MHB The function y = x is its own inverse?

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The function y = x is its own inverse because swapping the coordinates of its ordered pairs results in the same set, as each pair is of the form (x, x). This property is confirmed by the definition of an inverse function, which involves interchanging the coordinates. In contrast, other functions like y = x² do not share this property, as their ordered pairs change upon swapping. The discussion clarifies the concept through examples and definitions, leading to a better understanding of why y = x is unique in this regard. Overall, the explanation highlights the fundamental nature of inverse functions.
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The function y = x is its own inverse. Why?
 
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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?
 
Evgeny.Makarov said:
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
 
eleventhxhour said:
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 $\chi$ $\sigma$
 
eleventhxhour said:
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.
 
Evgeny.Makarov said:
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.
 
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.
 
Evgeny.Makarov said:
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! (:
 

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