Proving that a function is injective and strictly increasing I have recently started doing proofs, and I find it quite hard to construct one. I am not really clear on how proofs should generally look like. What information should it include? Are we always need to suppose something is true and then arrive at the conclusion that our assumption along with the reasonings we provided imply something else (what's asked from us to be proved?) More specifically, questions such as these two are perfect examples of questions that I am not sure what I am supposed to show in a format of a proof. 


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*Let I ⊆ ℝ be an interval. We say that a function f: I → ℝ is strictly increasing if whenever $a,b ∈ I$ satisfy $a < b$, then $f(a) < f(b)$. Show that a strictly increasing function is injective.


That's what I have done: 
Given that a function is increasing for some a,b ∈ R, assume a≠b. Since the function is strictly increasing, then $a < b$, which would also imply that $f(a)<f(b)$. Therefore, this shows that $f(a)≠f(b)$ and that the function is injective as needed to be shown.


*Suppose f: I → ℝ is strictly increasing if whenever $a,b ∈ I$ satisfy $a < b$, then $f(a) < f(b)$. Show that $f^{-1}$ is also strictly increasing.


For this question, should I apply the same reasoning as that above, and add that taking the inverse of that function wouldn't change the fact that $a < b$? Would that be sufficient?

You have the right ideas, but you need to be more rigorous in how you make the arguments.
Let's start with the definition of being injective. A function is injective when for all $a,b$ in the domain, $a\neq b\to f(a)\neq f(b)$. So, let $a\neq b$. By the law of trichotomy, $a<b$ or $a>b$. Without loss of generality let $a<b$. Then, by assumption, we have that $f(a)<f(b)\to f(a)\neq f(b)$, which is what we desired.
For the second proof, we write down the definition of a strictly increasing function. For any $a,b$ in the domain where $a<b$, $f(a)<f(b)$. So, consider $c,d$ in the domain of $f^{-1}$ such that $c<d$. By definition of the inverse, there exists $a,b$ in the domain of $f$ such that $f(a)=c,\;f(b)=d$. We seek to show that $a < b$. We know that $f^{-1}$ is a bijection, so $a\neq b$. By the law of trichotomy, $a<b$ or $a>b$. If $a>b$, then by assumption, $f(c)>f(d)$, a contradiction. Hence, $a<b$ as desired.
