Tomorrow_Farewell

joined 8 months ago
[–] Tomorrow_Farewell@hexbear.net 1 points 3 months ago* (last edited 3 months ago)

Now, ask yourself this question, 'is 0.999..., or any real number for that matter, a series?'. The answer to that question is 'no'.

You seem to be extremely confused, and think that the terms 'series' and 'the sum of a series' mean the same thing. They do not. 0.999... is the sum of the series 9/10+9/100+9/1000+..., and not a series itself.

EDIT: Also, the author does abuse the notations somewhat when she says '1+1/2+1/4 = 2' is a geometric series, as the geometric series 1+1/2+1/4+... does not equal 2, because a series is either just a formal sum, a sequence of its terms, or, in German math traditions, a sequence of its partial sums. It is the sum of the series 1+1/2+1/4+... that is equal to 2. The confusion is made worse by the fact that sums of series and the series themselves are often denoted in the same way. However, again, those are different things.
Would you mind providing a snippet with the definition of the term 'series' that she provides?

EDIT 2: Notably, that document has no theorem that is called 'convergence theorem' or 'the convergence theorem'. The only theorem that is present there is the one on convergence and divergence of geometric series.

[–] Tomorrow_Farewell@hexbear.net 1 points 3 months ago (2 children)

Ok. In mathematical notation/context, it is more specific, as I outlined.

It is not. You will routinely find it used in cases where your explanation does not apply, such as to denote the contents of a matrix.

Furthermore, we can define real numbers without defining series. In such contexts, your explanation also doesn't work until we do defines series of rational numbers.

Ok. Never said 0.999... is not a real number

In which case it cannot converge to anything on account of it not being a function or any other things that can be said to converge.

because solving the equation it truly represents, a geometric series, results in 1

A series is not an equation.

This solution is obtained using what is called the convergence theorem

What theorem? I have never heard of 'the convergence theorem'.

0.424242... solved via the convergence theorem simply results in itself

What do you mean by 'solving' a real number?

0.999... does not again result in 0.999..., but results to 1

In what way does it not 'result in 0.999...' when 0.999... = 1?

You seem to not understand what decimals are, because while decimals (which are representations of real numbers) '0.999...' and '1' are different, they both refer to the same real number. We can use expressions '0.999...' and '1' interchangeably in the context of base 10. In other bases, we can easily also find similar pairs of digital representations that refer to the same numbers.

I meant what I said: "know patterns of repeating numbers after the decimal point."

What we have after the decimal point are digits. OTOH, sure, we can treat them as numbers, but still, this is not a common terminology. Furthermore, 'repeating number' is not a term in any sort of commonly-used terminology in this context.

The actual term that you were looking for is 'repeating decimal'.

Perhaps I should have also clarified known finite patterns to further emphasize the difference between rational and irrational numbers

No irrational number can be represented by a repeating decimal.

[–] Tomorrow_Farewell@hexbear.net 1 points 3 months ago* (last edited 3 months ago) (4 children)

The explanation I've seen is that ... is notation for something that can be otherwise represented as sums of infinite series

The ellipsis notation generally refers to repetition of a pattern. Either ad infinitum, or up to some terminus. In this case we have a non-terminating decimal.

In the case of 0.999..., it can be shown to converge toward 1

0.999... is a real number, and not any object that can be said to converge. It is exactly 1.

So there you go, nothing gained from that other than seeing that 0.999... is distinct from other known patterns of repeating numbers after the decimal point

In what way is it distinct?
And what is a 'repeating number'? Did you mean 'repeating decimal'?

[–] Tomorrow_Farewell@hexbear.net 4 points 3 months ago

The decimals '0.999...' and '1' refer to the real numbers that are equivalence classes of Cauchy sequences of rational numbers (0.9, 0.99, 0.999,...) and (1, 1, 1,...) with respect to the relation R: (aRb) <=> (lim(a_n-b_n) as n->inf, where a_n and b_n are the nth elements of sequences a and b, respectively).

For a = (1, 1, 1,...) and b = (0.9, 0.99, 0.999,...) we have lim(a_n-b_n) as n->inf = lim(1-sum(9/10^k) for k from 1 to n) as n->inf = lim(1/10^n) as n->inf = 0. That means that (1, 1, 1,...)R(0.9, 0.99, 0.999,...), i.e. that these sequences belong to the same equivalence class of Cauchy sequences of rational numbers with respect to R. In other words, the decimals '0.999...' and '1' refer to the same real number. QED.

