Why is this not 0 on 50g or Prime?

[Joe Horn]

\[\frac { 1 }{ \infty +\cfrac { 1 }{ 0 } } =\frac { 1 }{ \infty +\infty } =\frac { 1 }{ \infty } =0\]The above is correct, no? Why then do both the HP 50g (in exact mode) and the HP Prime (in CAS view) say that the first expression, 1/(inf+1/0), is "undefined"?

[Voldemar]

http://www.wolframalpha.com/input/?i=1%2...B+(1%2F0))

[ijabbott]

(07-16-2018 03:35 PM)Joe Horn Wrote:  \[\frac { 1 }{ \infty +\cfrac { 1 }{ 0 } } =\frac { 1 }{ \infty +\infty } =\frac { 1 }{ \infty } =0\]The above is correct, no? Why then do both the HP 50g (in exact mode) and the HP Prime (in CAS view) say that the first expression, 1/(inf+1/0), is "undefined"?

Because \( \frac{1}{0} \) is undefined? Or failing that, because \( \infty + \frac{1}{0} = \frac{\infty \cdot 0 + 1}{0} \) and \(\infty \cdot 0\) is undefined?

[BartDB]

try the following:

\[\frac { 1 }{ \infty +\left | \cfrac { 1 }{ 0 } \right | } \]

Also in WA:
http://www.wolframalpha.com/input/?i=1%2...bs(1%2F0))


EDIT:
The reason for the WA answer is that 1/0 gives "complex infinity" which cannot be added to real infinity (definition of Complex Infinity: a complex quantity with infinite magnitude but indeterminate phase) .

The 50G also recognises 2 infinities: see AUR p3-289
"The calculator recognizes two kinds of infinity: signed and unsigned. Evaluating '1/0' gives an unsigned infinity. Selecting infinity from the keyboard ... returns '+inf' and the sign can be changed. Calculations with the unsigned infinity return unsigned infinity or ? as their result. Calculations with the signed infinity can return ordinary numeric results, as in the example. Positive infinity and unsigned infinity are equal if tested with ==, but are not identical if tested with SAME."

[Valentin Albillo]

(07-16-2018 03:35 PM)Joe Horn Wrote:  \[\frac { 1 }{ \infty +\cfrac { 1 }{ 0 } } =\frac { 1 }{ \infty +\infty } =\frac { 1 }{ \infty } =0\]The above is correct, no?

No.

Quote:Why then do both the HP 50g (in exact mode) and the HP Prime (in CAS view) say that the first expression, 1/(inf+1/0), is "undefined"?

Because it is. Mathematically, 1/0 is undefined, period,

Very, very informally you can see why this way:

The value of 1/0 depends on the sign of 0.

• In case of +0, 1/+0 would be +Inf and your denominator would be Inf + Inf = Inf. So far so good.
• In case of -0, 1/-0 would be -Inf and your denominator would be Inf - Inf = Undefined.

So, without knowing the sign of 0 the denominator is Undefined and the calculator is Ok.

Very informally. It can be made mora rigorous by taking limits, etc. but it would be overkill.

Regards.
V.
.

[Joe Horn]

Fascinating. I'm always learning. Thanks, guys!

[rprosperi]

(07-16-2018 05:38 PM)Valentin Albillo Wrote:  The value of 1/0 depends on the sign of 0.

Excluding limits when approaching from positive or negative, isn't 0 actually neither positive or negative by definition?

Maybe I've just been happily uninformed for many years...

(07-16-2018 05:38 PM)Valentin Albillo Wrote:  

• In case of +0, 1/+0 would be +Inf and your denominator would be Inf + Inf = Inf. So far so good.
• In case of -0, 1/-0 would be -Inf and your denominator would be Inf - Inf = Undefined.

I've never had a belief or opinion about these... are these simply defined as allowed and not defined respectively as a convention?

[KeithB]

My HP71 has +/- 0.

