Inopeki
TuringTest, what now?
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saifoo.khan
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Party time! :D
TuringTest
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lol
I dunno...
Inopeki
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What can you teach me?
TuringTest
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a lot, there's a fair amount to algebra, but what's the next logical step?
FoolForMath
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Teach about polynomial.
TuringTest
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I have already to an extent, but where to go next...
Inopeki
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Ratios?
FoolForMath
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In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?
TuringTest
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That's a good point FFM so lets do simplification that you know, but a little more intense:
I think I need to make sure you can simplify bigger things:\[\frac{x^2y^5z}{xy^9z^3}\]
saifoo.khan
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TuringTest
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awww, you're making me think of my cat in the hospital :(
Inopeki
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How did it go with her?
TuringTest
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5
broken leg...
waiting for an operation
FoolForMath
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You beat you cat Turing?!!!
Inopeki
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aww :(
saifoo.khan
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turing sat on her leg. :(
TuringTest
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5
she fell off the roof.
how pessimistic
saifoo.khan
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awww, sorry i was jking..
TuringTest
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It's all good :P
FoolForMath
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Due to Turing's intense physics stress his cat decided to commit suicide :P
Inopeki
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(x^2)*(y^5)*(z)
------------ = (x)*(y^5/9)*(z^1/3)?
(x)*(y^9)*(z^3)
saifoo.khan
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GT
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Cats are cool.
FoolForMath
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why \( y^{5/9} \) ?
saifoo.khan
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no doubt. ;)
TuringTest
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you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{a-b}\]always...
Inopeki
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Oh right
Inopeki
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(x^2)*(y^5)*(z)
------------ = (x)*(y^5-9)*(z^1-3)?
(x)*(y^9)*(z^3)
TuringTest
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yes, simplify...
saifoo.khan
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Inopeki
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(x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?
saifoo.khan
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cute^
TuringTest
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yes, do you know another way to write\[x^{-a}\]???
Inopeki
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No
TuringTest
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\[x^{-a}=\frac{1}{x^a}\]so it's probably nicer to rewrite your expression with all positive exponents in this way.
Inopeki
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Oh right, all negative exponents make the "total number" divided by one.
FoolForMath
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Btw who can explain why x^0 = 1?
Inopeki
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(x)*(1/y^4)*(1/z^2)
TuringTest
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Oh dear...
I have my answers about x^0=1 (not true for x=0), but I'm sure FFM would not approve of them :/
@inopeki, yes now rewrite it as 1 fraction...
FoolForMath
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Turing that's a very important yet fundamental question.
FoolForMath
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another one is why \( (a^b)^c = a^{bc} \) ?
TuringTest
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well I have a very simple proof of it
and I can show why it is not true for x=0
what more do I need in your opinion?
TuringTest
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the last rule is easier to explain, perhaps I should...
Inopeki
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(x)*(1/y^4)*(1/z^2)
(x)*(1/y^4*z^2)?
TuringTest
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yes, now write it as a fraction
what goes on the bottom?
what goes on the top?
TuringTest
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@FFMactually that's not so easy to explain now that I think about it.
FoolForMath
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Can you explain the second rule intuitively?
Inopeki
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1
x* --------
y^4*z^2
TuringTest
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no, I was thinking more of how easy it is to show
x^ax^b=x^(a+b)
intuitively, the other is tricky to me.
TuringTest
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@ inopeki
you can put the x on top, it means the same thing and looks nicer
Inopeki
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x*1
--------
y^4*z^2
FoolForMath
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Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?
Inopeki
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Whats that f?
TuringTest
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@Inopeki yes, no need to write the 1 though
@FFM, I'd have to think about that :/
TuringTest
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unspecified functions
he's asking how far the rule about exponents can be extended..
Inopeki
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x
-------- Oh right
y^4*z^2
FoolForMath
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Actually we can't but I believe that is true for exponential functions though.
Inopeki
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Umm, ok?
TuringTest
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it must be for simple ones\[2^x2^y=2^{x+y}\]of course
FoolForMath
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Tha's whats Inopeki is using right ?
TuringTest
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basically
Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like\[x^ax^b\]?
Inopeki
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Yeah, but does that mean that it becomes
x
-------- ?
yz^6
TuringTest
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no because y and z are different bases
notice the rule above had both base x.
Inopeki
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YEah but then i dont see the connection..
TuringTest
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just that dividing and multiplying are inverse operations
x^a=x*x*x*...*x (a times)
x^b=x*x*x*...*x (b times)
so if we have
a=3 b=2 we get
x^a
---
x^b
x*x*x
=-----= x
x*x
which is x^(a-b)
if we have
(x^a)(x^b) we get
(x*x*x)(x*x)=x*x*x*x*x=x^(a+b)
so the rules for exponents here come directly from their definitions. You can count the x and see that this relationship holds.
