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kc_kennylau

  • one year ago

Prove that \(y=x^3\) where \(x,y\in\mathbb R\) is a bijection.

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  1. UsukiDoll
    • one year ago
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    x and y belong in a set of real numbers... if we need to prove bijection, then surjection (onto) and injection (one to one) must hold true

  2. kc_kennylau
    • one year ago
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    Is injection a subset of surjection?

  3. kc_kennylau
    • one year ago
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    Well, we know that the domain and the co-domain (or image) are \(\mathbb R\)

  4. UsukiDoll
    • one year ago
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    the problem is I'm kind of stupid when it comes to these proofs...I had a horrible professor so I only know parts... not the full topic in detail.

  5. UsukiDoll
    • one year ago
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    bijection is a surjection and injection which means that surjection is onto and injection is one to one.. so if we have to prove bijection, we have to prove that the surjection and injection has to hold true for that problem.

  6. kc_kennylau
    • one year ago
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    I'm sorry I don't want to sound stupid but what is surjection?

  7. kc_kennylau
    • one year ago
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    And is injection a subset of surjection?

  8. kc_kennylau
    • one year ago
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    all functions are surjections?

  9. kc_kennylau
    • one year ago
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    @UsukiDoll and thank you for helping meeee

  10. UsukiDoll
    • one year ago
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    I'm also new at this...let me think XD! oh gawd I rather prove in set theory

  11. kc_kennylau
    • one year ago
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    lol

  12. UsukiDoll
    • one year ago
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    The function is surjective (onto) if every element of the codomain is mapped to by at least one element of the domain. (That is, the image and the codomain of the function are equal.) A surjective function is a surjection.

  13. UsukiDoll
    • one year ago
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    an injective function or injection or one-to-one function is a function that preserves distinctness: it never maps distinct elements of its domain to the same element of its codomain. In other words, every element of the function's codomain is the image of at most one element of its domain.

  14. UsukiDoll
    • one year ago
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    bijection is BOTH OF THOSE GUYS ^ ^

  15. UsukiDoll
    • one year ago
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    so your x and y better satisfy surjection and injection a.k.a those definitions otherwise it's not a bijection.

  16. kc_kennylau
    • one year ago
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    https://proofwiki.org/wiki/Existence_of_Positive_Root

  17. kc_kennylau
    • one year ago
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    https://proofwiki.org/wiki/Uniqueness_of_Positive_Root

  18. FibonacciChick666
    • one year ago
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    for that is injective a subset of surjective, nah. In the world of non functions, you can have onto with out having one-to-one they are mutually exclusive classifications

  19. freckles
    • one year ago
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    A function is onto if every element of the codomain gets hit. A function is 1-to-1 if every element of the codomain get his 0 or 1 times. That is pretend we have \[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by} f(x)=x^2 \] |dw:1434257538006:dw| This function is not onto on the real numbers because there is no such negative real number that equals x^2. Example: f(x)=-1 doesn't happen for any real x. This function is also not one-to-one. Example: f(1)=f(-1)=1. 1 gets hit more than 1 times.

  20. freckles
    • one year ago
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    his is meant to be the word hit

  21. UsukiDoll
    • one year ago
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    @freckles should've been my professor for Intro To Advanced Mathematics... I actually understood that...unlike my show off professor -_-

  22. FibonacciChick666
    • one year ago
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    oohooh, easy proof! (if this thm is what I remember) so if you have proved that x^3 is a continuous function, then you can use the MVT(or is it IVT) to prove onto and state one to one is by definition of continuous function

  23. kc_kennylau
    • one year ago
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    Continuous is \(\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\)?

  24. kc_kennylau
    • one year ago
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    I think the limit is easy to prove, by using delta-epsilon def

  25. FibonacciChick666
    • one year ago
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    but listen to freckles, he got me through abstract alg, basic analysis, pre-advanced calculus, number theory, and all the other proofy stuff I'd rather forget

