$\forall$ a,b $\in$ $\mathbb{R}$ satisfying a < b, $\exists$ n $\in$ $\mathbb{N}$ such that a + $\frac{1}{n}$ < b. My aim is to negate the statement.
Attempt:
$\exists$ a,b $\in$ $\mathbb{R}$ satisfying a $\geq$ b, $\forall$ $n \in \mathbb{N}$ we have that $a + \frac{1}{n} \geq b$.
When I went to search for the answer in the solution manual, I found that the author had the following:
$\exists a,b \in \mathbb{R}$ satisfying $a < b$ such that $\forall n \in \mathbb{N}$ we have $a + \frac{1}{n} \geq b$.
I don’t understand why the author didn’t negate the condition that $a < b$. Is this an error, or do I have a misunderstanding?
Thanks!

You misunderstand. The word "satisfying" implies a property of the numbers that is independent of the statement itself, much like a number can be prime, or divisible by 4. Therefore negation will have  no effect on this; the set of numbers is still the same.
You can also think of it like this:
The original statement can be reduced to this:
$a,b \in \mathbb{R}$ satisfying $a < b$ is equal to picking elements from the set $S = \{(a,b) \in \mathbb{R}^2 |  a< b\}$ and therefore the original statement can be written: $\forall a,b \in S, \exists n \in \mathbb{N}, a+\frac{1}{n}< b$
Applying negation to each part of this statement, you get:
$\exists a,b \in S, \forall n \in \mathbb{N}, a+\frac{1}{n} \geq b$
Hence the result.
Note that this negated statement is always false, a result of the Archemedian property of the Real numbers
