kc_kennylau
  • kc_kennylau
Prove that \(y=x^3\) where \(x,y\in\mathbb R\) is a bijection.
Mathematics
chestercat
  • chestercat
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UsukiDoll
  • UsukiDoll
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
kc_kennylau
  • kc_kennylau
Is injection a subset of surjection?
kc_kennylau
  • kc_kennylau
Well, we know that the domain and the co-domain (or image) are \(\mathbb R\)

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UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
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.
kc_kennylau
  • kc_kennylau
I'm sorry I don't want to sound stupid but what is surjection?
kc_kennylau
  • kc_kennylau
And is injection a subset of surjection?
kc_kennylau
  • kc_kennylau
all functions are surjections?
kc_kennylau
  • kc_kennylau
@UsukiDoll and thank you for helping meeee
UsukiDoll
  • UsukiDoll
I'm also new at this...let me think XD! oh gawd I rather prove in set theory
kc_kennylau
  • kc_kennylau
lol
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
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.
UsukiDoll
  • UsukiDoll
bijection is BOTH OF THOSE GUYS ^ ^
UsukiDoll
  • UsukiDoll
so your x and y better satisfy surjection and injection a.k.a those definitions otherwise it's not a bijection.
kc_kennylau
  • kc_kennylau
https://proofwiki.org/wiki/Existence_of_Positive_Root
kc_kennylau
  • kc_kennylau
https://proofwiki.org/wiki/Uniqueness_of_Positive_Root
FibonacciChick666
  • FibonacciChick666
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
freckles
  • freckles
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.
freckles
  • freckles
his is meant to be the word hit
UsukiDoll
  • UsukiDoll
@freckles should've been my professor for Intro To Advanced Mathematics... I actually understood that...unlike my show off professor -_-
FibonacciChick666
  • FibonacciChick666
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
kc_kennylau
  • kc_kennylau
Continuous is \(\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\)?
kc_kennylau
  • kc_kennylau
I think the limit is easy to prove, by using delta-epsilon def
FibonacciChick666
  • FibonacciChick666
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
UsukiDoll
  • UsukiDoll
number theory is awesome ^__^
freckles
  • freckles
\[\text{ Prove } g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=2x-1 \text{ is a bijection. } \\ \text{ Proof: (First \let's try to prove it is one-to-one)} \\ \text{ we want to try the following } f(a)=f(b) \implies \text{ only } a=b \\ \\f(a)=f(b) \\ 2a-1=2b-1 \\ 2a=2b \\ a=b \\ g \text{ is one-to-one } \\ \ \text{ Now we want to prove it is surjective } \\ \text{ we want to show } \\ \text{ for every } y \text{ in the codomain } \text{ there is } x \text{ in the domain } \\ \text{ such that } f(x)=y \\ \text{ so hmm... there exists } \\ \text{ so } 2x-1=y \text{ solve for } y \\ 2x=y+1 \\ x=\frac{y+1}{2} \\ \text{ so we have chosen an } x \text{ such that } \text{ we will have for every } y \\ \text{ there exist } x \text{ such that } f(x)=y \\ \text{ since } f(\frac{y+1}{2})=2(\frac{y+1}{2})-1=y+1-1=y\] This means g is a bijection.
freckles
  • freckles
That was just an example. You can do something very similar with your function.
kc_kennylau
  • kc_kennylau
My question would not be simple, because you would be using the result if you prove it in your manner
freckles
  • freckles
What do you mean?
kc_kennylau
  • kc_kennylau
\(\Large x=y\implies \sqrt[3]x=\sqrt[3]y\) is using the result (circular)
FibonacciChick666
  • FibonacciChick666
I'd use mean value theorem if you have it under your belt to prove onto. (after the quick showing of continuous)
kc_kennylau
  • kc_kennylau
I assure you, @FibonacciChick666 , continuous is not quick at all
kc_kennylau
  • kc_kennylau
\[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\]
freckles
  • freckles
I'm not sure what you are talking about. Are you having issues with the one-to-one part or the onto part?