  26. UsukiDoll
    • one year ago
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    number theory is awesome ^__^

  27. freckles
    • one year ago
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    \[\text{ Prove } g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=2x-1 \text{ is a bijection. } \\ \text{ Proof: (First \let's try to prove it is one-to-one)} \\ \text{ we want to try the following } f(a)=f(b) \implies \text{ only } a=b \\ \\f(a)=f(b) \\ 2a-1=2b-1 \\ 2a=2b \\ a=b \\ g \text{ is one-to-one } \\ \ \text{ Now we want to prove it is surjective } \\ \text{ we want to show } \\ \text{ for every } y \text{ in the codomain } \text{ there is } x \text{ in the domain } \\ \text{ such that } f(x)=y \\ \text{ so hmm... there exists } \\ \text{ so } 2x-1=y \text{ solve for } y \\ 2x=y+1 \\ x=\frac{y+1}{2} \\ \text{ so we have chosen an } x \text{ such that } \text{ we will have for every } y \\ \text{ there exist } x \text{ such that } f(x)=y \\ \text{ since } f(\frac{y+1}{2})=2(\frac{y+1}{2})-1=y+1-1=y\] This means g is a bijection.

  28. freckles
    • one year ago
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    That was just an example. You can do something very similar with your function.

  29. kc_kennylau
    • one year ago
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    My question would not be simple, because you would be using the result if you prove it in your manner

  30. freckles
    • one year ago
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    What do you mean?

  31. kc_kennylau
    • one year ago
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    \(\Large x=y\implies \sqrt[3]x=\sqrt[3]y\) is using the result (circular)

  32. FibonacciChick666
    • one year ago
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    I'd use mean value theorem if you have it under your belt to prove onto. (after the quick showing of continuous)

  33. kc_kennylau
    • one year ago
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    I assure you, @FibonacciChick666 , continuous is not quick at all

  34. kc_kennylau
    • one year ago
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    \[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\]

  35. freckles
    • one year ago
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    I'm not sure what you are talking about. Are you having issues with the one-to-one part or the onto part?

  36. FibonacciChick666
    • one year ago
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    onto freckles

  37. kc_kennylau
    • one year ago
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    I guess both

  38. freckles
    • one year ago
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    Suppose you have:\[v: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } v(x)=x^3\] for 1-1 you want to suppose v(a)=v(b) and arrive at the implication of only a=b

  39. FibonacciChick666
    • one year ago
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    for continuous, btw. The product of continuous functions is cont. SO x^3=x*x*x right? So you only have to do the proof of x being continuous.

  40. freckles
    • one year ago
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    Like with my f function above it wouldn't have worked because a^2=b^2 means we have a^2-b^2=0 which means we have a=b or a=-b

  41. freckles
    • one year ago
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    that actually does show f isn't 1-1

  42. kc_kennylau
    • one year ago
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    \[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x-c|<\epsilon\right)\]

  43. kc_kennylau
    • one year ago
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    oops too long

  44. kc_kennylau
    • one year ago
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    \[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x-c|<\epsilon\right)\]

  45. FibonacciChick666
    • one year ago
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    you forgot to define delta

  46. kc_kennylau
    • one year ago
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    \[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x^3-c^3|<\epsilon\right)\]

  47. kc_kennylau
    • one year ago
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    I didn't forget xd

  48. FibonacciChick666
    • one year ago
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    you have to say delta = epsilon

  49. kc_kennylau
    • one year ago
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    @freckles Then how do we know that \(\Large x^3=y^3\implies x=y\)?

  50. FibonacciChick666
    • one year ago
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    you are doing continuous, not limit

  51. freckles
    • one year ago
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    you play with it

  52. freckles
    • one year ago
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    \[x^3-y^3=0 \\ (x-y)(x^2+xy+y^2)=0\]

  53. kc_kennylau
    • one year ago
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    Without using this result?

  54. kc_kennylau
    • one year ago
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    \[\Large\begin{array}{lrcl} (1):&x^3&=&y^3\\(2):&x^3-y^3&=&0\\(3):&(x-y)(x^2+xy+y^2)&=&0 \end{array}\]

  55. freckles
    • one year ago
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    set both equal to 0 and you will see one equation is not possible over the reals anyways

  56. FibonacciChick666
    • one year ago
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    That breakdown allows you to say that since x can't equal y, x^2+xy+y^2 must equal zero

  57. UsukiDoll
    • one year ago
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    (x-y)(x^2+xy+y^2)=0 (x-y) = 0 and then x^2+xy+y^2 = 0 but for x-y=0 wouldn't we have just x = y ?