[–] Tomorrow_Farewell@hexbear.net 1 points 3 months ago (2 children)

I ran $ curl -s "https://archlinux.org/mirrorlist/?country=FR&country=GB&protocol=https&use_mirror_status=on" | sed -e 's/^#Server/Server/' -e '/^#/d' | rankmirrors -n 5 -. Where should I see an option to update to KDE 6.1?

 

Basically the title.

Also, in case somebody could answer the question: is it possible to enable the Cube effect in a Virtualbox machine? I already have kdeplasma-addons, qt6-quick3d and qt6-quick3dphysics installed but I have no Cube effect option in window effects.

[–] Tomorrow_Farewell@hexbear.net 1 points 4 months ago

Yeah, I was trying to do stuff like sudo install. I was not aware that it is used in combination with other commands like that.

[–] Tomorrow_Farewell@hexbear.net 1 points 4 months ago (2 children)

I was wondering what I was doing wrong with sudo.

[–] Tomorrow_Farewell@hexbear.net 1 points 4 months ago (4 children)

Try Settings > Window Management > Desktop Effects > "Windows Management" section > Overview and configure it

The problem is that there is no such option there, which is why I'm asking.

It seems that I need to install kdeplasma-addons. I will need to look into how people usually use pacman, as it keeps asking me to login as root.

[–] Tomorrow_Farewell@hexbear.net 1 points 4 months ago (6 children)

Exactly

Didn't work.

Read my other comment. Maybe this is the breeze setting that does not have any way to change layout at the logging screen

This issue seems to be resolved now, but the only layouts that I can switch between are the ones I manually set in the settings of the OS after I managed to change the login screen to a different one that allowed me to input the password in English.

This is weird, considering that on the previous installation, without having to manually set multiple layouts, I was able to switch between different layouts, but only outside of the 'bad' login screen. For clarity: the 'bad' login screen currently does allow me to switch between different layouts.

Also, in case you wouldn't mind helping me with this other thing that is outside the scope of the initial complaint: where do I find the KDE cube options? It doesn't seem to be in the Window management options, nor do I see a downloadable version of the such. The KDE plasma version is 6.0.5.

[–] Tomorrow_Farewell@hexbear.net 1 points 4 months ago (8 children)

It seems like VirtualBox doesn't automatic removed the iso file from the boot options. Try change this configuration at the VirtualBox and select the boot to be at the virtual driver

Not sure what you are suggesting, but it seems that the hard drive is the last active option in the boot order for that VM. I assume you are suggesting to make it the priority boot device?

At the archinstall script you must've set the root password, right?

I did, and I remember it.

At the SDDM login screen, you must press Ctrl+Alt+F3 to enter the TTY3 (a big whole full screen terminal) asking for a login. Type root, press enter and then it will ask for your password

Alright, I did that and run the command.

I still don't know how to switch between the keyboard layouts, and with this installation I opted for the US keyboard layout to be the default one, but have since added one more layout to the list. I couldn't place the keyboard layout widget for some reason, and there is no indication of what layout is chosen on the SDDM screen at all.

Right now, the user is blocked due to too many attempts to enter password, so I will have to wait.

[–] Tomorrow_Farewell@hexbear.net 1 points 4 months ago* (last edited 4 months ago) (12 children)

What happened?

I don't know beyond the fact that only Grub got installed and, apparently, nothing else. Upon starting the virtual machine, I was still offered to boot into the Arch installation environment. I have run archinstall again to make an installation without Grub, and am running an installation on a clone of that virtual machine where I have opted into using Grub.

Are you familiar with the syntax of Linux commands?

Depends on what you mean. I am not a frequent CLI user in general.

Currently, I am not sure how I am supposed to login as root.

EDIT: also, Ctrl+Alt+F2 just produces a black screen. Ctrl+Alt+F1 works, but the other problem persists.

EDIT 2: I still do not know how to switch between keyboard layouts in SDDM, and I can't find information regarding that.

 

Working inside a VirtualBox virtual machine. Installed Arch Linux using the Archinstall method. Chose KDE as the DE and SDDM as the greeting screen.

Managed to log in once, but after adding a language and rebooting, I didn't seem to be able to change the keyboard layout, making entering the correct password impossible. I have tried looking up how to switch the keyboard layout, but am coming up with nothing. What is the key combination to change the keyboard layout in SDDM?

Setting up a different virtual machine now.

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