[Vtile]

This is interesting topic. Also the answer do depend if we allow hyper real definitions.

My stupid engineer head claims that 1/0 = inf. That by purely logical reasoning because 1/(1+inf) = 0. Also 0/0=1. So 0 is logical operator for something so nonexistently insignificant that you just don't bother and inf is something so enormous that you just don't have imagination and the 1 is the swap point of the numerical continuum.

Of course this is just personal view and as such a worthless little fallacy. Which I throw out of the window if it gives wrong solution for the task in hand.

Interesting is also how all the primes (P) do have equivalents in
1/P form, if I have understood correctly.

[Claudio L.]

This is quite a subject.

The IEEE standard decided to create the signed zero, so they could map 1/0 to either +Inf or -Inf. That was their solution to avoid the "Undefined".
What if you don't want to have signed zero?
In reality, 1/0 should be "complex infinity", which is a circle, not a point (all points at an infinite distance from zero, coming from any direction). If you define a complex infinity (Wolfram and any serious math package seem to prefer this), then you don't need a signed zero and problem solved. Or not? How do you even handle a complex infinity? The only operation you can define with it is 1/Inf = 0. Anything else you try will have to be "Undefined", since you have more than a point as a result.

Now if you don't want to deal with the concept of complex numbers, then you could take only the real axis, and define "unsigned infinity", which is 2 points at the intersection of the complex infinity circle with the +x and -x real axis. But in practice this is the same as the complex infinity: because it represents 2 points, any operation that's not 1/Inf will be undefined.
Then you may need to use the individual "values", +Inf and -Inf that you can actually define a few more operations on.

So now we have 4 different types of infinity (complex, unsigned, +Inf and -Inf) to make a real mess of everything.

For newRPL I didn't know what to do so I opted for:

0 means 0, there's no signed zero. I implemented the complex infinity but not the unsigned infinity.
Then for practical reasons, when the user disables the complex mode flag, 1/0= +Inf.
Why? Because VTile said it, take it up on him :-)
EDIT: If you disable complex mode, Joe Horn gets his expected result of 0 (granted, there will be an error saying "Infinite result" but you can ignore that), but if you activate complex mode, Valentin gets his more academically correct "Undefined".

[rprosperi]

(07-17-2018 01:20 PM)KeithB Wrote:  My HP71 has +/- 0.

Oh, it appears my HP71 has it too.

Though I can happily report that I've never encountered it, most likely because I'm a Mechanical Engineer and such notions never seem to crop up in the real world of measuring and calculating physical things.

Wonder if there's a flag to disable that, to ensure I don't encounter it in the future as well....

[Valentin Albillo]

.
Hi, Bob:

rprosperi Wrote:Oh, it appears my HP71 has it too.

Mine too, +0 and -0 (which played an essential part in one of my S&SMCs), plus +Inf, -Inf, signaling NaNs and silent (i.e.: unsignaling) NaNs, all of which can be the real and/or imaginary part of a complex number.

Regards.
V.
.

[rprosperi]

(07-18-2018 12:45 AM)Valentin Albillo Wrote:  Mine too, +0 and -0 (which played an essential part in one of my S&SMCs), plus +Inf, -Inf, signaling NaNs and silent (i.e.: unsignaling) NaNs, all of which can be the real and/or imaginary part of a complex number.

Yes, I explored most of these when I first got my 71 in the 80's, particularly finding it interesting to fabricate programs to generate NaNs (both kinds), but also of course creating programs that could handle (and even DISPLAY/PRINT) answers that include Infinity, thus proving my Numerical Methods Professor wrong, having claimed that any good program spends as much code trapping and preventing such cases to avoid crashes, as the actual intended calculations!

But I somehow never noticed the -0 capability, or possibly did but then seeing no use for it, successfully forgot it over the years.

Anyhow, I am thrilled to continue to learn new things about my favorite machine, so thanks to you guys for this interesting discussion.

[KeithB]

(07-17-2018 10:33 PM)Vtile Wrote:  . Also 0/0=1.

