A community for students.
Here's the question you clicked on:
 0 viewing
kc_kennylau
 one year ago
Prove that \(y=x^3\) where \(x,y\in\mathbb R\) is a bijection.
kc_kennylau
 one year ago
Prove that \(y=x^3\) where \(x,y\in\mathbb R\) is a bijection.

This Question is Open

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1x 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
 one year ago
Best ResponseYou've already chosen the best response.1Is injection a subset of surjection?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Well, we know that the domain and the codomain (or image) are \(\mathbb R\)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.1bijection 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
 one year ago
Best ResponseYou've already chosen the best response.1I'm sorry I don't want to sound stupid but what is surjection?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1And is injection a subset of surjection?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1all functions are surjections?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@UsukiDoll and thank you for helping meeee

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I'm also new at this...let me think XD! oh gawd I rather prove in set theory

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1an injective function or injection or onetoone 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
 one year ago
Best ResponseYou've already chosen the best response.1bijection is BOTH OF THOSE GUYS ^ ^

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so your x and y better satisfy surjection and injection a.k.a those definitions otherwise it's not a bijection.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2for that is injective a subset of surjective, nah. In the world of non functions, you can have onto with out having onetoone they are mutually exclusive classifications

freckles
 one year ago
Best ResponseYou've already chosen the best response.11A function is onto if every element of the codomain gets hit. A function is 1to1 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 onetoone. Example: f(1)=f(1)=1. 1 gets hit more than 1 times.

freckles
 one year ago
Best ResponseYou've already chosen the best response.11his is meant to be the word hit

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1@freckles should've been my professor for Intro To Advanced Mathematics... I actually understood that...unlike my show off professor _

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2oohooh, 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
 one year ago
Best ResponseYou've already chosen the best response.1Continuous is \(\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\)?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I think the limit is easy to prove, by using deltaepsilon def

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2but listen to freckles, he got me through abstract alg, basic analysis, preadvanced calculus, number theory, and all the other proofy stuff I'd rather forget

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1number theory is awesome ^__^

freckles
 one year ago
Best ResponseYou've already chosen the best response.11\[\text{ Prove } g: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } g(x)=2x1 \text{ is a bijection. } \\ \text{ Proof: (First \let's try to prove it is onetoone)} \\ \text{ we want to try the following } f(a)=f(b) \implies \text{ only } a=b \\ \\f(a)=f(b) \\ 2a1=2b1 \\ 2a=2b \\ a=b \\ g \text{ is onetoone } \\ \ \text{ Now we want to prove it is surjective } \\ \text{ we want to show } \\ \text{ for every } y \text{ in the codomain } \text{ there is } x \text{ in the domain } \\ \text{ such that } f(x)=y \\ \text{ so hmm... there exists } \\ \text{ so } 2x1=y \text{ solve for } y \\ 2x=y+1 \\ x=\frac{y+1}{2} \\ \text{ so we have chosen an } x \text{ such that } \text{ we will have for every } y \\ \text{ there exist } x \text{ such that } f(x)=y \\ \text{ since } f(\frac{y+1}{2})=2(\frac{y+1}{2})1=y+11=y\] This means g is a bijection.

freckles
 one year ago
Best ResponseYou've already chosen the best response.11That was just an example. You can do something very similar with your function.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1My question would not be simple, because you would be using the result if you prove it in your manner

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\(\Large x=y\implies \sqrt[3]x=\sqrt[3]y\) is using the result (circular)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2I'd use mean value theorem if you have it under your belt to prove onto. (after the quick showing of continuous)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I assure you, @FibonacciChick666 , continuous is not quick at all

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.11I'm not sure what you are talking about. Are you having issues with the onetoone part or the onto part?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2onto freckles

freckles
 one year ago
Best ResponseYou've already chosen the best response.11Suppose you have:\[v: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by } v(x)=x^3\] for 11 you want to suppose v(a)=v(b) and arrive at the implication of only a=b

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.11Like with my f function above it wouldn't have worked because a^2=b^2 means we have a^2b^2=0 which means we have a=b or a=b

freckles
 one year ago
Best ResponseYou've already chosen the best response.11that actually does show f isn't 11

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:xc<\delta\impliesxc<\epsilon\right)\]

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:xc<\delta\impliesxc<\epsilon\right)\]

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you forgot to define delta

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:xc<\delta\impliesx^3c^3<\epsilon\right)\]

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I didn't forget xd

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you have to say delta = epsilon

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles Then how do we know that \(\Large x^3=y^3\implies x=y\)?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you are doing continuous, not limit

freckles
 one year ago
Best ResponseYou've already chosen the best response.11\[x^3y^3=0 \\ (xy)(x^2+xy+y^2)=0\]