FibonacciChick666
  • FibonacciChick666
onto freckles
kc_kennylau
  • kc_kennylau
I guess both
freckles
  • freckles
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
FibonacciChick666
  • FibonacciChick666
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.
freckles
  • freckles
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
freckles
  • freckles
that actually does show f isn't 1-1
kc_kennylau
  • kc_kennylau
\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x-c|<\epsilon\right)\]
kc_kennylau
  • kc_kennylau
oops too long
kc_kennylau
  • kc_kennylau
\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x-c|<\epsilon\right)\]
FibonacciChick666
  • FibonacciChick666
you forgot to define delta
kc_kennylau
  • kc_kennylau
\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x^3-c^3|<\epsilon\right)\]
kc_kennylau
  • kc_kennylau
I didn't forget xd
FibonacciChick666
  • FibonacciChick666
you have to say delta = epsilon
kc_kennylau
  • kc_kennylau
@freckles Then how do we know that \(\Large x^3=y^3\implies x=y\)?
FibonacciChick666
  • FibonacciChick666
you are doing continuous, not limit
freckles
  • freckles
you play with it
freckles
  • freckles
\[x^3-y^3=0 \\ (x-y)(x^2+xy+y^2)=0\]
kc_kennylau
  • kc_kennylau
Without using this result?
kc_kennylau
  • kc_kennylau
\[\Large\begin{array}{lrcl} (1):&x^3&=&y^3\\(2):&x^3-y^3&=&0\\(3):&(x-y)(x^2+xy+y^2)&=&0 \end{array}\]
freckles
  • freckles
set both equal to 0 and you will see one equation is not possible over the reals anyways
FibonacciChick666
  • FibonacciChick666
That breakdown allows you to say that since x can't equal y, x^2+xy+y^2 must equal zero
UsukiDoll
  • UsukiDoll
(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 ?
kc_kennylau
  • kc_kennylau
@freckles I guess the fact that \(\Large x^3\) is increasing suffice to prove injection
freckles
  • freckles
@FibonacciChick666 you are trying to show f(a)=f(b) only gives a=b and yes @UsukiDoll since the other equation gives: \[x=\frac{-y \pm \sqrt{-3y^2}}{2} \text{ which isn't a real number }\]
FibonacciChick666
  • FibonacciChick666
no, not really
FibonacciChick666
  • FibonacciChick666
and I was doing pf by contradiction. At beginning say \(x\not =y\)
UsukiDoll
  • UsukiDoll
@freckles not a real number because of that negative number in the radical... produces imaginary numbers.
kc_kennylau
  • kc_kennylau
\[(f(x)=f(y)\implies x=y)\iff(x\ne y\implies f(x)\ne f(y))\]
FibonacciChick666
  • FibonacciChick666
you get a therefore since that's not real and by multiplicative property of zero x-y=0 so x=y
kc_kennylau
  • kc_kennylau
The second statement follows from \(x^3\) being increasing
FibonacciChick666
  • FibonacciChick666
That's not enough
UsukiDoll
  • UsukiDoll
x = y and x = that giant thing which isn't a real number
kc_kennylau
  • kc_kennylau
If \(x\ne y\), WLOG let \(x
FibonacciChick666
  • FibonacciChick666
the ceiling function is increasing on 0, infty too
freckles
  • freckles
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
kc_kennylau
  • kc_kennylau
Then, since \(x^3\) is strictly increasing, \(x^3
kc_kennylau
  • kc_kennylau
@FibonacciChick666 come on, you know I mean strictly increasing
kc_kennylau
  • kc_kennylau
@freckles How to prove surjection?