0/0 does not equal 1, it is undefined.

[Vtile]

(07-18-2018 01:41 PM)KeithB Wrote:  

(07-17-2018 10:33 PM)Vtile Wrote:  . Also 0/0=1.

0/0 does not equal 1, it is undefined.

First of yes, that is the right consensus and mathematically solid definition. Hence my note of my personal rather meaningless philosophical little thinking game.

However if 1/inf = 0 and (a/b)/(a/b) = 1 then it would be only reasonable to say that (1/inf)/(1/inf) = 1 and 0/0 = 1. However it is not allowed to declare inf = inf, but inf <> inf is only allowed. For mathematical rigor this (0/0=1) of course is not the case since there is these 'special' rules and blu tack for certain things.

[ijabbott]

(07-18-2018 04:05 PM)Vtile Wrote:  

(07-18-2018 01:41 PM)KeithB Wrote:  0/0 does not equal 1, it is undefined.

First of yes, that is the right consensus and mathematically solid definition. Hence my note of my personal rather meaningless philosophical little thinking game.

However if 1/inf = 0 and (a/b)/(a/b) = 1 then it would be only reasonable to say that (1/inf)/(1/inf) = 1 and 0/0 = 1. However it is not allowed to declare inf = inf, but inf <> inf is only allowed. For mathematical rigor this (0/0=1) of course is not the case since there is these 'special' rules and blu tack for certain things.

So if \(0/0 = 1\), does \(2\cdot 0/0 = 0/0 = 1\) or does \(2\cdot 0/0 = 2\cdot 1 = 2\)? Things are undefined for a reason!

[Vtile]

(07-18-2018 05:18 PM)ijabbott Wrote:  

(07-18-2018 04:05 PM)Vtile Wrote:  First of yes, that is the right consensus and mathematically solid definition. Hence my note of my personal rather meaningless philosophical little thinking game.

However if 1/inf = 0 and (a/b)/(a/b) = 1 then it would be only reasonable to say that (1/inf)/(1/inf) = 1 and 0/0 = 1. However it is not allowed to declare inf = inf, but inf <> inf is only allowed. For mathematical rigor this (0/0=1) of course is not the case since there is these 'special' rules and blu tack for certain things.

So if \(0/0 = 1\), does \(2\cdot 0/0 = 0/0 = 1\) or does \(2\cdot 0/0 = 2\cdot 1 = 2\)? Things are undefined for a reason!

Number 2 would be solution, for symbolic state (2a/b)*(b/a) = 2, where a=1 and b=inf.

.. But what if +inf + n = -inf

Mind you personal view and small amusing thinking game, like I said earlier if something is not working I throw it out of the window.

[klesl]

According to me
(2a/b)*(b/a) = (2a/a)*(b/b) = 2*something because the b/b is undefined (inf/inf)

[Vtile]

(07-18-2018 06:34 PM)klesl Wrote:  According to me
(2a/b)*(b/a) = (2a/a)*(b/b) = 2*something because the b/b is undefined (inf/inf)

In my little thinking bubble (inf/inf) would be 1. All this naturally arises from the nature of infinity and also the inverse of it as zero(, which would lead to something odd like 1/Real as a fraction, known as irrational numbers algebraic fraction, another blu tack of math, since by first defined rules of fraction such shouldn't exist) both have properties of numbers and symbolic operators in academic view of mathematics. Which my little game (which started btw. when the limits and infinity were force-fed to me long way back.) doesn't try to challenge, it just holds true in another dimension.

[BartDB]

(07-18-2018 04:05 PM)Vtile Wrote:  (a/b)/(a/b) = 1
...
(2a/b)*(b/a) = 2

Not for all values of a and b.

More generally \(\frac{x}{x}= 1\: for \: all \:x\neq 0\)

The discontinuities must always be excluded. For example in equations with fractions, the points where the denominator(s) = 0 must be excluded (have exception handling).