8:00 PM
what decides the order of the square root or cube?

@Argon $\blacksquare$ :)

@Charlie For you
@Jordan You mean 4th root

yes

@Jordan It doesn't matter

:) thanks

8:01 PM
I was trying to think of a way to say exponent that didnt meant square root

It's just easier in this order

oh

:) thanks, Aaron

@Jordan You could find the $(16^3)^{1/4}$

I thought it had to be the exponent first than the root

8:01 PM
@Charlie np :)
@Jordan They are both exponents and are interchangable

these are all the little rules I forget all the time

@willhunting hii

user19161
@Charlie Hi! Are you done with exams?

@WillHunting

user19161
@Argon My dear Aaron!

8:04 PM
@WillHunting i do not think so...

@WillHunting Jasper

user19161

thanks though

Aaron + Jasper = Jasperon

user19161
@Charlie Oh noes!

8:08 PM
$\text{Marilia+Jasper = Marilion}= 10^{6} \cdot \text{Marilia}$

Jasper + Charlie = Jaslie
@Argon Nice

user19161

Million, Billion, Marilion
@WillHunting $J+X^3$
$\text{JAXSXPXER}$

@WillHunting, good afternoon!

user19161
@Limitless I sent you an email.

8:12 PM
@Limitless You mean good morning, for him!

@WillHunting, I replied! :)

man I am so bad at algebra
I should probably quit school for a year to catch up
how do I do $\frac{3x^{3/2} y^3}{x^2 y^{-1/2}}^{-2}$
I am thinking I just flip the fraction and square everything right?

user19161
@Jordan May I recommend you Paul's online notes? Have you heard of them?

yes I used the mfor calculus 2 but they were too difficult to follow becuase he doesn't explain enough

@Jordan Recall $\frac{a^n}{a^m} = a^{n-m}$

8:15 PM
computer algebra system

user19161
@Jordan Have you tried revising from your school textbooks?

CAS?
@WillHunting I am doing that now but I am stuck on the algebra diagnostic tests of my calc book
I have to take calc 2 next semester so I have a lot to learn

user19161
@PeterSheldrick I don't think so. One needs to understand how it works, not do more problems mechanically.

practice makes perfect

user19161
@Jordan You need to understand how the algebraic manipulations really work.

8:17 PM
@WillHunting Like in Singapore?

user19161
@PeterSheldrick Only with understanding.

user19161
@Argon What happens here is mechanical learning, not true understanding, I said this a million times, don't be fooled by ratings and the news.

@WillHunting I know, that's what you said

user19161
Many things the world teaches us is untrue, that's why many of us grow more foolish as we age, not wiser. The world teaches us sick things.

I dont think I have time to relearn all the math I learned

user19161
8:19 PM
True wisdom comes from deeply observing and reflecting upon things oneself, not how many years you have lived, not learning all the stupid cliches from people that have been used a million times.

I am almost done with math anyways, just need to take calculus 2 again, calc 3 and like 3 more math classes and I am done with math

good that true wisdom isn't tested on his next exam
doing lots of problems is what gets him good scores

@Jordan If you don't like math, why are you taking it?

@Argon College requirement
I like math, I am just really bad at it

Oh. What are you doing in university?

8:21 PM
there are books full of questions with solutions but a CAS is more flexible since you look for your own problems at your level (as many as desired) - a bit like the CAS has also superseeded tables of integrals

Computer science

Compsci, nice

I am not very good at programming either but I like it

user19161

I have to learn python now actually
I learned a tiny bit of c++, scheme and now I need to learn python

8:23 PM
Java?

I need to learn java next semester

user19161
@PeterSheldrick I see you changed your picture again, now there are two Peters in this chat!

@PeterSheldrick Cash money, ain't nothin' funny!

@Will, yup - i suddenly had panic about using graphics from proprietary math software :/

man I suck

8:26 PM
big companies scare me

how do I do $(3x^{3/2})^2$?

user19161
@jordan If you are really bad at algebra you can think of many operations geometrically, for example, ab is just the area of a rectangle with sides a and b, so from that you get the intuitive idea why ab=ba.

I am even worse at algebra
err geometry

@Jordan $(a^b)^c = a^{bc}$

@WillHunting, dude, you probably mean well, but stuff like that does not help on tests

8:27 PM
@WillHunting I think that complicates stuff...

user19161
@PeterSheldrick Dude, one can do well for the tests but so what? After that, everything will fall apart. The rain will come and wash the house away...

what is 'area' anyway? it just goes down the philosophy spiral

I doubt I will need to know much math beyond basic algebra really

user19161
@Argon It's a better way of remembering ab=ba anyway.

user19161
@PeterSheldrick Well, we aren't discussing that now.

8:28 PM
@WillHunting I guess.....

man I am shit at algebra, still can't get this basic problem

it's fine to write oh tests are meaningless, but then please be consistent and don't look down on people with poor results on horrible mechanical test scores

this is math for 10 year olds and I cant do it

@Jordan If I were 10, I couldn't do it

If I was 25 I couldn't do it :P

8:33 PM
good evening!

is the exponent in this positive or negative $(y^{-1/2})^2$
I can't remember that algebra rule

Hi @Nimza

user19161
The thing is if you do the algebra without understanding, sooner or later you will forget the rules, or you may mess up because you will apply the rules wrongly.