  58. kc_kennylau
    • one year ago
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    @freckles I guess the fact that \(\Large x^3\) is increasing suffice to prove injection

  59. freckles
    • one year ago
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    @FibonacciChick666 you are trying to show f(a)=f(b) only gives a=b and yes @UsukiDoll since the other equation gives: \[x=\frac{-y \pm \sqrt{-3y^2}}{2} \text{ which isn't a real number }\]

  60. FibonacciChick666
    • one year ago
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    no, not really

  61. FibonacciChick666
    • one year ago
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    and I was doing pf by contradiction. At beginning say \(x\not =y\)

  62. UsukiDoll
    • one year ago
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    @freckles not a real number because of that negative number in the radical... produces imaginary numbers.

  63. kc_kennylau
    • one year ago
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    \[(f(x)=f(y)\implies x=y)\iff(x\ne y\implies f(x)\ne f(y))\]

  64. FibonacciChick666
    • one year ago
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    you get a therefore since that's not real and by multiplicative property of zero x-y=0 so x=y

  65. kc_kennylau
    • one year ago
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    The second statement follows from \(x^3\) being increasing

  66. FibonacciChick666
    • one year ago
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    That's not enough

  67. UsukiDoll
    • one year ago
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    x = y and x = that giant thing which isn't a real number

  68. kc_kennylau
    • one year ago
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    If \(x\ne y\), WLOG let \(x<y\)

  69. FibonacciChick666
    • one year ago
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    the ceiling function is increasing on 0, infty too

  70. freckles
    • one year ago
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    yes y^2 is positive or yeah 0 for all y and sqrt(negative) is imaginary and if y does equal 0 we have x=0 but that still falls under the y=x implication we already have

  71. kc_kennylau
    • one year ago
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    Then, since \(x^3\) is strictly increasing, \(x^3<y^3 \implies x^3\ne y^3\)

  72. kc_kennylau
    • one year ago
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    @FibonacciChick666 come on, you know I mean strictly increasing

  73. kc_kennylau
    • one year ago
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    @freckles How to prove surjection?

  74. FibonacciChick666
    • one year ago
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    still, you can have skips in strictly inc. You must have continuous to attempt that.

  75. kc_kennylau
    • one year ago
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    @FibonacciChick666 I proved continuous using limit

  76. FibonacciChick666
    • one year ago
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    but you also must know there is no lower bound

  77. FibonacciChick666
    • one year ago
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    no, you didn't

  78. kc_kennylau
    • one year ago
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    @FibonacciChick666 That can also be proved using limits

  79. FibonacciChick666
    • one year ago
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    continuous and limit are not the same thing in analysis

  80. freckles
    • one year ago
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    Well we want to show for all y in the codomain there exist x in the domain such that f(x)=y so you could just solve x^3=y for x and use that as your chosen element in the domain to prove that you have f(x)=y for all y

  81. FibonacciChick666
    • one year ago
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    you need to use a different set up

  82. kc_kennylau
    • one year ago
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    @FibonacciChick666 What is the definition of continuous?

  83. FibonacciChick666
    • one year ago
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    @freckles That has never made sense to me. I thought we had to prove the existence of an inverse?

  84. kc_kennylau
    • one year ago
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    @freckles The existence of an inverse uses the result itself

  85. freckles
    • one year ago
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    Well we want to show for all y in the codomain there exist x in the domain such that f(x)=y The key word here is there exist it it is kind of like an existential proof

  86. kc_kennylau
    • one year ago
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    \[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x-c|<\epsilon\right)\]which holds when \(\delta = \epsilon\)

  87. freckles
    • one year ago
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    you are choosing an element from the domain such that you have f(x)=y for all y

  88. FibonacciChick666
    • one year ago
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    uh, h/o it's been a while. But it is suuuuupppper close to ep delta limit def. I wanna say only difference is that you use \[|g(x)-g(y)|<\epsilon\]

  89. FibonacciChick666
    • one year ago
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    You have to get the terminology right

  90. freckles
    • one year ago
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    \[x^3=y \\ x=\sqrt[3]{y} \in \mathbb{R} \] you can still use this to show for every y there exist x such that f(x)=y plug in the x we chose and you will see this

  91. kc_kennylau
    • one year ago
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    @FibonacciChick666 What is the definition of continuous?

