## anonymous 4 years ago TuringTest, what now?

1. saifoo.khan

Party time! :D

2. TuringTest

lol I dunno...

3. anonymous

What can you teach me?

4. TuringTest

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

5. anonymous

6. TuringTest

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

7. anonymous

Ratios?

8. anonymous

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

9. TuringTest

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

10. saifoo.khan

11. TuringTest

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

12. anonymous
13. anonymous

How did it go with her?

14. TuringTest

broken leg... waiting for an operation

15. anonymous

You beat you cat Turing?!!!

16. anonymous

aww :(

17. saifoo.khan

turing sat on her leg. :(

18. TuringTest

she fell off the roof. how pessimistic

19. saifoo.khan

awww, sorry i was jking..

20. TuringTest

It's all good :P

21. anonymous

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

22. anonymous

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

23. saifoo.khan

24. anonymous

Cats are cool.

25. anonymous

why $$y^{5/9}$$ ?

26. saifoo.khan

no doubt. ;)

27. TuringTest

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

28. anonymous
29. anonymous

Oh right

30. anonymous

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

31. TuringTest

yes, simplify...

32. saifoo.khan

33. anonymous

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

34. anonymous
35. saifoo.khan

cute^

36. TuringTest

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

37. anonymous

No

38. anonymous
39. TuringTest

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

40. anonymous

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

41. anonymous

Btw who can explain why x^0 = 1?

42. anonymous

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

43. TuringTest

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...

44. anonymous

Turing that's a very important yet fundamental question.

45. anonymous

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

46. TuringTest

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?

47. TuringTest

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

48. anonymous

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

49. TuringTest

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

50. TuringTest

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

51. anonymous

Can you explain the second rule intuitively?

52. anonymous

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

53. TuringTest

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.

54. TuringTest

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

55. anonymous

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

56. anonymous

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

57. anonymous

Whats that f?

58. TuringTest

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

59. TuringTest

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

60. anonymous

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

61. anonymous

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

62. anonymous

Umm, ok?

63. TuringTest

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

64. anonymous

Tha's whats Inopeki is using right ?

65. TuringTest

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

66. anonymous

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

67. TuringTest

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

68. anonymous

YEah but then i dont see the connection..

69. TuringTest

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.

70. anonymous

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

71. anonymous

x^10/x^5=x^5

72. anonymous

Basic

73. TuringTest

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

74. anonymous

Turing you have a heck of patience :D

75. TuringTest

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

76. anonymous

:-)

77. anonymous

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?

78. TuringTest

yes exactly

79. anonymous

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

80. TuringTest

yep :) great!

81. anonymous

Really? :D

82. anonymous

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

83. TuringTest

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)

84. anonymous

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

85. anonymous

a^2+ab-ba-b^2

86. TuringTest

simplify

87. anonymous

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

88. anonymous

a^2-b^2?

89. Akshay_Budhkar

it doesnt @ffm i disagree

90. anonymous

Foolformath, im having trouble finding books here in sweden

91. TuringTest

and what is the name of that form? remember?

92. anonymous

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

93. TuringTest

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

94. Akshay_Budhkar

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

95. TuringTest

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

96. anonymous

Oh right

97. TuringTest

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

98. TuringTest

so remember that we can run this backwards, and I can say 'factor'$x^2-4$using difference of squares wanna try?

99. TuringTest

run the FOIL backwards I meant*

100. anonymous

x(x-4)?

101. anonymous

Oh

102. TuringTest

no look at the form$a^2-b^2=(a-b)(a+b)$so what is a and b in$x^2-4$???

103. anonymous

(x-2)(x+2)?

104. TuringTest

there ya go :)

105. TuringTest

how about$x^4-y^4$(this is a favorite question on OS)

106. anonymous

(x^2-y^2)(x^2+y^2)

107. anonymous

Why?

108. TuringTest

good, now is that all we can do with it though? (I've seen a lot of tutors stop here too ;-)

109. TuringTest

hint:look at the first set of parentheses

110. anonymous

Ummm

111. TuringTest

what is the first sett of parentheses?

112. anonymous

(x^2-y^2)?

113. TuringTest

yes, and can you factor that?

114. anonymous

GCF of x^2 is x GCF of y^2 is y xy(x^2/xy)-xy(y^2/xy)?

115. TuringTest

no you're over-thinking x^2-y^2 is difference of squares again!

116. TuringTest

$a^2-b^2=(a-b)(a+b)$

117. anonymous

So (x-y)(x+y)?

118. TuringTest

yes! so now factor$p^4-q^4$completely!!

119. TuringTest

you will apply difference of squares twice

120. anonymous

(p^2-q^2)(p^2+q^2) (p-q)(p+q)(p+q)+(p+q)?

121. anonymous

Without the +

122. anonymous

between the parenthesis

123. TuringTest

(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

124. anonymous

(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

125. TuringTest

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...

126. anonymous

Ohhh