The reason I am posting this is that these seem too trivial and my "proofs" feel like I am doing nothing other than stating definitions, not even manipulating them.
Here is how functions equality is defined in the book:
Two functions $f : X → Y$ ,$g: X→ Y$ with the same domain and range are said to be equal, $f=g$, if and only if $f(x) = g(x)$ for all $x ∈ X$. (If $f(x)$ and $g(x)$ agree for
some values of $x$, but not others, then we do not consider $f$ and $g$ to be equal.)
My work:
Reflexive:
Let $f : X → Y$ be a function then by definition for all $x ∈ X$ we have $f(x)=f(x)$ so $f$ is equal to itself.
Symmetric:
Let $f : X → Y$ and $g: X→ Y$ be two functions, if we have $f=g$ then for all $x ∈ X$ we have $f(x)=g(x)$, if every input $x$ of $f$ gives the same output as when this input is put in $g$ then conversely every input $x$ of $g$ gives the same output as when this input is put in $f$, formally, $x ∈ X$ we have $g(x)=f(x)$, thus $g=f$ and the notion of function equality is symmetric.
Transitive:
Let $f : X → Y$ and $g: X→ Y$ and $h: X→ Y$ be three functions if $f=g$ and $g=h$, for all $x ∈ X$ we have $f(x)=g(x)$ and for all $x ∈ X$ we have $g(x)=h(x)$ similarly, we can essentially replace $g$ by $f$ as they are equal, so we get $f=h$, thus the notion of function equality is transitive.
Can someone let me know is what I did is correct and if not how to improve it?
Best Answer
For transitivity, you could directly invoke the transitivity of equality. That looks more elegant to me.
You may also want to improve the grammar by splitting up the various run-on sentences into separate bits.