  92. kc_kennylau
    • one year ago
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    @freckles cube root is just a name. how to prove that it exists?

  93. freckles
    • one year ago
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    it exists because y is real and x is real

  94. FibonacciChick666
    • one year ago
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    same as ep delta for limit except you use g(x)-g(y) KC

  95. FibonacciChick666
    • one year ago
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    freckles, can we prove it another way? I just don't buy that enough

  96. freckles
    • one year ago
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    you can chose x to be cube root of y \[f(\sqrt[3]{y})=(\sqrt[3]{y})^3=y \] see we have chosen an x such that we have all y

  97. kc_kennylau
    • one year ago
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    \[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\] https://en.wikipedia.org/wiki/Continuous_function Wikipedia says I'm correct

  98. FibonacciChick666
    • one year ago
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    THAT is not the analysis definition

  99. kc_kennylau
    • one year ago
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    @freckles You have to prove that \(\sqrt[3]x\) exists

  100. FibonacciChick666
    • one year ago
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    trust me, I got it sooooooo wrong for trying that

  101. freckles
    • one year ago
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    it exists lol how can you say it does not?

  102. kc_kennylau
    • one year ago
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    what is \(g\) in your formula @FibonacciChick666

  103. kc_kennylau
    • one year ago
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    @freckles what the hell, you're like telling me "of course god exists, you can't disprove me"

  104. kc_kennylau
    • one year ago
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    @freckles I'm trying to prove that it exists, and you're using a tautology here

  105. freckles
    • one year ago
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    So you have never seen that function before?

  106. UsukiDoll
    • one year ago
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    wikipedia is a bad source.

  107. kc_kennylau
    • one year ago
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    @FibonacciChick666 what is g, what is x, what is y

  108. FibonacciChick666
    • one year ago
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    look at the weirstrass definition

  109. kc_kennylau
    • one year ago
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    @FibonacciChick666 what is the full form of the formula

  110. FibonacciChick666
    • one year ago
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    OMG you should know this if you can use it

  111. freckles
    • one year ago
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    what do you mean show it exists? what does that mean? I'm not understanding that... The only thing I can think it means it you never seen the function before.

  112. FibonacciChick666
    • one year ago
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    I do not have the time to explain the stupid mumblings of Newton and Weirstrass

  113. kc_kennylau
    • one year ago
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    @freckles I am trying to prove that for every y, there is an x such that x^3=y

  114. kc_kennylau
    • one year ago
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    @FibonacciChick666 okay i got the formula from wiki

  115. freckles
    • one year ago
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    we have done that

  116. UsukiDoll
    • one year ago
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    I feel bad for freckles and FibonacciChick666... x_X

  117. freckles
    • one year ago
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    There exists means I can pick an x , any x

  118. freckles
    • one year ago
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    as long as that x is a real number

  119. freckles
    • one year ago
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    I chose x to be cube root of y

  120. kc_kennylau
    • one year ago
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    @freckles I'm trying to prove that you can pick a number

  121. freckles
    • one year ago
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    this gets me what I want

  122. kc_kennylau
    • one year ago
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    such that x^3 = y

  123. FibonacciChick666
    • one year ago
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    ie I pick ne number that works and the rest is solid

  124. freckles
    • one year ago
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    I did pick a number cube root of y

  125. kc_kennylau
    • one year ago
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    That is just a name given to a function

  126. kc_kennylau
    • one year ago
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    whose existence is what we're trying to prove

  127. FibonacciChick666
    • one year ago
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    I get it now @freckles, I missed the there exists for way too long.

  128. freckles
    • one year ago
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    no we are trying to prove for all y THERE EXISTS an x such that f(x)=y

  129. UsukiDoll
    • one year ago
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    kc, you have to use correct definitions to fit this situation... I thought we're not supposed to pick numbers when writing a proof? I know if I let x =a number and y = a number I will get points knocked off.

