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

- kc_kennylau

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

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

Is injection a subset of surjection?

- kc_kennylau

Well, we know that the domain and the co-domain (or image) are \(\mathbb R\)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

I'm sorry I don't want to sound stupid but what is surjection?

- kc_kennylau

And is injection a subset of surjection?

- kc_kennylau

all functions are surjections?

- kc_kennylau

@UsukiDoll and thank you for helping meeee

- UsukiDoll

I'm also new at this...let me think XD! oh gawd I rather prove in set theory

- kc_kennylau

lol

- 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

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

bijection is BOTH OF THOSE GUYS ^ ^

- UsukiDoll

so your x and y better satisfy surjection and injection a.k.a those definitions otherwise it's not a bijection.

- kc_kennylau

https://proofwiki.org/wiki/Existence_of_Positive_Root

- kc_kennylau

https://proofwiki.org/wiki/Uniqueness_of_Positive_Root

- 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

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

his is meant to be the word hit

- UsukiDoll

@freckles should've been my professor for Intro To Advanced Mathematics... I actually understood that...unlike my show off professor -_-

- 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

Continuous is \(\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\)?

- kc_kennylau

I think the limit is easy to prove, by using delta-epsilon def

- 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

number theory is awesome ^__^

- 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

That was just an example. You can do something very similar with your function.

- kc_kennylau

My question would not be simple, because you would be using the result if you prove it in your manner

- freckles

What do you mean?

- kc_kennylau

\(\Large x=y\implies \sqrt[3]x=\sqrt[3]y\) is using the result (circular)

- 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

I assure you, @FibonacciChick666 , continuous is not quick at all

- kc_kennylau

\[\displaystyle\large\forall c\in\mathbb R:\lim_{x\to c}x^3=c^3\]

- 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

onto freckles

- kc_kennylau

I guess both

- 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

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

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

that actually does show f isn't 1-1

- 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

oops too long

- 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

you forgot to define delta

- 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

I didn't forget xd

- FibonacciChick666

you have to say delta = epsilon

- kc_kennylau

@freckles Then how do we know that \(\Large x^3=y^3\implies x=y\)?

- FibonacciChick666

you are doing continuous, not limit

- freckles

you play with it

- freckles

\[x^3-y^3=0 \\ (x-y)(x^2+xy+y^2)=0\]

- kc_kennylau

Without using this result?

- 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

set both equal to 0
and you will see one equation is not possible over the reals anyways

- FibonacciChick666

That breakdown allows you to say that since x can't equal y, x^2+xy+y^2 must equal zero

- 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

@freckles I guess the fact that \(\Large x^3\) is increasing suffice to prove injection

- 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

no, not really

- FibonacciChick666

and I was doing pf by contradiction. At beginning say \(x\not =y\)

- UsukiDoll

@freckles not a real number because of that negative number in the radical... produces imaginary numbers.

- kc_kennylau

\[(f(x)=f(y)\implies x=y)\iff(x\ne y\implies f(x)\ne f(y))\]

- FibonacciChick666

you get a therefore since that's not real and by multiplicative property of zero x-y=0 so x=y

- kc_kennylau

The second statement follows from \(x^3\) being increasing

- FibonacciChick666

That's not enough

- UsukiDoll

x = y and x = that giant thing which isn't a real number

- kc_kennylau

If \(x\ne y\), WLOG let \(x

- FibonacciChick666

the ceiling function is increasing on 0, infty too

- 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

Then, since \(x^3\) is strictly increasing, \(x^3

- kc_kennylau

@FibonacciChick666 come on, you know I mean strictly increasing

- kc_kennylau

@freckles How to prove surjection?

- FibonacciChick666

still, you can have skips in strictly inc. You must have continuous to attempt that.

- kc_kennylau

@FibonacciChick666 I proved continuous using limit

- FibonacciChick666

but you also must know there is no lower bound

- FibonacciChick666

no, you didn't

- kc_kennylau

@FibonacciChick666 That can also be proved using limits

- FibonacciChick666

continuous and limit are not the same thing in analysis

- 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

you need to use a different set up

- kc_kennylau

@FibonacciChick666 What is the definition of continuous?

- FibonacciChick666

@freckles That has never made sense to me. I thought we had to prove the existence of an inverse?

- kc_kennylau

@freckles The existence of an inverse uses the result itself

- 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

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

- freckles

you are choosing an element from the domain such that you have f(x)=y for all y

- 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

You have to get the terminology right

- 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

@FibonacciChick666 What is the definition of continuous?

- kc_kennylau

@freckles cube root is just a name. how to prove that it exists?

