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I think I see where you're hung up. If I change it to:

    x = 1e20 + 1 - 1e20
    y = 1e20 - 1e20 + 1
    assert x == y

There is only 1 digit, and it's wrong. You don't even need 6-8. I probably should've used this as my example in the first place.

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You're making the same mistake but now it's less obvious because the change of scale. When you compare floating point numbers, simple == is not usually what you want; you need to compare them with a tolerance. Choosing the tolerance can be difficult, but in general when working with small numbers you need a small value and with large numbers you need a large value. This dataset involves datapoint(s) at 1e20; at that magnitude, whatever you're measuring, the error in your measurements is going to be way more than 1, so a choice of tolerance ≤ 1 is a mistake.

Ugh, you're preaching to the choir. I wasn't trying to make a point about the equality operator, I was trying to make a point about x and y being completely different. I must be really bad at communicating with people.

That's the thing: x and y are not "completely different"; they are within the tolerance one would use to compare them correctly.

So umm, more than a "difference in the realm of 6-8 significant digits"?

That construction can turn any residual N digits out into a first digit difference. It wouldn't matter without making comparisons with tolerance way under the noise floor of the dataset. But yes, you have technically invented a situation that differs from that real-world anecdote in regard to that property, in an extremely literal interpretation.

And here I was, worried I might be tedious or pedantic, trying to argue that floating point is just not that simple. You've really outdone me in that regard.

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