Negative times negative makes positive

[sethoflagos]

1 hour ago, Genady said:

I'd say that the basic operations are ascend/descend-by-one. Then to ascend/descend-by-two you just ascend/descend-by-one twice, to ascend/descend-by-three you ascend/descend-by-one three times, etc. To ascend/descend-by-zero you then ascend/descend-by-one zero times, i.e., you don't move. I think it is not difficult to swallow that to ascend/descend-by-(-1) you descend/ascend-by-one, to ascend/descend-by-(-2) you descend/ascend-by-two, etc.

Pretty much it. Main point is kids understand stairs so there's a clear physical and intuitive link to each operation that you can leverage. Give boys half a chance to compete at who can take stairs two or three at a time and they'll be doing it at playtime too.

[joigus]

14 hours ago, Eise said:

Shouldn't the sum of all spins of the fermions in my body sometimes add to n + 1/2 so I am a fermion as a whole? Then by turning around 2 times I should look in the opposite direction. I try several times per day, but it never happened. Physics is wrong!

Not exactly. You might be made up of an even number of fermions and thereby operate as a boson. But this is a very interesting point, that I've thought about many times.

Let's assume for the sake of argument that you're made up of an odd number of fermions. Your wave function, after being rotated 360º (which for just one point in space looks like a sequence of two reflections) would be minus your original wave function.

This has nothing to do with ordinary space. 'You' haven't been reflected at all. This all happens in the space of states (quantum amplitudes). So how do we know your wave function has changed its phase to produce a global minus sign?

The only way to do it is to prepare a high number of indistinguisable Eises with the same number of fermions, and make that number be odd by design. Then make sure this number doesn't change. And lastly conceive of a way to make all these Eises interfere with each other (like for example throwing them through a double-slit screen). Something like that. Forgive me if the details of the experiment are not watertight.  

[Eise]

9 hours ago, joigus said:

Forgive me if the details of the experiment are not watertight.  

Oh, I can wait. I trust you that there is no danger in the experiment. Or? Joigus?

[joigus]

49 minutes ago, Eise said:

Oh, I can wait. I trust you that there is no danger in the experiment. Or? Joigus?

I would never harm a senior member on purpose. But I'm not to be trusted with experimental equipment.

I wouldn't harm any member, actually.

[studiot]

Until Genady posted this I had never really thought about the issue.
I just followed the rules.

One very important thing that has come out of my share dealing example is this.

 

It is no use whatsoever finding a single example of a quantity that can be measured/signed as positive or negative.

You need two separate quantities.

And these quantities must be connected by a multplicative connection.

 

 

[Genady]

33 minutes ago, studiot said:

It is no use whatsoever finding a single example of a quantity that can be measured/signed as positive or negative.

You need two separate quantities.

And these quantities must be connected by a multiplicative connection.

This is a good summary. +1

[swansont]

As I recall what I was taught was in terms of the number line. Positive to the right. Negative meant “opposite” so it was to the left. Addition/subtraction was “jumps” and multiplication was like an expansion or magnification, but the negative meant in the opposite direction.

[John Cuthber]

Repeatedly taking away a debt is the same as repeatedly adding a credit.

[joigus]

On 10/28/2024 at 5:02 PM, studiot said:

Yes that's not a bad explanation (but not proof) for beginners.

Consider a transformation T such that T² (-1) = 1

We could also compare a transformation T² (-1) = -1

 

The first transformation might actually be a reflection.

The second one will be equivalent to introducing i.

You're right. There's nothing a priori that says what (-1)² should be. I suppose it's rather a question of how far you can go with a definition like this and not find that it's inconvenient for certain purposes. As you well know, multiplication by complex numbers is better suited to represent rotations. Spatial reflections are a better embodiment of complex conjugation really. At least in 2D.

Ultimately, I don't think one can prove that (-1)*(-1)=1, and one must decide what the suitable definition for the purposes of extending the system in a useful way.

How many times have we been told something like 'you will understand later'?

Btw, I think your example for i would be the first one rather. Wouldn't it?

[studiot]

27 minutes ago, joigus said:

Btw, I think your example for i would be the first one rather. Wouldn't it?

I don't think I got it wrong.

T² (i) = (i)  x  (i)  = -1

[joigus]

44 minutes ago, studiot said:

I don't think I got it wrong.

T² (i) = (i)  x  (i)  = -1

Ah. Ok. You wrote T²(-1)=-1

So I understood something like (i)x(i)(-1)=-1, which didn't seem right.