Help me please, if $\mu$ is a signed measure such that $|\mu|$ is finite is it true that $\left| \int f(x) \mu(dx) \right| \leqslant \int |f(x)| \, |\mu|(dx)$?
@Argon hi!

what is to understand about the rules? they are pretty basic and they exist because they work
there is no understanding

8:34 PM
@Jordan $(a^b)^c=a^{bc}$

I can't conceptualize why a negative fraction exponent does a specific thing to a number, I just memorize that it does

@Jordan $2^1 2^1 = 2^{1+1}$
right?

user19161
@Jordan Well, that is not true. For example, why is a to the b times a to the c equal to a to the b plus c?

$2^{-1} 2^1 = 2^{-1+1}= 2^0 = 1$

I don't know why, it just works

8:36 PM
Then $\frac{1}{2}=2^{-1}$
Does that make sense?

no

user19161
@Jordan That's the problem, that's why you are still having problems with algebra. If you learn by understanding first, it will be slower at first but faster later on.

@Jordan $2^{a} = 2\cdot 2 \cdots 2$, $a$ times

I understand that is works, but I don't see any bigger picture to it, it works because it does

Right?

8:38 PM
yes
but $2^{1/3} = 2 * 1/3$?

So, $2^a 2^b = (2\cdot 2 \cdots 2 ) (2\cdot 2 \cdots 2 )$
And now there are $a+b$ $2$s
$2^a 2^b = 2^{a+b}$
Makes sense?

no

$2^a 2^b = \overbrace{(2 \cdot 2 \cdots 2)}^n \overbrace{(2 \cdot 2 \cdots 2)}^m = \overbrace{2 \cdot 2 \cdots 2}^{n+m}$

isn't it n * m?

Which is $2^{n+m}$

8:41 PM
it's not even rendering the latex properly on my browser

@Jordan Say I have $2^2 = 2\cdot 2$ and $2^3 = 2\cdot 2\cdot 2$

it does on the actual site, just not in the chat

this is why I can't learn math in school, I learn way too slowly and I never learn anything in class and I spend an immense amount of time on homework only to learn nothing and fall behind

Then $2^2 \cdot 2^3 = \overbrace{2\cdot 2\cdot 2\cdot 2\cdot 2}^5$
Makes sense?

@Kalima - why did you delete the Diophantine question? - you should have left it, as other visitors might have found it interesting.

8:43 PM
yes that makes sense

Does this make sense now:
3 mins ago, by Argon
$2^a 2^b = \overbrace{(2 \cdot 2 \cdots 2)}^n \overbrace{(2 \cdot 2 \cdots 2)}^m = \overbrace{2 \cdot 2 \cdots 2}^{n+m}$

I get it now, I was thinking something weird
yes

Ok. Now, of course, this makes sense only for nonzero, natural numbers, right?
Assume it holds true for all rational numbers too.
So now we may say
$2^{1/2}2^{1/2} = 2^1$

writing the equations in plain text does have advantages to

@Jordan Makes sense so far?

8:46 PM
yes

@all, ciao! I'm out to the hospital to check on my ears.

So, $2^{1/2}2^{1/2} = (2^{1/2})^2 = 2$
@Limitless Bye!

@Limitless, bye get well soon

Or in other words, $2^{1/2}$ is a number that, when squared, equals $2$, i.e. $\sqrt 2$
Now, we can use the same logic to show
$2^0 2^1 = 2^1$

I have too much to learn to ask questions here, I will just make poeple angry

8:49 PM
So
$2^0 = 1$
Similarly,
$2^{-1}2^{1}= 2^0 = 1$
thus
$2^{-1} = \frac{1}{2}$
etc.
Makes sense?

why is something to the zero power always 1? I know that it works, and I know that I can work backwards to it but that is all i know

definition

like 2*1 = 2 I get that it needs to start at 1

@Jordan $a^0 a^1 = a^{1+0} = a$

like 0! = 1

8:52 PM
@PeterSheldrick Let's not get too confusing!
@Jordan What do you mean?

it helps you write stuff like exp(x) = x^0/0!+x^1/1!+x^2/2!+...

$2 ^ 1 = 2$ $2^2 = 2* 2$

@Jordan $2 \neq 2^2$

I don't know if this makes sense to anyone but me but if you keep dividing down from higher powers by the base number you get the previous power

$2^1 = 2^{-1} 2^2$
@Jordan Sure, if you mean $a^b/a=a^{b-1}$

8:54 PM
yes

This is true, and can be generalized to $\frac{a^b}{a^c} = a^{b-c}$
because $\frac{a^b}{a^c} = a^b \frac{1}{a^c} = a^{b}a^{-c}=a^{b-c}$

that makes sense
I think I need to quit school for a year to learn all of this

school is boring enough if you just keep going