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Without using this result?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large\begin{array}{lrcl} (1):&x^3&=&y^3\\(2):&x^3y^3&=&0\\(3):&(xy)(x^2+xy+y^2)&=&0 \end{array}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.11set both equal to 0 and you will see one equation is not possible over the reals anyways

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2That breakdown allows you to say that since x can't equal y, x^2+xy+y^2 must equal zero

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1(xy)(x^2+xy+y^2)=0 (xy) = 0 and then x^2+xy+y^2 = 0 but for xy=0 wouldn't we have just x = y ?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles I guess the fact that \(\Large x^3\) is increasing suffice to prove injection

freckles
 one year ago
Best ResponseYou've already chosen the best response.11@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
 one year ago
Best ResponseYou've already chosen the best response.2no, not really

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2and I was doing pf by contradiction. At beginning say \(x\not =y\)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1@freckles not a real number because of that negative number in the radical... produces imaginary numbers.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[(f(x)=f(y)\implies x=y)\iff(x\ne y\implies f(x)\ne f(y))\]

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you get a therefore since that's not real and by multiplicative property of zero xy=0 so x=y

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1The second statement follows from \(x^3\) being increasing

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2That's not enough

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1x = y and x = that giant thing which isn't a real number

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1If \(x\ne y\), WLOG let \(x<y\)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2the ceiling function is increasing on 0, infty too

freckles
 one year ago
Best ResponseYou've already chosen the best response.11yes 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
 one year ago
Best ResponseYou've already chosen the best response.1Then, since \(x^3\) is strictly increasing, \(x^3<y^3 \implies x^3\ne y^3\)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 come on, you know I mean strictly increasing

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles How to prove surjection?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2still, you can have skips in strictly inc. You must have continuous to attempt that.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 I proved continuous using limit

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2but you also must know there is no lower bound

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2no, you didn't

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 That can also be proved using limits

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2continuous and limit are not the same thing in analysis

freckles
 one year ago
Best ResponseYou've already chosen the best response.11Well 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
 one year ago
Best ResponseYou've already chosen the best response.2you need to use a different set up

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 What is the definition of continuous?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2@freckles That has never made sense to me. I thought we had to prove the existence of an inverse?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles The existence of an inverse uses the result itself

freckles
 one year ago
Best ResponseYou've already chosen the best response.11Well 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\displaystyle\large\left(\forall c\in\mathbb R:\lim_{x\to c}x=c\right)\iff\\\left(\forall c\in\mathbb R:\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:xc<\delta\impliesxc<\epsilon\right)\]which holds when \(\delta = \epsilon\)

freckles
 one year ago
Best ResponseYou've already chosen the best response.11you are choosing an element from the domain such that you have f(x)=y for all y

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2uh, 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
 one year ago
Best ResponseYou've already chosen the best response.2You have to get the terminology right

freckles
 one year ago
Best ResponseYou've already chosen the best response.11\[x^3=y \\ x=\sqrt[3]{y} \in \mathbb{R} \] 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
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 What is the definition of continuous?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles cube root is just a name. how to prove that it exists?

freckles
 one year ago
Best ResponseYou've already chosen the best response.11it exists because y is real and x is real

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2same as ep delta for limit except you use g(x)g(y) KC

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2freckles, can we prove it another way? I just don't buy that enough

freckles
 one year ago
Best ResponseYou've already chosen the best response.11you 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
 one year ago
Best ResponseYou've already chosen the best response.1\[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\] https://en.wikipedia.org/wiki/Continuous_function Wikipedia says I'm correct

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2THAT is not the analysis definition

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles You have to prove that \(\sqrt[3]x\) exists

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2trust me, I got it sooooooo wrong for trying that

freckles
 one year ago
Best ResponseYou've already chosen the best response.11it exists lol how can you say it does not?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1what is \(g\) in your formula @FibonacciChick666

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles what the hell, you're like telling me "of course god exists, you can't disprove me"

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles I'm trying to prove that it exists, and you're using a tautology here

freckles
 one year ago
Best ResponseYou've already chosen the best response.11So you have never seen that function before?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1wikipedia is a bad source.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 what is g, what is x, what is y

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2look at the weirstrass definition

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 what is the full form of the formula

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2OMG you should know this if you can use it

freckles
 one year ago
Best ResponseYou've already chosen the best response.11what 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
 one year ago
Best ResponseYou've already chosen the best response.2I do not have the time to explain the stupid mumblings of Newton and Weirstrass

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles I am trying to prove that for every y, there is an x such that x^3=y

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 okay i got the formula from wiki