[sethoflagos]

On 10/29/2024 at 11:58 PM, joigus said:

You're right. There's nothing a priori that says what (-1)² should be 

Ultimately, I don't think one can prove that (-1)*(-1)=1

So does -1 stand to 1 as i does to -1, no more no less?

[joigus]

13 hours ago, sethoflagos said:

So does -1 stand to 1 as i does to -1, no more no less?

I don't know what to say, Seth.  I would try not to look for paralellism between -1/1 and i/-1. I think they're different couples. But I'm guessing you're thinking about something having to do with the square root... Am I right?

My usual approach for these properties (that try to explore the limits of what we (as students) previously knew about the numbers) is to try to find an intuitive hookup first, and then try to convince the students that it's a good thing to try to extend the idea so that one doesn't have to change the rules. (Mind you, I've used the word 'try' 5 times in the same paragraph).

Example: Why is n⁰=1 for n>0 or n<0? Because we wish to preserve the property n^k+j=n^kn^j. So pick k=0, and j=anything, and there you are.

[Genady]

37 minutes ago, joigus said:

I would try not to look for paralellism between -1/1 and i/-1. I think they're different couples.

I agree. For example, why i/-1 and not -i/-1?

[sethoflagos]

43 minutes ago, joigus said:

I don't know what to say, Seth.  I would try not to look for paralellism between -1/1 and i/-1. I think they're different couples. But I'm guessing you're thinking about something having to do with the square root... Am I right?

I'm sure I missed most of the subtlety of @studiot's point, but if as he seems to be saying, (-1)*(-1) only has a value of +1 by convention, then reading it backwards, -1 is simply a label for the 'other' root of +1, isn't it?. This instantly reminded me of my picture of +/-i being axiomatic labels for the roots of -1.

As someone who habitually conceptualises ideas in geometrical terms, the paralellism is certainly apparent when the two relationships are expressed as rotations of a unit vector in the complex plane: two rotations of 0 or pi radians restore the unit vector; two rotations of +/-i*pi/2 give the negative unit vector.

1 hour ago, joigus said:

Example: Why is n⁰=1 for n>0 or n<0? Because we wish to preserve the property n^k+j=n^kn^j.

... or to preserve the state of the unit vector under zero rotations.

NB this is far from the OP objective: just trying to better explain what seems to have been seen as a daft idea.

[Genady]

10 minutes ago, sethoflagos said:

-1 is simply a label for the 'other' root of +1, isn't it?

Regardless of being a root of +1, -1 is a label for the additive inverse of +1. How come we use the same label?

[sethoflagos]

3 minutes ago, Genady said:

Regardless of being a root of +1, -1 is a label for the additive inverse of +1. How come we use the same label?

My first wild stab in the dark would be their shared initial value and symmetry about zero. = common end point?

[Genady]

If we accept that -1 and +1 are additive inverses, and if we want to keep the distributive property of multiplication, then it appears that we don't have a choice but make (-1)×(-1)=+1. Here it goes:

(-1)×(-1)=(1-2)×(-1)=1×(-1)-2×(-1)=(-1)-(-1)-(-1)=(-1)+(+1)+(+1)=+1

QED

[studiot]

7 hours ago, sethoflagos said:

I'm sure I missed most of the subtlety of @studiot's point, but if as he seems to be saying, (-1)*(-1) only has a value of +1 by convention, then reading it backwards, -1 is simply a label for the 'other' root of +1, isn't it?. This instantly reminded me of my picture of +/-i being axiomatic labels for the roots of -1.

As someone who habitually conceptualises ideas in geometrical terms, the paralellism is certainly apparent when the two relationships are expressed as rotations of a unit vector in the complex plane: two rotations of 0 or pi radians restore the unit vector; two rotations of +/-i*pi/2 give the negative unit vector.

You got it. +1

7 hours ago, Genady said:

If we accept that -1 and +1 are additive inverses, and if we want to keep the distributive property of multiplication, then it appears that we don't have a choice but make (-1)×(-1)=+1. Here it goes:

(-1)×(-1)=(1-2)×(-1)=1×(-1)-2×(-1)=(-1)-(-1)-(-1)=(-1)+(+1)+(+1)=+1

QED

Haven't you assumed that which was to be proven ?

[Genady]

1 hour ago, studiot said:

 

Haven't you assumed that which was to be proven ?

I have assumed only that multiplication is distributive (on the step you have marked):

(a-b)×c=a×c-b×c

In the case above, a=1, b=2, c=-1.

Edited 4 hours ago by Genady