The function y = x is its own inverse?

[eleventhxhour]

The function y = x is its own inverse. Why?

 

[Evgeny.Makarov]

To answer this question you need to know the definition of the inverse function (in general). Do you belong to the set of people who have this knowledge?

 

[eleventhxhour]

To answer this question you need to know the definition of the inverse function (in general). Do you belong to the set of people who have this knowledge?

Yup, I know what an inverse function is. I just don't understand how it can be its own inverse

 

[chisigma]

The function y = x is its own inverse. Why?

A trivial explanation of that is $\displaystyle y = x \implies x = y$. Also the function $\displaystyle y= \frac{1}{x}$ is its own inverse because $\displaystyle y = \frac{1}{x} \implies x= \frac{1}{y}$. Other functions don't have this property, for example $\displaystyle y= x^{2} \implies x = \sqrt{y}$... Kind regards $\chi$ $\sigma$

 

[Evgeny.Makarov]

Yup, I know what an inverse function is. I just don't understand how it can be its own inverse

Let's see the definition that you are using and we'll determine if $y=x$ fits it.

 

[eleventhxhour]

Let's see the definition that you are using and we'll determine if $y=x$ fits it.

The inverse of a function is a relation formed by interchanging the coordinates within each ordered pair of the original function.

 

[Evgeny.Makarov]

Great. Let's assume that the domain and codomain of the function $y=x$ are $D$. Then the function is $\{(x,x)\mid x\in D\}$. Obviously, if you swap the elements of each pair, you get the same set since each pair does not change.

 

[eleventhxhour]

Great. Let's assume that the domain and codomain of the function $y=x$ are $D$. Then the function is $\{(x,x)\mid x\in D\}$. Obviously, if you swap the elements of each pair, you get the same set since each pair does not change.

Ohh, okay. That makes sense. Thanks! (: