[AnthonyMouse]

You keep referring to the magnitude of the vector itself rather than the magnitude of its components.

> Vectors with larger magnitudes have larger magnitudes than vectors with smaller magnitudes do?

Vectors with more dimensions have larger magnitudes than vectors with fewer components, for the same average magnitude of the components. The distance between the origin and (1,1) is less than the distance between the origin and (1,1,1) even though the components in both cases all have magnitude 1.

[thaumasiotes]

> Vectors with more dimensions have larger magnitudes than vectors with fewer components, for the same average magnitude of the components.

Is this related to something that's been said so far?

>> [sidethread] The next step is them doing a black knight and pretending they didn't put in the requirement by hand.

Obviously, I didn't. It was already there before I made my first comment. Look up:

>>> Two complex numbers can have the same magnitude & be very far apart.

The only thing we've ever been discussing is what can happen between vectors of the same magnitude. But if you want to discuss what can happen between vectors of different magnitudes... everything I said is still true! It's easy to construct low-dimensional vectors with high magnitudes, and in fact the construction that I already gave, of interpreting large vectors within a space defined partially by themselves, will do the job.

[AnthonyMouse]

> Is this related to something that's been said so far?

Are you considering what all of this is supposed to be an analogy for?

Suppose autism has different components, something like this:

https://getgoally.com/blog/autism-spectrum-wheel/

You rate someone on each factor using the same scale, e.g. a real number from 0 to 1, or a scale of 1 to 10. The scale is arbitrary but consistent.

Then someone whose "average" rating is 0.5 on a scale of 0 to 1 can be farther away from someone else whose "average" rating is 0.5 when there are more factors. On a linear scale two people both at 0.5 have distance zero. On a two dimensional scale, you could have one at (0, 1) and one at (1, 0) and then each of their averages is still 0.5 but their distance is ~1.4.

That's what we're talking about.

[SiempreViernes]

I think their point boils down to the fact that you can require that all vectors have the same magnitude, irrespective of the dimensionality of the space, which is of course true.

The next step is them doing a black knight and pretending they didn't put in the requirement by hand.

[thaumasiotes]

Here's what you said:

> Your last comment is completely incorrect, [random gibberish]

Here's what you were referring to:

>> If you stick to the first quadrant / octant / whatever n-dimensional division of space where all coordinates are positive... I don't think the number of dimensions makes any difference there either? Any two vectors define a plane (or a line, or, if they're both zero, a point), so two vectors in a 500-dimensional space can't be farther apart from each other than is possible for two vectors in a 2-dimensional space. Those 500-dimensional vectors are already embedded in a 2-dimensional space.

All of those statements are, obviously, true. What did you think was incorrect?