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I feel bad for freckles and FibonacciChick666... x_X

freckles
 one year ago
Best ResponseYou've already chosen the best response.11There exists means I can pick an x , any x

freckles
 one year ago
Best ResponseYou've already chosen the best response.11as long as that x is a real number

freckles
 one year ago
Best ResponseYou've already chosen the best response.11I chose x to be cube root of y

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles I'm trying to prove that you can pick a number

freckles
 one year ago
Best ResponseYou've already chosen the best response.11this gets me what I want

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2ie I pick ne number that works and the rest is solid

freckles
 one year ago
Best ResponseYou've already chosen the best response.11I did pick a number cube root of y

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1That is just a name given to a function

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1whose existence is what we're trying to prove

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2I get it now @freckles, I missed the there exists for way too long.

freckles
 one year ago
Best ResponseYou've already chosen the best response.11no we are trying to prove for all y THERE EXISTS an x such that f(x)=y

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1kc, 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
 one year ago
Best ResponseYou've already chosen the best response.11This is an existential proof all you have to do is define a number x that fits our criteria to get the job done

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2@UsukiDoll when you define it in terms of variables, you need to include the WLOG provided that is the case

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 \[\Large\forall\epsilon>0:\exists\delta>0:\forall x\in\mathbb R:xc<\delta\impliesx^3c^3<\epsilon\]

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles I was expecting something like the intermediate value theorem

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2ie, since we can prove that cube root of x has an all reals domain, it is a completely possible choice

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you don't even know the definition of continuous, I doubt you are allowed IVT or MVT

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 The ability to choose is given by the fact that it is a surjection...

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 Come on, different people use different definitions of continuity

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2In the proof you are missing the step where you set delta equal to something

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2and not in analysis

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2if you haven't hit that definition of continuous, you can't have proved IVT or MVT so you cannot use either

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1The condition for \(\Large y=x^3\) to be continuous is: \[\Large\forall\epsilon>0:\exists\delta>0:\forall x\in[c\delta\,.\,.c+\delta]: c^3\epsilon<x^3<c^3+\epsilon\]

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1which is satisfied when \(\delta = ...\)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 this is what i meant, so you get why i didn't define delta

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2You still have to define delta

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I know, I'm thinking

freckles
 one year ago
Best ResponseYou've already chosen the best response.11http://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
 one year ago
Best ResponseYou've already chosen the best response.2You haven't covered it. LISTEN TO FRECKLES, that's the basic way prior to this crap

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 okay sorry

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2And read the first link on existential and this will be an ahhhh moment

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@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
 one year ago
Best ResponseYou've already chosen the best response.2read the link....

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2good grief, http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2You don't write the part where you calculate the inverse in the proof

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2that's scratch work

freckles
 one year ago
Best ResponseYou've already chosen the best response.11what 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
 one year ago
Best ResponseYou've already chosen the best response.1come on, cube root is defined as \(\Large \sqrt[3]x:=y\in\mathbb R:y^3=x\)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2If you read the link, you would find this a classic form of existence proof....

freckles
 one year ago
Best ResponseYou've already chosen the best response.11\[f(\sqrt[3]{y}))=(\sqrt[3]{y})^3 \text{ does this } x \text{ I chose \not give you } y ?\]

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1If we substitute the definition into your equation

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1\[\Large f(y\in\mathbb R:y^3=x)\ =\ (y\in\mathbb R:y^3=x)^3 = x\]

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2did you read the link?

freckles
 one year ago
Best ResponseYou've already chosen the best response.11so you think we aren't suppose to use f(x)=x^3?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1You haven't proved that the inverse of cube exists

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1(cube root is the inverse of cube)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2did you read the link? @kc_kennylau http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 Yes, it says construct a number \(x\) such that \(x^3=y\)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 And \(x=\sqrt[3]y\) is a valid construction, it says

freckles
 one year ago
Best ResponseYou've already chosen the best response.11http://www.mathwords.com/c/cube_root.htm this is the cube root function defined for you

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2yes, it is, and we don't need to prove it

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles How do you prove that the inverse of cube exist

freckles
 one year ago
Best ResponseYou've already chosen the best response.11\[\sqrt[3]{8}=2\] like maybe you remember this function from algebra? like that is equal to 2 because 8 is 2*2*2

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2(it's a function comes to mind)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1"It exists because I say it exists and you feel that it exists" "and because you've been using it since eternity"

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2prove why ln is the inverse of e.

freckles
 one year ago
Best ResponseYou've already chosen the best response.11so how does it not exist?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2can you prove it?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2NO it is the notation we chose to represent the inverse

