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4:23 AM
1 message moved to ­Trash
 
 
2 hours later…
6:44 AM
Hey everyone! Happy Friday 🙂
 
 
3 hours later…
10:12 AM
Jul 23 at 12:01, by F. Zer
@user21820, when doing Velleman I found the exercise (Chapter 3): Suppose that a and b are nonzero real numbers. Prove that if a < 1/a < b < 1/b then a < -1.
2 messages moved from Basic Mathematics
 
10:31 AM
For example, you can prove ¬∃x∈ℚ ( x·x = 2 ) by using (PA4).
 
 
1 hour later…
11:36 AM
2 messages moved from Basic Mathematics
 
 
5 hours later…
4:18 PM
Hi Ollie 🙂 How are you going?
 
‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ Hello @Rosie, have I seen you in here before?‬
 
Hi Vikas 🙂 How are you?
 
@Ollie lol
 
@Vikas ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮What's so funny? ;)‭‬‭‭‬
 
@Ollie are you bot?
 
4:25 PM
Nope.
 
Ah 23k reps
 
Yeah not many bots get that far.
 
so you're very fast at typing reverse
 
Kind of, yeah. But most of the time I just ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬type like this.‭‭‬
Or ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‭‬‭‭‬ ‭‮‬‬ ‬‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮‮I use Unicode characters.‮‭‬‭‭‬
 
5:05 PM
Hi Wolgwang 🙂 How are you going?
 
I am going to perform an experiment.
Any RO online?
 
I will unstar messages...
So one can star 20 messages.
 
Hi. I'm also room owner 😉😀
 
NVM experiment successful...
Conclusion: One can star 20 messages ....
 
5:11 PM
Do I have to do something?
 
in 24 hours?
 
I know nothing 🤣
 
@Vikas Nope, I thought I would not be able to unstar messages...
 
Ok
 
Hi Slate 🙂 How are you doing?
 
5:30 PM
Hi Wolgwang 🙂 How are you going?
Hi Wolgwang 🙂 How are you going?
 
@Rosie What can you do?
 
@Wolgwang I think so.
 
@Rosie hmm?
 
rosie is in development still
needs lot of work and I'm too lazy to fix it :(
 
:-(
IG there are 3 bots here....
 
5:36 PM
well one of them does not actually respond here (unless I make it)
but yes, @Rosie, @VyxalBot, @TryAPL are all bots
 
Which one doesn't respond?
@VyxalBot
 
Vyxal Bot
TryAPL responds to APL commands like ⎕←'Hello, World!'
 
@hyper-neutrino Hello, World!
 
Vyxal Bot serves a different chat room and only is active here if I swap its room here manually
 
@Wolgwang IG?
 
5:37 PM
@hyper-neutrino IK about this...I had tried it :-)
@Vikas I Guess
 
⎕←3 3 3 ⍴ ⍳ 27
 
@hyper-neutrino
 1  2  3
 4  5  6
 7  8  9

10 11 12
13 14 15
16 17 18

19 20 21
22 23 24
25 26 27
 
@Wolgwang wow what a short form 😅
@TryAPL hi
@TryAPL 3+3 = ?
@VyxalBot hi
 
@Vikas APL commands....
 
yeah
was just checking other possibilities
@Rosie ♦ hi
 
5:42 PM
@Vikas Hi, what's up?
 
⎕←2×3
 
@Wolgwang 6
 
@TryAPL ⎕←2×3
 
⎕←99999999*999999999999999999
 
⎕←2×3
 
5:46 PM
@Wolgwang
DOMAIN ERROR
      ⎕←99999999*1000000000000000000
              ∧
@Vikas Did you forget to add backticks around your code (`⎕←code`)? You can edit your message and I will edit my reply.
 
@TryAPL -_-
⎕←99*99
 
@Wolgwang 3.697296376E197
 
⎕←99*9999999
 
@Wolgwang
DOMAIN ERROR
      ⎕←99*9999999
        ∧
 
⎕←2×3
 
5:48 PM
@Vikas 6
 
⎕←(~R∊R∘.×R)/R←1↓ιR
 
@Wolgwang
NOT PERMITTED: Illegal token
      ⎕←(~R∊R∘.×R)/R←1↓ιR
     ^
 
⎕←2×3+5
 
@Vikas Did you forget to add backticks around your code (`⎕←code`)? You can edit your message and I will edit my reply.
 
⎕←2×3+5
 
5:48 PM
@Vikas 16
 
@TryAPL Incorrect
 
@TryAPL What?
11?
 
@Wolgwang lol bot failed
 
I don't think so....IDK about APL maybe...
 
⎕←(2×3)+5
 
5:50 PM
@Vikas 11
 
I also don't know apl
 
⎕←(~15∊15∘.×15)/15←1↓ι15
 
@Wolgwang
NOT PERMITTED: Illegal token
      ⎕←(~15∊15∘.×15)/15←1↓ι15
     ^
 
:-/
 
⎕←2^5
 
5:51 PM
@Vikas 10
 
Hi Wolgwang 🙂 What's up?
 
APL is right to left
 
well
 
1+2×3 is 7, 2×3+1 is 8
 
⎕←5+2×3
 
5:52 PM
@Vikas 11
 
good
 
6:21 PM
∀x,z∈ℝ ∀y,w∈ℝ[≠0] ( (x/y)·(z/w) = (x·z)/(y·w) ).
Given x,z ∈ ℝ:
Given y,w ∈ ℝ[≠ 0]:
∀x∈ℝ ∀y∈ℝ[≠0] ( x = y·(x/y) )
x = y·(x/y)
z = w·(z/w)
(x/y)·(z/w) = (y·(x/y)/y)·(w·(z/w)/w)
...
∀x,z∈ℝ ∀y,w∈ℝ[≠0] ( (x/y)·(z/w) = (x·z)/(y·w) )
@user21820, please delete the above post (and this one).
 
6:52 PM
2 messages moved from Basic Mathematics
 

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