Question

TuringTest, what now?

Answer 1

Party time! :D

Answer 2

lol
I dunno...

Answer 3

What can you teach me?

Answer 4

a lot, there's a fair amount to algebra, but what's the next logical step?

Answer 5

Teach about polynomial.

Answer 6

I have already to an extent, but where to go next...

Answer 7

Ratios?

Answer 8

In maths the fundamentals concepts are limited but the problems are not, so how about practicing what you have already learned?

Answer 9

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:\[\frac{x^2y^5z}{xy^9z^3}\]

Answer 10

awww, you're making me think of my cat in the hospital :(

Answer 11

http://a4.sphotos.ak.fbcdn.net/hphotos-ak-snc6/271016_253886774622476_100000034671401_1128907_4380462_n.jpg

Answer 12

How did it go with her?

Answer 13

broken leg...
waiting for an operation

Answer 14

You beat you cat Turing?!!!

Answer 15

aww :(

Answer 16

turing sat on her leg. :(

Answer 17

she fell off the roof.
how pessimistic

Answer 18

awww, sorry i was jking..

Answer 19

It's all good :P

Answer 20

Due to Turing's intense physics stress his cat decided to commit suicide :P

Answer 21

(x^2)*(y^5)*(z)
------------ = (x)*(y^5/9)*(z^1/3)?
(x)*(y^9)*(z^3)

Answer 22

Cats are cool.

Answer 23

why \( y^{5/9} \) ?

Answer 24

no doubt. ;)

Answer 25

you have inconsistent rules above inopeki\[\frac{x^a}{x^b}=x^{a-b}\]always...

Answer 26

http://i.imgur.com/48wOw.jpg

Answer 27

Oh right

Answer 28

(x^2)*(y^5)*(z)
------------ = (x)*(y^5-9)*(z^1-3)?
(x)*(y^9)*(z^3)

Answer 29

yes, simplify...

Answer 30

(x)*(y^5-9)*(z^1-3)=(x)*(y^-4)*(z^-2)?

Answer 31

http://i.imgur.com/6uzXf.jpg

Answer 32

cute^

Answer 33

yes, do you know another way to write\[x^{-a}\]???

Answer 34

No

Answer 35

http://i.imgur.com/eqTIb.jpg

Answer 36

\[x^{-a}=\frac{1}{x^a}\]so it's probably nicer to rewrite your expression with all positive exponents in this way.

Answer 37

Oh right, all negative exponents make the "total number" divided by one.

Answer 38

Btw who can explain why x^0 = 1?

Answer 39

(x)*(1/y^4)*(1/z^2)

Answer 40

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...

Answer 41

Turing that's a very important yet fundamental question.

Answer 42

another one is why \( (a^b)^c = a^{bc} \) ?

Answer 43

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?

Answer 44

the last rule is easier to explain, perhaps I should...

Answer 45

(x)*(1/y^4)*(1/z^2)
(x)*(1/y^4*z^2)?

Answer 46

yes, now write it as a fraction
what goes on the bottom?
what goes on the top?

Answer 47

@FFMactually that's not so easy to explain now that I think about it.

Answer 48

Can you explain the second rule intuitively?

Answer 49

1
x* --------
y^4*z^2

Answer 50

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.

Answer 51

@ inopeki
you can put the x on top, it means the same thing and looks nicer

Answer 52

x*1
--------
y^4*z^2

Answer 53

Can we say that all power function obeys the functional equation \( f(x)f(y)= f(x+y) \)?

Answer 54

Whats that f?

Answer 55

@Inopeki yes, no need to write the 1 though
@FFM, I'd have to think about that :/

Answer 56

unspecified functions
he's asking how far the rule about exponents can be extended..

Answer 57

x
-------- Oh right
y^4*z^2

Answer 58

Actually we can't but I believe that is true for exponential functions though.

Answer 59

Umm, ok?

Answer 60

it must be for simple ones\[2^x2^y=2^{x+y}\]of course

Answer 61

Tha's whats Inopeki is using right ?

Answer 62

basically
Inopeki do you see the connection between what you are doing and the rule for multiplying exponents like\[x^ax^b\]?

Answer 63

Yeah, but does that mean that it becomes
x
-------- ?
yz^6

Answer 64

no because y and z are different bases
notice the rule above had both base x.

Answer 65

YEah but then i dont see the connection..

Answer 66

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.

Answer 67

I know about that, like x^2*x^8=x^10

Answer 68

x^10/x^5=x^5

Answer 69

Basic

Answer 70

right, I want you to see how that and the rule for division are inverses of each other for a reason...

Answer 71

Turing you have a heck of patience :D

Answer 72

eh it's easy FFM, doesn't require too much concentration.
good practice too I learn by teaching ;)
ok factor
\[3x^3y^3+6xy+9x^2y\]

Answer 73

:-)

Answer 74

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?

Answer 75

yes exactly

Answer 76

3xy(3x^3*y^3/3xy)+3xy(6xy/3xy)+3xy(9x^2*y/3xy)=3xy(x^2*y^2+2+3x)?

Answer 77

yep :)
great!

Answer 78

Really? :D

Answer 79

Note to Inopeki: Turing is a great resource, learn new things from him and practice as much as you can on your own :-)

Answer 80

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)

Answer 81

When hes not here i try to go on purplemath and khan :)

Answer 82

a^2+ab-ba-b^2

Answer 83

simplify

Answer 84

Books books books!! online learning has it's own limitation :-)

Answer 85

a^2-b^2?

Answer 86

it doesnt @ffm i disagree

Answer 87

Foolformath, im having trouble finding books here in sweden

Answer 88

and what is the name of that form? remember?

Answer 89

The fundamental theorem of algebra? Or want that the one with the multiplex?

Answer 90

Multiplex?
no I think you're thinking of 'multiplicity'
but this is the 'difference of squares'
\[a^2-b^2=(a-b)(a+b)\]

Answer 91

with ocw.mit , khan, purple math and openstudy and people like turing there is no limit to online education ^

Answer 92

Thanks but you get what you put in, I think is true with all this stuff.

Answer 93

Oh right