main thing that equates the cardinalities of the set of integers and the set of rationals is that both types of numbers are countable by some progressive sequence that tags each element of the set with successive natural numbers. They are infinite, but both "countably infinite".

There may be more than one way to count them, but the important point is that you can indeed progressively count them (although you'll never reach the end of course because they are endless).

For example, here is one possible way to progressively order the integers: 0, 1, -1, 2, -2, 3, -3, ....

And here is one way to progressively order the rationals (but ignore the ones in parentheses as they are not in simplest form and serve merely to show the pattern being followed): 0, 1/1, -1/1, 1/2, -1/2, 2/1, -2/1, 1/3, -1/3, 3/1, -3/1, 2/3, -2/3, 3/2, -3/2, 1/4, -1/4, 4/1, -4/1, (2/4, -2/4, 4/2, -4/2, ) 3/4, -3/4, 4/3, -4/3, 1/5, -1/5, ....

(In both cases, I haven't shown the natural numbers used to tag the elements, but you can just use the position of the element in the sequence as the tag.)

I haven't touched on the cardinality of the set of complex numbers versus reals as it's not quite as easy to illustrate, but as an extremely non-rigourous explanation, a complex number can be considered to be an ordered pair of real numbers (along with a set of arithmetic rules for combining with other such pairs) in the same way that a rational number can be considered as an ordered pair of integers (with common factors removed, and obeying their own rules of arithmetic), and you get the idea that the sets may be the same size (although I have by no means proved it here).

J-F Garnier
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### Re: Little Problem with Complex Mathematics on Calculators

J-F Garnier wrote:
Wed Feb 20, 2019 10:07 am
Walter wrote:
Sat Feb 16, 2019 10:16 am
My basic question was: why is i ∞ = 0 + i ∞ but it looks that (i ∞)^2 ≠ (0 + i ∞)^2 ?

Reasoning for the second equation:
(i ∞)^2 = -∞ while (0 + i ∞)^2 = 0 - ∞ + 2 i 0 ∞ (this expansion of the squared parenthesis doesn't assume anything IMO).
For making both sides equal, 0 × ∞ must be zero.
OTOH, WolframAlpha claims 0 × ∞ being undefined, and the 34S returns 0 × ∞ being 'not numeric'.
At the same time, WolframAlpha claims (0 + i ∞)^2 = -∞ while the 34S claims (0 + i ∞)^2 = -∞ + i × NaN; and I don't see that NaN = 0.
There must be a fundamental error in this reasoning but I can't find it.
If I remember a course I watched some time ago, the root cause is that when you add the Infinity quantity to the Real number set, you can't apply all the usual arithemtic rules to Infinity.
A trivial example, from 2+Inf = 3+Inf , you can't deduce 2 = 3. This is similar to the more familiar kind of expressions 2*0 = 3*0 from which we don't deduce 2=3 since both sides are 0.
So I think (but not 100% sure) that you can't deduce the 0*Inf value from (i ∞)^2 = (0 + i ∞)^2 since both sides are Infinity.
I was re-reading the article on the 71 Math ROM (worth to read - several times) in the HP Journal, Juy 1984, and found that the 71 Math ROM designers had this very problem with infinity and complex numbers, more than 35 years ago !
They solved it by ignoring the finite part when a infinity quantity is detected in a complex number, and indeed the 71 - with Math ROM - gives (0,Inf)^2 -> (-Inf,0) without any kind of 0*Inf warning, Nan, or so.
More details in the article p27.

J-F

Walter
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### Re: Little Problem with Complex Mathematics on Calculators

J-F Garnier wrote:
Mon Nov 04, 2019 9:37 am
I was re-reading the article on the 71 Math ROM (worth to read - several times) in the HP Journal, Juy 1984, and found that the 71 Math ROM designers had this very problem with infinity and complex numbers, more than 35 years ago !
They solved it by ignoring the finite part when a infinity quantity is detected in a complex number, and indeed the 71 - with Math ROM - gives (0,Inf)^2 -> (-Inf,0) without any kind of 0*Inf warning, Nan, or so.
More details in the article p27.
Thanks a lot for finding and sharing, Jean-Francois! Merci bien!

EDIT: The left column of p. 27 looks especially interesting.

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Pirx
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### Re: Little Problem with Complex Mathematics on Calculators

Walter, the main fact that you are missing is that "∞" is not a number.
You cannot do arithmetic with it.

Walter
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### Re: Little Problem with Complex Mathematics on Calculators

Pirx wrote:
Sun Nov 17, 2019 6:26 am
... "∞" is not a number.
You cannot do arithmetic with it.
Depends. IIRC, IEEE 754 takes ±∞ as numeric (and NaN as "not a number"). You can do some arithmetic with ±∞ but have to be careful in transforms of complex numbers. E.g. ∞ ∡ π/2 = 0 + i×∞ which you won't get by applying the standard polar to rectangular transformation formula.
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Walter
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### Re: Little Problem with Complex Mathematics on Calculators

J-F Garnier wrote:
Mon Nov 04, 2019 9:37 am
I was re-reading the article on the 71 Math ROM (worth to read - several times) in the HP Journal, Juy 1984, and found that the 71 Math ROM designers had this very problem with infinity and complex numbers, more than 35 years ago !
They solved it by ignoring the finite part when a infinity quantity is detected in a complex number, and indeed the 71 - with Math ROM - gives (0,Inf)^2 -> (-Inf,0) without any kind of 0*Inf warning, Nan, or so.
More details in the article p27.
The longer I look at it the more I can concur with Jean-Francois recommending this article. Merci encore une fois!
DM42 SN: 00041 Beta
WP 43S running on this device

HP-35, HP-45, ..., HP-50, WP 34S, WP 31S, DM16L