- freckles

it exists because y is real and x is real

- FibonacciChick666

same as ep delta for limit except you use g(x)-g(y) KC

- FibonacciChick666

freckles, can we prove it another way? I just don't buy that enough

- 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

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

- FibonacciChick666

THAT is not the analysis definition

- kc_kennylau

@freckles You have to prove that \(\sqrt[3]x\) exists

- FibonacciChick666

trust me, I got it sooooooo wrong for trying that

- freckles

it exists lol
how can you say it does not?

- kc_kennylau

what is \(g\) in your formula @FibonacciChick666

- kc_kennylau

@freckles what the hell, you're like telling me "of course god exists, you can't disprove me"

- kc_kennylau

@freckles I'm trying to prove that it exists, and you're using a tautology here

- freckles

So you have never seen that function before?

- UsukiDoll

wikipedia is a bad source.

- kc_kennylau

@FibonacciChick666 what is g, what is x, what is y

- FibonacciChick666

look at the weirstrass definition

- kc_kennylau

@FibonacciChick666 what is the full form of the formula

- FibonacciChick666

OMG you should know this if you can use it

- 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

I do not have the time to explain the stupid mumblings of Newton and Weirstrass

- kc_kennylau

@freckles I am trying to prove that for every y, there is an x such that x^3=y

- kc_kennylau

@FibonacciChick666 okay i got the formula from wiki

- freckles

we have done that

- UsukiDoll

I feel bad for freckles and FibonacciChick666... x_X

- freckles

There exists means I can pick an x , any x

- freckles

as long as that x is a real number

- freckles

I chose x to be cube root of y

- kc_kennylau

@freckles I'm trying to prove that you can pick a number

- freckles

this gets me what I want

- kc_kennylau

such that x^3 = y

- FibonacciChick666

ie I pick ne number that works and the rest is solid

- freckles

I did pick a number cube root of y

- kc_kennylau

That is just a name given to a function

- kc_kennylau

whose existence is what we're trying to prove

- FibonacciChick666

I get it now @freckles, I missed the there exists for way too long.

- freckles

no
we are trying to prove for all y THERE EXISTS an x such that f(x)=y

- 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

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

@UsukiDoll when you define it in terms of variables, you need to include the WLOG provided that is the case

- UsukiDoll

oh xD

- 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

@freckles I was expecting something like the intermediate value theorem

- FibonacciChick666

ie, since we can prove that cube root of x has an all reals domain, it is a completely possible choice

- FibonacciChick666

you don't even know the definition of continuous, I doubt you are allowed IVT or MVT

- kc_kennylau

@FibonacciChick666 The ability to choose is given by the fact that it is a surjection...

- kc_kennylau

@FibonacciChick666 Come on, different people use different definitions of continuity

- FibonacciChick666

In the proof you are missing the step where you set delta equal to something

- FibonacciChick666

and not in analysis

- 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

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

which is satisfied when \(\delta = ...\)

- kc_kennylau

@FibonacciChick666 this is what i meant, so you get why i didn't define delta

- FibonacciChick666

You still have to define delta

- kc_kennylau

I know, I'm thinking

- 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

You haven't covered it. LISTEN TO FRECKLES, that's the basic way prior to this crap

- kc_kennylau

@FibonacciChick666 okay sorry

- FibonacciChick666

And read the first link on existential and this will be an ahhhh moment

- 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

read the link....

- FibonacciChick666

good grief,
http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000

- FibonacciChick666

You don't write the part where you calculate the inverse in the proof

- FibonacciChick666

that's scratch work

- 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

come on, cube root is defined as \(\Large \sqrt[3]x:=y\in\mathbb R:y^3=x\)

- FibonacciChick666

If you read the link, you would find this a classic form of existence proof....

- freckles

\[f(\sqrt[3]{y}))=(\sqrt[3]{y})^3 \text{ does this } x \text{ I chose \not give you } y ?\]

- kc_kennylau

If we substitute the definition into your equation

- kc_kennylau

\[\Large f(y\in\mathbb R:y^3=x)\ =\ (y\in\mathbb R:y^3=x)^3 = x\]

- FibonacciChick666

did you read the link?

- freckles

so you think we aren't suppose to use f(x)=x^3?

- kc_kennylau

You haven't proved that the inverse of cube exists

- kc_kennylau

(cube root is the inverse of cube)

- FibonacciChick666

did you read the link? @kc_kennylau http://www.math.csusb.edu/notes/proofs/pfnot/node8.html#SECTION00044000000000000000

- kc_kennylau

@FibonacciChick666 Yes, it says construct a number \(x\) such that \(x^3=y\)

- kc_kennylau

@FibonacciChick666 And \(x=\sqrt[3]y\) is a valid construction, it says

- freckles

http://www.mathwords.com/c/cube_root.htm
this is the cube root function defined for you

- FibonacciChick666

yes, it is, and we don't need to prove it

- kc_kennylau

@freckles How do you prove that the inverse of cube exist

- 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

(it's a function comes to mind)

- 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

prove why ln is the inverse of e.