Inopeki
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I know about that, like x^2*x^8=x^10
Inopeki
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x^10/x^5=x^5
Inopeki
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Basic
TuringTest
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right, I want you to see how that and the rule for division are inverses of each other for a reason...
FoolForMath
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Turing you have a heck of patience :D
TuringTest
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eh it's easy FFM, doesn't require too much concentration.
good practice too I learn by teaching ;)
ok factor
\[3x^3y^3+6xy+9x^2y\]
FoolForMath
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:-)
Inopeki
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The GCF of 3,6,9 is 3
The GCF of x,x^2,x^3 is x
The GCF of y,y^2,y^3 is y
3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Middle step, right?
TuringTest
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yes exactly
Inopeki
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3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?
TuringTest
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yep :)
great!
Inopeki
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Really? :D
FoolForMath
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Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :-)
TuringTest
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Yes I've stressed the importance of studying on your own as well
and yes Inopeki really, FFM would have noticed a mistake I'm sure
ok quick, foil
(a-b)(a+b)
Inopeki
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When hes not here i try to go on purplemath and khan :)
Inopeki
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a^2+ab-ba-b^2
TuringTest
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simplify
FoolForMath
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Books books books!! online learning has it's own limitation :-)
Inopeki
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a^2-b^2?
Akshay_Budhkar
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it doesnt @ffm i disagree
Inopeki
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Foolformath, im having trouble finding books here in sweden
TuringTest
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and what is the name of that form? remember?
Inopeki
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The fundamental theorem of algebra? Or want that the one with the multiplex?
TuringTest
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Multiplex?
no I think you're thinking of 'multiplicity'
but this is the 'difference of squares'
\[a^2-b^2=(a-b)(a+b)\]
Akshay_Budhkar
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with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^
TuringTest
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Thanks but you get what you put in, I think is true with all this stuff.
Inopeki
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Oh right
TuringTest
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The fundamental theorem of algebra:
"the sum of the multiplicity of the roots of a polynomial is equal to its order"
difference of squares\[a^2-b^2=(a-b)(a+b)\]both important, but very different
TuringTest
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so remember that we can run this backwards, and I can say
'factor'\[x^2-4\]using difference of squares
wanna try?
TuringTest
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run the FOIL backwards I meant*
Inopeki
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x(x-4)?
Inopeki
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Oh
TuringTest
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no look at the form\[a^2-b^2=(a-b)(a+b)\]so what is a and b in\[x^2-4\]???
Inopeki
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(x-2)(x+2)?
TuringTest
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there ya go :)
TuringTest
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how about\[x^4-y^4\](this is a favorite question on OS)
Inopeki
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(x^2-y^2)(x^2+y^2)
Inopeki
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Why?
TuringTest
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good, now is that all we can do with it though?
(I've seen a lot of tutors stop here too ;-)
TuringTest
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hint:look at the first set of parentheses
Inopeki
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Ummm
TuringTest
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what is the first sett of parentheses?
Inopeki
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(x^2-y^2)?
TuringTest
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yes, and can you factor that?
Inopeki
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GCF of x^2 is x
GCF of y^2 is y
xy(x^2/xy)-xy(y^2/xy)?
TuringTest
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5
no you're over-thinking
x^2-y^2 is difference of squares again!
TuringTest
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\[a^2-b^2=(a-b)(a+b)\]
Inopeki
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So (x-y)(x+y)?
TuringTest
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yes!
so now factor\[p^4-q^4\]completely!!
TuringTest
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you will apply difference of squares twice
Inopeki
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(p^2-q^2)(p^2+q^2)
(p-q)(p+q)(p+q)+(p+q)?
Inopeki
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Without the +
Inopeki
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between the parenthesis
TuringTest
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(p^2-q^2)(p^2+q^2)
stop here
we cannot factor the second set of parentheses, it is not a /difference/ of squares, it is a /sum/ of squares, and we have no formula for that.
check that
(a+b)(a+b) is not a^2+b^2
so only factor the one that is a /difference/ of squares
Inopeki
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(p-q)(p+q)(p^2+q^2)
You said i had to do difference in squares 2 times?
Btw, lets move to a new thread, this one is starting to lag a little bit
TuringTest
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That is above correct, you did difference of squares twice:
p^4-q^4
(p^2-q^2)(p^2+q^2) <-once
(p-q)(p+q)(p^2+q^2) <-twice
ok new thread...
Inopeki
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Ohhh