  130. freckles
    • one year ago
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    This is an existential proof all you have to do is define a number x that fits our criteria to get the job done

  131. FibonacciChick666
    • one year ago
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    @UsukiDoll when you define it in terms of variables, you need to include the WLOG provided that is the case

  132. UsukiDoll
    • one year ago
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    oh xD

  133. kc_kennylau
    • one year ago
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    @FibonacciChick666 \[\Large\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x^3-c^3|<\epsilon\]

  134. kc_kennylau
    • one year ago
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    @freckles I was expecting something like the intermediate value theorem

  135. FibonacciChick666
    • one year ago
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    ie, since we can prove that cube root of x has an all reals domain, it is a completely possible choice

  136. FibonacciChick666
    • one year ago
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    you don't even know the definition of continuous, I doubt you are allowed IVT or MVT

  137. kc_kennylau
    • one year ago
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    @FibonacciChick666 The ability to choose is given by the fact that it is a surjection...

  138. kc_kennylau
    • one year ago
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    @FibonacciChick666 Come on, different people use different definitions of continuity

  139. FibonacciChick666
    • one year ago
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    In the proof you are missing the step where you set delta equal to something

  140. FibonacciChick666
    • one year ago
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    and not in analysis

  141. FibonacciChick666
    • one year ago
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    if you haven't hit that definition of continuous, you can't have proved IVT or MVT so you cannot use either

  142. kc_kennylau
    • one year ago
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    The condition for \(\Large y=x^3\) to be continuous is: \[\Large\forall\epsilon>0:\exists\delta>0:\forall x\in[c-\delta\,.\,.c+\delta]: c^3-\epsilon<x^3<c^3+\epsilon\]

  143. kc_kennylau
    • one year ago
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    which is satisfied when \(\delta = ...\)

  144. kc_kennylau
    • one year ago
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    @FibonacciChick666 this is what i meant, so you get why i didn't define delta

  145. FibonacciChick666
    • one year ago
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    You still have to define delta

  146. kc_kennylau
    • one year ago
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    I know, I'm thinking

  147. freckles
    • one year ago
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    http://www.math.csusb.edu/notes/proofs/pfnot/node4.html this gives a good example of existential proof another example: http://www.math.csusb.edu/notes/proofs/bpf/node5.html

  148. FibonacciChick666
    • one year ago
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    You haven't covered it. LISTEN TO FRECKLES, that's the basic way prior to this crap

  149. kc_kennylau
    • one year ago
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    @FibonacciChick666 okay sorry

  150. FibonacciChick666
    • one year ago
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    And read the first link on existential and this will be an ahhhh moment

  151. kc_kennylau
    • one year ago
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    @freckles you're using the result to prove the result... "y=x^3 is surjective because cbrt(x) exists because y=x^3 is surjective"

  152. FibonacciChick666
    • one year ago
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    read the link....

  153. FibonacciChick666
    • one year ago
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    good grief, http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000

  154. FibonacciChick666
    • one year ago
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    You don't write the part where you calculate the inverse in the proof

  155. FibonacciChick666
    • one year ago
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    that's scratch work

  156. freckles
    • one year ago
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    what do you mean I'm using the result to prove the result. I'm not using that it is surjective to prove it is surjective. I'm picking an x to show it is surjective, how is that using the result that is is surjective?

  157. kc_kennylau
    • one year ago
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    come on, cube root is defined as \(\Large \sqrt[3]x:=y\in\mathbb R:y^3=x\)

  158. FibonacciChick666
    • one year ago
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    If you read the link, you would find this a classic form of existence proof....

  159. freckles
    • one year ago
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    \[f(\sqrt[3]{y}))=(\sqrt[3]{y})^3 \text{ does this } x \text{ I chose \not give you } y ?\]

  160. kc_kennylau
    • one year ago
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    If we substitute the definition into your equation

  161. kc_kennylau
    • one year ago
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    \[\Large f(y\in\mathbb R:y^3=x)\ =\ (y\in\mathbb R:y^3=x)^3 = x\]

  162. FibonacciChick666
    • one year ago
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    did you read the link?

  163. freckles
    • one year ago
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    so you think we aren't suppose to use f(x)=x^3?