- freckles

so how does it not exist?

- FibonacciChick666

can you prove it?

- FibonacciChick666

NO it is the notation we chose to represent the inverse

- 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

that still proves nada

- kc_kennylau

I have written the proof here:
https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Natural_Logarithm

- FibonacciChick666

It's exponents for cripes sake. You are saying that we can't multiply exponents right now.

- kc_kennylau

@freckles "God exists because you can't prove that it doesn't"

- freckles

so you are trying to prove definitions?

- freckles

you don't prove definitions

- freckles

you use definitions to prove other things

- FibonacciChick666

Hint your proof is circular too

- kc_kennylau

@FibonacciChick666 \(\ln x\) is a function because \(e^x\) is bijective

- kc_kennylau

Because \(e^x\) is strictly increasing and continuous

- FibonacciChick666

hahahah

- FibonacciChick666

you are allowing an inverse log, but not an inverse exponent. seriously....

- FibonacciChick666

you know logs undo a base to an exponent right?

- kc_kennylau

@FibonacciChick666 Come on, \(e^x\) is defined as the limit to a sequence

- freckles

I still don't understand
It is like you are asking me why the number 2 exists
it just does

- kc_kennylau

@freckles No, I'm asking "why must there be a number whose cube is 8"

- FibonacciChick666

because exponents are shorthand for multiplication and multiplication is a closed group on the set of reals?

- freckles

because 2 is that number

- freckles

and 2 exists

- freckles

I'm very sure it does

- kc_kennylau

@FibonacciChick666 No, \(x^r\) is defined as \(e^{r\ln x}\)

- kc_kennylau

@freckles How do you obtain the number 2 from the number 8

- freckles

you can take the cube root of 8

- kc_kennylau

@FibonacciChick666 when r is a real number.

- FibonacciChick666

.... exp[onents are defined as shorthand for multiplication....

- UsukiDoll

What the heck did I just came back to?

- FibonacciChick666

THAT IS THEIR DEFININTION

- kc_kennylau

@freckles You're like saying "I obtain 2 from 8 because 2^3=8"

- freckles

he doesn't believe the cube function exists

- FibonacciChick666

ie x^r=x*x*x*....x r times

- kc_kennylau

@FibonacciChick666 that definition is for integer r

- kc_kennylau

@freckles Not cube, inverse of cube

- FibonacciChick666

would you wait.

- kc_kennylau

@freckles Prove that the inverse of cube exists

- FibonacciChick666

WE DEFINE THE FUNCTION THAT SENDS EXPONENTS BACK(udoes them) as fractional exponents. It's a definition

- 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

@FibonacciChick666 Pink unicorn is defined as a unicorn that is pink. Doesn't mean it exists.

- kc_kennylau

THAT IS THE DEFINITION OF A PINK UNICORN

- FibonacciChick666

then do a bloody contradiction proof to make yourself happy.

- 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

which verifies they are inverses

- 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

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

@freckles Inverse of cube exists because cube is injection.
I accept this proof. Thank you.

- freckles

because all I was trying to do was find a x such that we had f(x)=y
*

- UsukiDoll

who is trying to prove a definition? that's crazy @____@

- freckles

for a sec I thought you never seen the cube root function before

- FibonacciChick666

that's what it sounded like

- kc_kennylau

@freckles No, I'm just challenging the foundations of maths

- UsukiDoll

that can get you into trouble man

- kc_kennylau

"WHO ARE YOU TO EVEN CHALLENGE MATHS"

- UsukiDoll

*facepalm*

- FibonacciChick666

And being annoyingly ridiculous because it's not relevent

- ganeshie8

i don't challenge math, math challenges i ;p

- kc_kennylau

Can someone be fair here

- freckles

no
it is not fun to be fair :p

- FibonacciChick666

FReckles has been more than patient with you. He deserves more than a medal...

- UsukiDoll

it's fun to be UNFAIR >:D

- kc_kennylau

okay, got to go now, thanks everyone, have a good day :D

- UsukiDoll

he runnin... get em!

- kc_kennylau

@freckles Sorry for not being comprehensable

- freckles

good luck with your pink unicorns

- freckles

I'm a freckled unicorn myself

- UsukiDoll

I'm a Hatsune Miku Unicorn

- FibonacciChick666

I'm a skeletal unicorn, similar to a threstral.

- 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

@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

I had fun reading this entire discussion lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.