This discussion board is a place to discuss our Daily Challenges and the math and science
related to those challenges. Explanations are more than just a solution — they should
explain the steps and thinking strategies that you used to obtain the solution. Comments
should further the discussion of math and science.
When posting on Brilliant:
Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.
Markdown
Appears as
*italics* or _italics_
italics
**bold** or __bold__
bold
- bulleted - list
bulleted
list
1. numbered 2. list
numbered
list
Note: you must add a full line of space before and after lists for them to show up correctly
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
Math
Appears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3
2×3
2^{34}
234
a_{i-1}
ai−1
\frac{2}{3}
32
\sqrt{2}
2
\sum_{i=1}^3
∑i=13
\sin \theta
sinθ
\boxed{123}
123
Comments
a∈Z,pa=⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧pa=1pa=ppa=1pa=1pa=0 if a is positive, and even if a is positive, and odd if a is zero if a is negative, and even if a is negative, and odd
If you understand these, you are almost mastered this.
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
a∈Z,pa=⎩⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎧pa=1pa=ppa=1pa=1pa=0 if a is positive, and even if a is positive, and odd if a is zero if a is negative, and even if a is negative, and odd
If you understand these, you are almost mastered this.
Log in to reply
Okay so are these facts about p or properties of proper numbers(p*0≠0) ?
Log in to reply
I think I should first look up what proper numbers are then I'll come back here to get a better understanding of what's going on.
Of, course.
But, anyway, can you teach let me to use the simultaneous symbol? ({)
I used them in the way this Wikipedia page, but I cannot know them.
Log in to reply
{ This bracket can be written by adding a backslash(\) before ({).
Log in to reply
{tanxcosx} ?
Ohh did you mean these bigger ones? For egThe latex code for it is as follows:
\ (\left{ \dfrac{cosx}{tanx}\right})
I.e. adding \left before the text will work.
These are the links to a few latex guides on brilliant that I found useful:
1.Pall Marton's latex guide
2.Percy Jackson's latex guide
Log in to reply
cases.
{1/2.
Log in to reply
X(m,n)=⎩⎪⎨⎪⎧x(n),x(n−1)x(n−1)
Do u wanna write something like this ↑
Log in to reply
Log in to reply
{n/2(3n+1)/2if n≡0if n≡1(mod2)
How about this ↑?
Reqd code:
I couldn't find any other than this on stack exchange.
I will prove those.
a∈N,p2n=1
a∈N,p2n+1=p
@Agent T, You will understand these because I mentioned them before.
p0=1.
I'm not sure, but I think that you will understand.
a∈N,p−2n=p2n1=1
a∈N,p−(2n+1)=p−2n−1=p2n+11=p1=0.
I hope you will understand.
And do you understand my explanation of my problem?
Log in to reply
Yes ,it all makes sense now. Sometimes it's just us who limit our thinking and refuse to grab a concept even if we are familiar with it:^)
Log in to reply
Thank you for the explanations ,I really appreciate them!
Yo! Have a great day :)
Log in to reply
It is a proper number, Pr.
And M which is mixed number, C+Pr.
A proper number is a number that gets multiplied with zero, which is not equal to zero.
A proper unit is p.
p×0=1 which I want to create.
And, if ap×0, you have to calculate a×p×0, p×0 first.
So, ap×0=a.
Basic calculations.
p±α cannot be calculated like an irrational number calculation like 2+elog32+π3!×e.
p×α,α=0 is equal to αp, which is equal to multiplication of irrational numbers.
p÷0 is equal to 1 because I said 0×p=1→0=1÷p→p×1÷p=1.
p÷α,α=0,p is equal to αp, same as calculating irrational numbers.
Upgraded calculations.
p×p=p×1÷0=p÷0=1.
p×p×p=1×p=p.
pp=1.
n∈N,p2n=1.
n∈N,p2n+1=p.
p=p2n=1,n∈N.
Now, let us talk about mixed numbers.
A normal form is a+bi+cp, where a,b are real numbers, and c is complex number.
Expansion of mixed number.
(a+bi+cp)(d+ei+fp)=ad+aei+afp+bdi+bei2+bfip+cdp+ceip+cfp2=ad+aei+afp+bdi−be+bfip+cdp+ceip+cf=(ad−be+cf)+(ae+bd)i+(af+cd)p+(bf+ce)ip.
The end.
Log in to reply
Normally, I wanted to define ÷0.
It seems interesting,what topic does it come under? I wanna explore it more!
Log in to reply
The ball is in all our court, including me.
Log in to reply
Btw did u mean that this topic is very much related to our life or something? Eeeeeeeee I wanna knowww!
Log in to reply
this my problem, I will tell you more about the number.
If you solvedLog in to reply
Log in to reply
If you understand my explanation, then it doesn't matter.
By the way, I promised you, so I will keep that.
As the replies are too long, I will make the new line.
Log in to reply
I just found out that I already knew complex fields and real numbers(idk what u must be thinking XP)
But I wonder why have you used the word "proper number" rather than real numbers and "mixed" for complex??
And I guess there's an error in the part where you divided p by 0 ,kindly check that once .
@. . you might love these if u wanna go beyond complex 8)