Magnus Anderson wrote:https://en.wikipedia.org/wiki/Carry_(arithmetic) wrote:In elementary arithmetic, a carry is a digit that is transferred from one column of digits to another column of more significant digits. It is part of the standard algorithm to add numbers together by starting with the rightmost digits and working to the left. For example, when 6 and 7 are added to make 13, the "3" is written to the same column and the "1" is carried to the left. When used in subtraction the operation is called a borrow.

That's a Wikipedia article (not a "Math for Kids" website) describing a very specific way to find the sum of two quantities. It's definitely not the only way. You can add "6" and "7" together by placing six apples and seven oranges in front of yourself and then counting all of them. That's not the same as the above.

Addition is a much more general concept, and saying that I'm talking about addition in general is not the same as saying that I'm talking about some specific algorithm for adding numbers. I did not the do the former, I did the latter.

Thanks for the masterclass on carrying and borrowing - you could have stuck to your kid's website though - it's that basic. No wonder you're more qualified than professional mathematicians to determine these issues.

Are these processes how you're distinguishing the standard algorithm from addition "in general"?

You realise that carrying and borrowing are only superficial accomodations for numerical systems that use positional notation? Without positional notation, let's say we're using a conceptual numerical system that had an infinite number of digits, you wouldn't need to carry or borrow, because they don't actually do anything to the quantity itself that is being denoted. The standard algorithm is addition with superficial decoration - your argument that they're different is like saying a house is different to a house with graffiti on it.

This is probably indicative of a major difference between mathematicians and non-mathematicians: that mathematicians understand quantity at a deeper level since they traverse across multiple different representations depending on the job at hand and the best way to go about it. They don't get hung up on the extraneous and incidental, they don't confuse appearance for essence. This will be why non-mathematicians get so bewildered by \(1=0.\dot9\) because their intuitions lack adaptability to different representations of the same thing.

Magnus Anderson wrote:Silhouette wrote:By contrast, you even quoted one of my explanations on the last page.

Yes, and it's wrong, and I explained why, even though it's unnecessary. (One doesn't have to prove more than it's necessary.)

It's not wrong and as mathematicians will tell you - explaining why is always necessary.

Simply saying it's wrong backed by a vague objection and saying it's unnecessary to prove it more is a far cry from mathematical proofs that can require obnoxious amounts of backing just to be complete.

Your main objection is that \(\frac{9.\dot9}{10}\neq0.\dot9\) right?

As above, you're getting "carried" away by the superficial - thinking that \(\frac{9.\dot9}{10}\) literally shifts all the digits 1 space to the right (or the decimal point 1 space to the left) instead of that just being appearance, meaning there's an extra "unmatched 9" at the "end" of the endless sequence of 9s compared to the "end" of the endless sequence of 9s in \(0.\dot9\)

The whole reason I explicitly used this notion for \(\frac{s}{10}=\frac{9/{10^0}+9/{10^1}+9/{10^2}+...}{10}=\frac{9/1+9/{10}+9/{100}+...}{10}=9/{10^1}+9/{10^2}+9/{10^3}+...=\sum_{x=1}^\infty \frac9{10^x}\) was to emphasise the one-to-one correspondence for "each decimal place" in both \(\frac{s}{10}\) and \(s\) but by representing each as a fraction instead of a decimal place - to try and help you from getting confused about superficial appearance of decimals.

This matches the \(9/{10^1}\) in \(s\) with the \(\frac{9/{10^0}}{10}\) in \(\frac{s}{10}\) (i.e. each 9 in the tenths column together for both s and s/10)

It does NOT match the \(9/{10^1}\) in \(s\) with the \(\frac{9/{10^1}}{10}\) in \(\frac{s}{10}\) (i.e. not the 0.9 in s with the 0.9/10=0.09 in s/10)

No doubt this still went over your head and you still got confused by the appearance of \(9/{10^1}\) looking the same in either, even though one was divided by 10 in \(\frac{s}{10}\) (obviously) and the other in \(s\) wasn't (obviously).

Magnus Anderson wrote:Silhouette wrote:Quite obviously I know what I'm talking about, so maybe you should just stop insisting I don't, what do you think?

This is one of the most pointless things you can say in a forum discussion.

This was exactly why it was a rephrasing of something you said to me, that "You have no clue what you're talking about, so maybe you should just stop it and instead stick to the topic, what do you think?"

The whole point of me saying it was to demonstrate to you that what you just said was "one of the most pointless things you can say in a forum discussion."

God, you miss literally everything...

You wonder why I'm getting so frustrated?

And you think I'm proud of it?!

I couldn't be further from proud of the fact that even the most clear and obvious explanations mean nothing to you.

I love how you've just brought up Hilbert's Hotel too, because it's just another one of those thought experiments that demonstrates the indeterminate sizelessness of infinity (the hotel with infinite rooms in it) and therefore disproves all your nonsense about sizes of infinity. There's still only 1 infinity (the hotel) whatever happens to it, and regardless of any finite operations/adjustments that you make to its rooms, it's counterintuitively still full whilst also able to have spaces. Operating on infinity does nothing definite to the fact that it's still infinite.

This is what indeterminate/undefinable/infinite/endless amounts to.

You can emphasise that, no matter how many 9s in \(0.\dot9\) fill the gap between it and \(1\), there's always an infinite number of smaller infinitesimal gaps that could be filled, and I can emphasise that infinite recurring 9s will always fill these gaps and both/neither of us would be fully right, because infinity is undefinable, which is why things like \(\frac1\infty\) are invalid because it's the same as saying \(\frac1{undefinable}\).

And since there's nothing definite about how many 9s there are in \(0.\dot9\) (including any "spare 9s" compared to \(9.\dot9\)), this is the quality of endlessness that changes all your finite understandings of progressions like \(0.9+0.1=1\) and \(0.09+0.01=1\) and so on such that \(1=0.\dot9\) - there's no gap because the indeterminacy of endlessness/infinity means there's no way to determine any gap.

Yet another set of explanations that really really should put this topic to rest.

No doubt it won't. Disappoint me further.