FibonacciChick666
  • FibonacciChick666
still, you can have skips in strictly inc. You must have continuous to attempt that.
kc_kennylau
  • kc_kennylau
@FibonacciChick666 I proved continuous using limit
FibonacciChick666
  • FibonacciChick666
but you also must know there is no lower bound
FibonacciChick666
  • FibonacciChick666
no, you didn't
kc_kennylau
  • kc_kennylau
@FibonacciChick666 That can also be proved using limits
FibonacciChick666
  • FibonacciChick666
continuous and limit are not the same thing in analysis
freckles
  • freckles
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
FibonacciChick666
  • FibonacciChick666
you need to use a different set up
kc_kennylau
  • kc_kennylau
@FibonacciChick666 What is the definition of continuous?
FibonacciChick666
  • FibonacciChick666
@freckles That has never made sense to me. I thought we had to prove the existence of an inverse?
kc_kennylau
  • kc_kennylau
@freckles The existence of an inverse uses the result itself
freckles
  • freckles
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
kc_kennylau
  • kc_kennylau
\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x-c|<\epsilon\right)\]which holds when \(\delta = \epsilon\)
freckles
  • freckles
you are choosing an element from the domain such that you have f(x)=y for all y
FibonacciChick666
  • FibonacciChick666
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 \[|g(x)-g(y)|<\epsilon\]
FibonacciChick666
  • FibonacciChick666
You have to get the terminology right
freckles
  • freckles
\[x^3=y \\ x=\sqrt[3]{y} \in \mathbb{R} \] 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
kc_kennylau
  • kc_kennylau
@FibonacciChick666 What is the definition of continuous?
kc_kennylau
  • kc_kennylau
@freckles cube root is just a name. how to prove that it exists?
freckles
  • freckles
it exists because y is real and x is real
FibonacciChick666
  • FibonacciChick666
same as ep delta for limit except you use g(x)-g(y) KC
FibonacciChick666
  • FibonacciChick666
freckles, can we prove it another way? I just don't buy that enough
freckles
  • freckles
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
kc_kennylau
  • kc_kennylau
\[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\] https://en.wikipedia.org/wiki/Continuous_function Wikipedia says I'm correct
FibonacciChick666
  • FibonacciChick666
THAT is not the analysis definition
kc_kennylau
  • kc_kennylau
@freckles You have to prove that \(\sqrt[3]x\) exists
FibonacciChick666
  • FibonacciChick666
trust me, I got it sooooooo wrong for trying that
freckles
  • freckles
it exists lol how can you say it does not?
kc_kennylau
  • kc_kennylau
what is \(g\) in your formula @FibonacciChick666
kc_kennylau
  • kc_kennylau
@freckles what the hell, you're like telling me "of course god exists, you can't disprove me"
kc_kennylau
  • kc_kennylau
@freckles I'm trying to prove that it exists, and you're using a tautology here
freckles
  • freckles
So you have never seen that function before?
UsukiDoll
  • UsukiDoll
wikipedia is a bad source.
kc_kennylau
  • kc_kennylau
@FibonacciChick666 what is g, what is x, what is y
FibonacciChick666
  • FibonacciChick666
look at the weirstrass definition
kc_kennylau
  • kc_kennylau
@FibonacciChick666 what is the full form of the formula
FibonacciChick666
  • FibonacciChick666
OMG you should know this if you can use it
freckles
  • freckles
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.
FibonacciChick666
  • FibonacciChick666
I do not have the time to explain the stupid mumblings of Newton and Weirstrass
kc_kennylau
  • kc_kennylau
@freckles I am trying to prove that for every y, there is an x such that x^3=y
kc_kennylau
  • kc_kennylau
@FibonacciChick666 okay i got the formula from wiki
freckles
  • freckles
we have done that
UsukiDoll
  • UsukiDoll
I feel bad for freckles and FibonacciChick666... x_X
freckles
  • freckles
There exists means I can pick an x , any x
freckles
  • freckles
as long as that x is a real number
freckles
  • freckles
I chose x to be cube root of y
kc_kennylau
  • kc_kennylau
@freckles I'm trying to prove that you can pick a number
freckles
  • freckles
this gets me what I want
kc_kennylau
  • kc_kennylau
such that x^3 = y
FibonacciChick666
  • FibonacciChick666
ie I pick ne number that works and the rest is solid
freckles
  • freckles
I did pick a number cube root of y
kc_kennylau
  • kc_kennylau
That is just a name given to a function
kc_kennylau
  • kc_kennylau
whose existence is what we're trying to prove
FibonacciChick666
  • FibonacciChick666
I get it now @freckles, I missed the there exists for way too long.
