Please excuse my ignorance, I am doing a Mathematics for IT paper and have not done this type of maths for over 20 years.
The problem I have is this:
Let: $$T = \{x\, \lvert x \in \mathbb{R}, -10 \le x \le 5\}$$ Therefore I have figured out that $$T = \{-10, -9, -8. -7. -6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$$
I have to use set builder notation to solve $T \cap {√2}$ .
The answer I keep coming back to is $\{\}$. Could I please get assistance as to if I am on the right track?
Many thanks
Note that $$T = \{x | x ∈ ℝ, -10 ≤ x ≤ 5\} = [-10,5]$$
is the set of all real numbers between $-10$ and $5$ inclusive.
Since $\sqrt 2$ is a member of $T$, we have $$T\cap \{\sqrt 2\} = \{\sqrt 2\} $$