Multiplication of a Range Say I have a number, "a".  0 < a < 5.
Let's say I have another number, "b".  0 < b < 10
Is ab < 50?
If you were to write this as a * b < 5 * b, you would find that the upper bound is < than 50, since b is strictly less than 10. 5 * <10 = < 50
If the upper bound is less than 50, doesn't a have to be less than a number that's less than 50, and not 50 itself?
 A: You use only one "less than" sign. 
Consider this as an analogy: Just as we would not say: $a = 5$, $b = 5$, so therefore $a + b$ "equals equals" $10$, but rather, we'd conclude that $a + b = 10$. In your case, we would not say $ab$ "is less than less than" $50$, but rather, $ab < 50$.
For any two real numbers, $x, y$, one and only one of the following is true: 


*

*$x < y\quad$ OR (x less than y)

*$x = y\quad$ OR (x equals y)

*$x > y\quad$ (x is greater than y)



EDIT following question edit:
To answer the question in the comment below: There is no largest number preceding 50: For every $k < 50$, there is a $j$ such that $k < j < 50$, and there is an $m$ such that $k < j < m < 50...$ and on and on and on...for every proposed "largest number $n$ preceding 50", there is a larger number than $n$ preceding 50. 
In this case, for example, take any possible choice of $ab < 50$. Then there exists $x =  \large \frac {50 + ab}{2}$ $> ab$ but nonetheless, $x < 50$. If we then let $ab = x$, there exists $y = \large\frac {50 + x}{2}$ $> x,$ with $y< 50$...and so on.
The least upper bound for $ab$, in this case, is not in the set of possible values for $ab$.  Think of it like this:  $$ 0 < a < 5,\;\;0 < b < 10, \;\; \implies \;\; ab \in (0, 50),$$ where $(0, 50)$ is the open interval of reals greater than $0$ but less than $50$ (the interval of all real numbers from $0$ to $50$, excluding $0$ and $50$:
$$ab \in \{x \in \mathbb{R}: 0 < x < 50\}$$
A: The product $ab$ is less than 50, but it's not "less than less than 50", because it can be as close to 50 as you want to make it: for any number $x$ at all, if $x<50$, then there is some $a$ and $b$ with $x < ab < 50$. 
For example, say $x = 50 - \frac1{1,000}$.  But then with $a = 5 - \frac1{1,000,000}$ and $b = 10-\frac1{1,000,000}$, you get $x < ab < 50$: $x = 49.999$, but $ab = 49.999985000001$, which is even closer to 50 than $x$ is.  And you can do this for any $x$ at all, no matter how close to 50 it is.
There is no "largest number preceding 50". For any $x < 50$, the number $25 + \frac x2$ is bigger than $x$ and still smaller than 50.
A: The basic definition of "<" for real numbers
is that $u < v$ means that
there is a positive real number $w$ such that
$u + w = v$.
If $0 < a < 5$, there is a positive number $p < 5$
such that $a+p = 5$.
Similarly,
if $0 < b < 10$,
there is a positive number $q < 10$
such that
$b+q = 10$.
So
$a b = (5-p)(10-q)
= 50 - 10p -5q + pq
$.
Since $p < 5$
and $q < 10$,
both $10p$ and $5q$
are greater than $pq$,
so $10p+5q-pq > 0$,
so $ab < 50$.
This clearly generalizes.
