How to understand that $\min(a,b) < c \Leftrightarrow a < c \text{ or } b < c$ and $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$?? In many probability related questions, I need to find the the probability of 
$P[\min(X_1,X_2) < c]$
where $X_1, X_2$ are random variables

However, I can never remember this basic relationship:
given any real numbers $a,b,c$, we have that $\min(a,b) < c \Leftrightarrow a < c \text{ or } b < c$.
Ditto, $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$.
I have looked up the explanation and people basically said this is just "logic". 

I guess my problem is not understanding what $\min(a,b) < c$ means. 
To me, it means pick a number less than $c$ that is either $a$ or $b$, and show me the smaller of the two.
Aha, that must be where the "or" come from. 
But wait a minute, it is also true that $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$.
In this case, we also have: pick a number less than $c$ that is either $a$ or $b$, and show me the larger of the two.
Now where would the "and" come from?

All in all, is there a more rigorous way of showing that this is true? 
 A: I think of $\min(a,b)<c$ to mean that "the smaller number of $a$ and $b$ is less than $c$. I think of $\max(a,b)<c$ to mean that "the largest number out of $a$ and $b$ is less than $c$. If you think of it this way, then it becomes quite clear: if the smallest number out of $a$ and $b$ is less than $c$ then this means at least one of them is less than $c$, so either $a<c$ or $b<c$. If the largest number out of $a$ and $b$ is less than $c$, then since it is the largest of the two, the other one is less than this maximum number, so they musts both be less that $c$.

More rigorously:
Suppose $a<b$. Then $\min(a,b)=a$, and since $\min(a,b)<c$, we have that $a<c$.
Alternatively, $b<a$. Then $\min(a,b)=b$, and since $\min(a,b)<c$, we have that $b<c$.
(Also, if $a=b$, then $\min(a,b)=a=b<c$, so this still holds)
So we have determined that if $\min(a,b)<c$, then either $a<c$ or $b<c$.
Using similar reasoning you can determine that the backward implication also holds.
For the $\max$ case, the proof is very similar, but the word I have in bold above, alternatively, becomes also.
A: Visualising numbers on an axis, it's obvious:


*

*The first assertion means the leftmost of $a$ and $b$ is left of $c$ if and only if one of $a$ and $b$ is left of $c$.

*The second assertion means the rightmost of $a$ and $b$ is left of $c$ if and only if both $a$ and $b$ are left of $c$.

A: 
I guess my problem is not understanding what $\min(a,b) < c$ means.
To me, it means pick a number less than $c$ that is either $a$ or $b$, and show me the smaller of the two.
Aha, that must be where the "or" come from.

Indeed, $\min\{a,b\}< c$ means that the lesser of the two values is less than $c$; so one of them is but the other might not be.   Therefore $a<c$ or $b<c$.
$$a\leq b<c,\\ a<c\leq b,\\ b<a<c,\\ b<c\leq a.$$

But wait a minute, it is also true that $\max(a,b) < c \Leftrightarrow a < c \text{ and } b < c$.
In this case, we also have: pick a number less than $c$ that is either $a$ or $b$, and show me the larger of the two.
Now where would the "and" come from?

$\max\{a,b\}<c$ means that the greater of the two values is less than $c$, and if the greater is then the lesser also must be too.   Hence, it means that both values are less than $c$   Therefore $a<c$ and $b<c$.
$$a\leq b<c,\\b<a<c.$$
