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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
@Argon Another nice song. youtube.com/watch?v=qzizhokLeV8
 
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
How about J+XXX? =)
 
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
If I learn programming I would start with Python!
 
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
 

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