Question

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

Answer 1

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

Answer 2

Is injection a subset of surjection?

Answer 3

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

Answer 4

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.

Answer 5

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.

Answer 6

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

Answer 7

And is injection a subset of surjection?

Answer 8

all functions are surjections?

Answer 9

@UsukiDoll and thank you for helping meeee

Answer 10

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

Answer 11

lol

Answer 12

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.

Answer 13

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.

Answer 14

bijection is BOTH OF THOSE GUYS ^ ^

Answer 15

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

Answer 16

https://proofwiki.org/wiki/Existence_of_Positive_Root

Answer 17

https://proofwiki.org/wiki/Uniqueness_of_Positive_Root

Answer 18

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

Answer 19

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
\[f: \mathbb{R} \rightarrow \mathbb{R} \text{ defined by} f(x)=x^2 \]

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.

Answer 20

his is meant to be the word hit

Answer 21

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

Answer 22

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