  164. kc_kennylau
    • one year ago
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    You haven't proved that the inverse of cube exists

  165. kc_kennylau
    • one year ago
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    (cube root is the inverse of cube)

  166. FibonacciChick666
    • one year ago
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    did you read the link? @kc_kennylau http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000

  167. kc_kennylau
    • one year ago
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    @FibonacciChick666 Yes, it says construct a number \(x\) such that \(x^3=y\)

  168. kc_kennylau
    • one year ago
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    @FibonacciChick666 And \(x=\sqrt[3]y\) is a valid construction, it says

  169. freckles
    • one year ago
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    http://www.mathwords.com/c/cube_root.htm this is the cube root function defined for you

  170. FibonacciChick666
    • one year ago
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    yes, it is, and we don't need to prove it

  171. kc_kennylau
    • one year ago
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    @freckles How do you prove that the inverse of cube exist

  172. freckles
    • one year ago
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    \[\sqrt[3]{8}=2\] like maybe you remember this function from algebra? like that is equal to 2 because 8 is 2*2*2

  173. FibonacciChick666
    • one year ago
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    (it's a function comes to mind)

  174. kc_kennylau
    • one year ago
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    "It exists because I say it exists and you feel that it exists" "and because you've been using it since eternity"

  175. FibonacciChick666
    • one year ago
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    prove why ln is the inverse of e.

  176. freckles
    • one year ago
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    so how does it not exist?

  177. FibonacciChick666
    • one year ago
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    can you prove it?

  178. FibonacciChick666
    • one year ago
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    NO it is the notation we chose to represent the inverse

  179. kc_kennylau
    • one year ago
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    @FibonacciChick666 There are two definitions of \(\ln\), one as the inverse of \(e^x\), one as \(\int_1^x \frac{dt}t\)

  180. FibonacciChick666
    • one year ago
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    that still proves nada

  181. kc_kennylau
    • one year ago
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    I have written the proof here: https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Natural_Logarithm

  182. FibonacciChick666
    • one year ago
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    It's exponents for cripes sake. You are saying that we can't multiply exponents right now.

  183. kc_kennylau
    • one year ago
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    @freckles "God exists because you can't prove that it doesn't"

  184. freckles
    • one year ago
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    so you are trying to prove definitions?

  185. freckles
    • one year ago
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    you don't prove definitions

  186. freckles
    • one year ago
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    you use definitions to prove other things

  187. FibonacciChick666
    • one year ago
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    Hint your proof is circular too

  188. kc_kennylau
    • one year ago
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    @FibonacciChick666 \(\ln x\) is a function because \(e^x\) is bijective

  189. kc_kennylau
    • one year ago
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    Because \(e^x\) is strictly increasing and continuous

  190. FibonacciChick666
    • one year ago
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    hahahah

  191. FibonacciChick666
    • one year ago
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    you are allowing an inverse log, but not an inverse exponent. seriously....

  192. FibonacciChick666
    • one year ago
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    you know logs undo a base to an exponent right?

  193. kc_kennylau
    • one year ago
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    @FibonacciChick666 Come on, \(e^x\) is defined as the limit to a sequence

  194. freckles
    • one year ago
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    I still don't understand It is like you are asking me why the number 2 exists it just does

  195. kc_kennylau
    • one year ago
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    @freckles No, I'm asking "why must there be a number whose cube is 8"

  196. FibonacciChick666
    • one year ago
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    because exponents are shorthand for multiplication and multiplication is a closed group on the set of reals?

  197. freckles
    • one year ago
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    because 2 is that number

  198. freckles
    • one year ago
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    and 2 exists

  199. freckles
    • one year ago
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    I'm very sure it does

  200. kc_kennylau
    • one year ago
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    @FibonacciChick666 No, \(x^r\) is defined as \(e^{r\ln x}\)

  201. kc_kennylau
    • one year ago
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    @freckles How do you obtain the number 2 from the number 8

  202. freckles
    • one year ago
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    you can take the cube root of 8

  203. kc_kennylau
    • one year ago
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    @FibonacciChick666 when r is a real number.

  204. FibonacciChick666
    • one year ago
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    .... exp[onents are defined as shorthand for multiplication....

  205. UsukiDoll
    • one year ago
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    What the heck did I just came back to?

  206. FibonacciChick666
    • one year ago
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    THAT IS THEIR DEFININTION

  207. kc_kennylau
    • one year ago
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    @freckles You're like saying "I obtain 2 from 8 because 2^3=8"

  208. freckles
    • one year ago
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    he doesn't believe the cube function exists

  209. FibonacciChick666
    • one year ago
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    ie x^r=x*x*x*....x r times

  210. kc_kennylau
    • one year ago
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    @FibonacciChick666 that definition is for integer r

  211. kc_kennylau
    • one year ago
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    @freckles Not cube, inverse of cube

  212. FibonacciChick666
    • one year ago
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    would you wait.

  213. kc_kennylau
    • one year ago
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    @freckles Prove that the inverse of cube exists

  214. FibonacciChick666
    • one year ago
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    WE DEFINE THE FUNCTION THAT SENDS EXPONENTS BACK(udoes them) as fractional exponents. It's a definition

  215. kc_kennylau
    • one year ago
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    @FibonacciChick666 Caps don't do you anything. It's just a name you give to a concept. Doesn't mean it exists.