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 There are two definitions of \(\ln\), one as the inverse of \(e^x\), one as \(\int_1^x \frac{dt}t\)

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2that still proves nada

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1I have written the proof here: https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Natural_Logarithm

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2It's exponents for cripes sake. You are saying that we can't multiply exponents right now.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles "God exists because you can't prove that it doesn't"

freckles
 one year ago
Best ResponseYou've already chosen the best response.11so you are trying to prove definitions?

freckles
 one year ago
Best ResponseYou've already chosen the best response.11you don't prove definitions

freckles
 one year ago
Best ResponseYou've already chosen the best response.11you use definitions to prove other things

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2Hint your proof is circular too

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 \(\ln x\) is a function because \(e^x\) is bijective

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Because \(e^x\) is strictly increasing and continuous

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you are allowing an inverse log, but not an inverse exponent. seriously....

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2you know logs undo a base to an exponent right?

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 Come on, \(e^x\) is defined as the limit to a sequence

freckles
 one year ago
Best ResponseYou've already chosen the best response.11I still don't understand It is like you are asking me why the number 2 exists it just does

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles No, I'm asking "why must there be a number whose cube is 8"

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2because exponents are shorthand for multiplication and multiplication is a closed group on the set of reals?

freckles
 one year ago
Best ResponseYou've already chosen the best response.11because 2 is that number

freckles
 one year ago
Best ResponseYou've already chosen the best response.11I'm very sure it does

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 No, \(x^r\) is defined as \(e^{r\ln x}\)

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles How do you obtain the number 2 from the number 8

freckles
 one year ago
Best ResponseYou've already chosen the best response.11you can take the cube root of 8

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 when r is a real number.

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2.... exp[onents are defined as shorthand for multiplication....

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1What the heck did I just came back to?

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2THAT IS THEIR DEFININTION

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles You're like saying "I obtain 2 from 8 because 2^3=8"

freckles
 one year ago
Best ResponseYou've already chosen the best response.11he doesn't believe the cube function exists

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2ie x^r=x*x*x*....x r times

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 that definition is for integer r

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles Not cube, inverse of cube

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2would you wait.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles Prove that the inverse of cube exists

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2WE DEFINE THE FUNCTION THAT SENDS EXPONENTS BACK(udoes them) as fractional exponents. It's a definition

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 Caps don't do you anything. It's just a name you give to a concept. Doesn't mean it exists.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@FibonacciChick666 Pink unicorn is defined as a unicorn that is pink. Doesn't mean it exists.

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1THAT IS THE DEFINITION OF A PINK UNICORN

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2then do a bloody contradiction proof to make yourself happy.

freckles
 one year ago
Best ResponseYou've already chosen the best response.11IF you want to prove the inverse exist: Since f:R>R defined by f=x^3 is onetoone 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
 one year ago
Best ResponseYou've already chosen the best response.11which verifies they are inverses

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2if 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
 one year ago
Best ResponseYou've already chosen the best response.11but ... 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
 one year ago
Best ResponseYou've already chosen the best response.1@freckles Inverse of cube exists because cube is injection. I accept this proof. Thank you.

freckles
 one year ago
Best ResponseYou've already chosen the best response.11because all I was trying to do was find a x such that we had f(x)=y *

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1who is trying to prove a definition? that's crazy @____@

freckles
 one year ago
Best ResponseYou've already chosen the best response.11for a sec I thought you never seen the cube root function before

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2that's what it sounded like

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles No, I'm just challenging the foundations of maths

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1that can get you into trouble man

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1"WHO ARE YOU TO EVEN CHALLENGE MATHS"

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2And being annoyingly ridiculous because it's not relevent

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i don't challenge math, math challenges i ;p

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1Can someone be fair here

freckles
 one year ago
Best ResponseYou've already chosen the best response.11no it is not fun to be fair :p

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2FReckles has been more than patient with you. He deserves more than a medal...

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1it's fun to be UNFAIR >:D

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1okay, got to go now, thanks everyone, have a good day :D

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1he runnin... get em!

kc_kennylau
 one year ago
Best ResponseYou've already chosen the best response.1@freckles Sorry for not being comprehensable

freckles
 one year ago
Best ResponseYou've already chosen the best response.11good luck with your pink unicorns

freckles
 one year ago
Best ResponseYou've already chosen the best response.11I'm a freckled unicorn myself

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1I'm a Hatsune Miku Unicorn

FibonacciChick666
 one year ago
Best ResponseYou've already chosen the best response.2I'm a skeletal unicorn, similar to a threstral.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.11@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
 one year ago
Best ResponseYou've already chosen the best response.0I had fun reading this entire discussion lol
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.