freckles
  • freckles
no we are trying to prove for all y THERE EXISTS an x such that f(x)=y
UsukiDoll
  • UsukiDoll
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.
freckles
  • freckles
This is an existential proof all you have to do is define a number x that fits our criteria to get the job done
FibonacciChick666
  • FibonacciChick666
@UsukiDoll when you define it in terms of variables, you need to include the WLOG provided that is the case
UsukiDoll
  • UsukiDoll
oh xD
kc_kennylau
  • kc_kennylau
@FibonacciChick666 \[\Large\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:|x-c|<\delta\implies|x^3-c^3|<\epsilon\]
kc_kennylau
  • kc_kennylau
@freckles I was expecting something like the intermediate value theorem
FibonacciChick666
  • FibonacciChick666
ie, since we can prove that cube root of x has an all reals domain, it is a completely possible choice
FibonacciChick666
  • FibonacciChick666
you don't even know the definition of continuous, I doubt you are allowed IVT or MVT
kc_kennylau
  • kc_kennylau
@FibonacciChick666 The ability to choose is given by the fact that it is a surjection...
kc_kennylau
  • kc_kennylau
@FibonacciChick666 Come on, different people use different definitions of continuity
FibonacciChick666
  • FibonacciChick666
In the proof you are missing the step where you set delta equal to something
FibonacciChick666
  • FibonacciChick666
and not in analysis
FibonacciChick666
  • FibonacciChick666
if you haven't hit that definition of continuous, you can't have proved IVT or MVT so you cannot use either
kc_kennylau
  • kc_kennylau
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
kc_kennylau
  • kc_kennylau
which is satisfied when \(\delta = ...\)
kc_kennylau
  • kc_kennylau
@FibonacciChick666 this is what i meant, so you get why i didn't define delta
FibonacciChick666
  • FibonacciChick666
You still have to define delta
kc_kennylau
  • kc_kennylau
I know, I'm thinking
freckles
  • freckles
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
FibonacciChick666
  • FibonacciChick666
You haven't covered it. LISTEN TO FRECKLES, that's the basic way prior to this crap
kc_kennylau
  • kc_kennylau
@FibonacciChick666 okay sorry
FibonacciChick666
  • FibonacciChick666
And read the first link on existential and this will be an ahhhh moment
kc_kennylau
  • kc_kennylau
@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"
FibonacciChick666
  • FibonacciChick666
read the link....
FibonacciChick666
  • FibonacciChick666
good grief, http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000
FibonacciChick666
  • FibonacciChick666
You don't write the part where you calculate the inverse in the proof
FibonacciChick666
  • FibonacciChick666
that's scratch work
freckles
  • freckles
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?
kc_kennylau
  • kc_kennylau
come on, cube root is defined as \(\Large \sqrt[3]x:=y\in\mathbb R:y^3=x\)
FibonacciChick666
  • FibonacciChick666
If you read the link, you would find this a classic form of existence proof....
freckles
  • freckles
\[f(\sqrt[3]{y}))=(\sqrt[3]{y})^3 \text{ does this } x \text{ I chose \not give you } y ?\]
kc_kennylau
  • kc_kennylau
If we substitute the definition into your equation
kc_kennylau
  • kc_kennylau
\[\Large f(y\in\mathbb R:y^3=x)\ =\ (y\in\mathbb R:y^3=x)^3 = x\]
FibonacciChick666
  • FibonacciChick666
did you read the link?
freckles
  • freckles
so you think we aren't suppose to use f(x)=x^3?