  216. kc_kennylau
    • one year ago
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    @FibonacciChick666 Pink unicorn is defined as a unicorn that is pink. Doesn't mean it exists.

  217. kc_kennylau
    • one year ago
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    THAT IS THE DEFINITION OF A PINK UNICORN

  218. FibonacciChick666
    • one year ago
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    then do a bloody contradiction proof to make yourself happy.

  219. freckles
    • one year ago
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    IF you want to prove the inverse exist: Since f:R->R defined by f=x^3 is one-to-one then its inverse exist and it is g=cuberoot(x) And if you do the following: \[g(f(x))=\sqrt[3]{x^3}=x=(\sqrt[3]{x})^3=f(g(x))\]

  220. freckles
    • one year ago
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    which verifies they are inverses

  221. FibonacciChick666
    • one year ago
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    if there is no incerse of x^3 then no number multiplies 3 times to equal x^3. Counter ex. x^3=8 2*2*2=8 disproved voila

  222. freckles
    • one year ago
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    but ... I didn't really care about that in my proof because all I was trying to do was fun a x such that we had f(x)=y

  223. kc_kennylau
    • one year ago
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    @freckles Inverse of cube exists because cube is injection. I accept this proof. Thank you.

  224. freckles
    • one year ago
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    because all I was trying to do was find a x such that we had f(x)=y *

  225. UsukiDoll
    • one year ago
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    who is trying to prove a definition? that's crazy @____@

  226. freckles
    • one year ago
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    for a sec I thought you never seen the cube root function before

  227. FibonacciChick666
    • one year ago
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    that's what it sounded like

  228. kc_kennylau
    • one year ago
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    @freckles No, I'm just challenging the foundations of maths

  229. UsukiDoll
    • one year ago
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    that can get you into trouble man

  230. kc_kennylau
    • one year ago
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    "WHO ARE YOU TO EVEN CHALLENGE MATHS"

  231. UsukiDoll
    • one year ago
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    *facepalm*

  232. FibonacciChick666
    • one year ago
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    And being annoyingly ridiculous because it's not relevent

  233. ganeshie8
    • one year ago
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    i don't challenge math, math challenges i ;p

  234. kc_kennylau
    • one year ago
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    Can someone be fair here

  235. freckles
    • one year ago
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    no it is not fun to be fair :p

  236. FibonacciChick666
    • one year ago
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    FReckles has been more than patient with you. He deserves more than a medal...

  237. UsukiDoll
    • one year ago
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    it's fun to be UNFAIR >:D

  238. kc_kennylau
    • one year ago
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    okay, got to go now, thanks everyone, have a good day :D

  239. UsukiDoll
    • one year ago
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    he runnin... get em!

  240. kc_kennylau
    • one year ago
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    @freckles Sorry for not being comprehensable

  241. freckles
    • one year ago
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    good luck with your pink unicorns

  242. freckles
    • one year ago
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    I'm a freckled unicorn myself

  243. UsukiDoll
    • one year ago
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    I'm a Hatsune Miku Unicorn

  244. FibonacciChick666
    • one year ago
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    I'm a skeletal unicorn, similar to a threstral.

  245. anonymous
    • one year ago
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    @freckles Is this a valid proof to show one to one? $$ \Large{ \text{Assume: }\\ f(x)=f(y) \\ \text{Then } \\ x^3 = y^3 \\ \sqrt[3]{x^3} = \sqrt[3]{y^3} \\ x = y }$$

  246. freckles
    • one year ago
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    @jayzdd That sounds fine to me but I think the OP only believes cube root function can only exist if you show it is the inverse of the cube function.

  247. geerky42
    • one year ago
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    I had fun reading this entire discussion lol

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is replying to Can someone tell me what button the professor is hitting...

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