kc_kennylau
  • kc_kennylau
You haven't proved that the inverse of cube exists
kc_kennylau
  • kc_kennylau
(cube root is the inverse of cube)
FibonacciChick666
  • FibonacciChick666
did you read the link? @kc_kennylau http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000
kc_kennylau
  • kc_kennylau
@FibonacciChick666 Yes, it says construct a number \(x\) such that \(x^3=y\)
kc_kennylau
  • kc_kennylau
@FibonacciChick666 And \(x=\sqrt[3]y\) is a valid construction, it says
freckles
  • freckles
http://www.mathwords.com/c/cube_root.htm this is the cube root function defined for you
FibonacciChick666
  • FibonacciChick666
yes, it is, and we don't need to prove it
kc_kennylau
  • kc_kennylau
@freckles How do you prove that the inverse of cube exist
freckles
  • freckles
\[\sqrt[3]{8}=2\] like maybe you remember this function from algebra? like that is equal to 2 because 8 is 2*2*2
FibonacciChick666
  • FibonacciChick666
(it's a function comes to mind)
kc_kennylau
  • kc_kennylau
"It exists because I say it exists and you feel that it exists" "and because you've been using it since eternity"
FibonacciChick666
  • FibonacciChick666
prove why ln is the inverse of e.
freckles
  • freckles
so how does it not exist?
FibonacciChick666
  • FibonacciChick666
can you prove it?
FibonacciChick666
  • FibonacciChick666
NO it is the notation we chose to represent the inverse
kc_kennylau
  • kc_kennylau
@FibonacciChick666 There are two definitions of \(\ln\), one as the inverse of \(e^x\), one as \(\int_1^x \frac{dt}t\)
FibonacciChick666
  • FibonacciChick666
that still proves nada
kc_kennylau
  • kc_kennylau
I have written the proof here: https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Natural_Logarithm
FibonacciChick666
  • FibonacciChick666
It's exponents for cripes sake. You are saying that we can't multiply exponents right now.
kc_kennylau
  • kc_kennylau
@freckles "God exists because you can't prove that it doesn't"
freckles
  • freckles
so you are trying to prove definitions?
freckles
  • freckles
you don't prove definitions
freckles
  • freckles
you use definitions to prove other things
FibonacciChick666
  • FibonacciChick666
Hint your proof is circular too
kc_kennylau
  • kc_kennylau
@FibonacciChick666 \(\ln x\) is a function because \(e^x\) is bijective
kc_kennylau
  • kc_kennylau
Because \(e^x\) is strictly increasing and continuous
FibonacciChick666
  • FibonacciChick666
hahahah
FibonacciChick666
  • FibonacciChick666
you are allowing an inverse log, but not an inverse exponent. seriously....
FibonacciChick666
  • FibonacciChick666
you know logs undo a base to an exponent right?
kc_kennylau
  • kc_kennylau
@FibonacciChick666 Come on, \(e^x\) is defined as the limit to a sequence
freckles
  • freckles
I still don't understand It is like you are asking me why the number 2 exists it just does
kc_kennylau
  • kc_kennylau
@freckles No, I'm asking "why must there be a number whose cube is 8"
FibonacciChick666
  • FibonacciChick666
because exponents are shorthand for multiplication and multiplication is a closed group on the set of reals?
freckles
  • freckles
because 2 is that number
freckles
  • freckles
and 2 exists
freckles
  • freckles
I'm very sure it does
kc_kennylau
  • kc_kennylau
@FibonacciChick666 No, \(x^r\) is defined as \(e^{r\ln x}\)
kc_kennylau
  • kc_kennylau
@freckles How do you obtain the number 2 from the number 8
freckles
  • freckles
you can take the cube root of 8
kc_kennylau
  • kc_kennylau
@FibonacciChick666 when r is a real number.
FibonacciChick666
  • FibonacciChick666
.... exp[onents are defined as shorthand for multiplication....
UsukiDoll
  • UsukiDoll
What the heck did I just came back to?
FibonacciChick666
  • FibonacciChick666
THAT IS THEIR DEFININTION
kc_kennylau
  • kc_kennylau
@freckles You're like saying "I obtain 2 from 8 because 2^3=8"
freckles
  • freckles
he doesn't believe the cube function exists
FibonacciChick666
  • FibonacciChick666
ie x^r=x*x*x*....x r times
kc_kennylau
  • kc_kennylau
@FibonacciChick666 that definition is for integer r
kc_kennylau
  • kc_kennylau
@freckles Not cube, inverse of cube
FibonacciChick666
  • FibonacciChick666
would you wait.
kc_kennylau
  • kc_kennylau
@freckles Prove that the inverse of cube exists
FibonacciChick666
  • FibonacciChick666
WE DEFINE THE FUNCTION THAT SENDS EXPONENTS BACK(udoes them) as fractional exponents. It's a definition
kc_kennylau
  • kc_kennylau
@FibonacciChick666 Caps don't do you anything. It's just a name you give to a concept. Doesn't mean it exists.
kc_kennylau
  • kc_kennylau
@FibonacciChick666 Pink unicorn is defined as a unicorn that is pink. Doesn't mean it exists.
kc_kennylau
  • kc_kennylau
THAT IS THE DEFINITION OF A PINK UNICORN
FibonacciChick666
  • FibonacciChick666
then do a bloody contradiction proof to make yourself happy.
freckles
  • freckles
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: \[g(f(x))=\sqrt[3]{x^3}=x=(\sqrt[3]{x})^3=f(g(x))\]
freckles
  • freckles
which verifies they are inverses
FibonacciChick666
  • FibonacciChick666
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
freckles
  • freckles
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
kc_kennylau
  • kc_kennylau
@freckles Inverse of cube exists because cube is injection. I accept this proof. Thank you.
freckles
  • freckles
because all I was trying to do was find a x such that we had f(x)=y *
UsukiDoll
  • UsukiDoll
who is trying to prove a definition? that's crazy @____@
freckles
  • freckles
for a sec I thought you never seen the cube root function before
FibonacciChick666
  • FibonacciChick666
that's what it sounded like
kc_kennylau
  • kc_kennylau
@freckles No, I'm just challenging the foundations of maths
UsukiDoll
  • UsukiDoll
that can get you into trouble man
kc_kennylau
  • kc_kennylau
"WHO ARE YOU TO EVEN CHALLENGE MATHS"
UsukiDoll
  • UsukiDoll
*facepalm*
FibonacciChick666
  • FibonacciChick666
And being annoyingly ridiculous because it's not relevent
ganeshie8
  • ganeshie8
i don't challenge math, math challenges i ;p
kc_kennylau
  • kc_kennylau
Can someone be fair here
freckles
  • freckles
no it is not fun to be fair :p
FibonacciChick666
  • FibonacciChick666
FReckles has been more than patient with you. He deserves more than a medal...
UsukiDoll
  • UsukiDoll
it's fun to be UNFAIR >:D
kc_kennylau
  • kc_kennylau
okay, got to go now, thanks everyone, have a good day :D
UsukiDoll
  • UsukiDoll
he runnin... get em!
kc_kennylau
  • kc_kennylau
@freckles Sorry for not being comprehensable
freckles
  • freckles
good luck with your pink unicorns
freckles
  • freckles
I'm a freckled unicorn myself
UsukiDoll
  • UsukiDoll
I'm a Hatsune Miku Unicorn
FibonacciChick666
  • FibonacciChick666
I'm a skeletal unicorn, similar to a threstral.
anonymous
  • anonymous
@freckles Is this a valid proof to show one to one? $$ \Large{ \text{Assume: }\\ f(x)=f(y) \\ \text{Then } \\ x^3 = y^3 \\ \sqrt[3]{x^3} = \sqrt[3]{y^3} \\ x = y }$$
freckles
  • freckles
@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.
geerky42
  • geerky42
I had fun reading this entire discussion lol

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