Nonempty sets Discrete Mathematics Textbook Question:

Textbook Solution:

For part a) and d)  :


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*What is the mathematical proof to show that the set is an empty set or nonempty set ?

 A: When we write $\{x\in S: P(x)\}$ what we mean is "the set of all numbers, $x$, such that $x$ is an element of $x$ and property $P$ holds for $x$. This set is non-empty if there is at least one such number. In your case, the properties are always either equations, or equations combined with the assertion that the number falls in a certain set (though more generally they don't have to be), and so the set is non-empty precisely when there is a solution to the equation.
The first set is the set of all $x$ such that $x$ is in $\mathbb{N}=\{0,1,2,\ldots\}$ and $2x+7=3$. We can solve the equation using algebra to find out that the only solution is $x=-2$, which is not a natural number. Thus there are no numbers, x, in $\mathbb{N}$ such that $2x+7=3$. Therefore $\{x|x\in\mathbb{N},2x+7=3\}=\emptyset$.
The fourth set is the set of all $x$ such that $x$ is in $\mathbb{R}$ and $x^2+4=6$. Again doing algebra tells us that there are two solutions, $x=\sqrt{2}$ and $x=-\sqrt{2}$. Both of these are real numbers, and so $\{x\in\mathbb{R}|x^2+4=6\}=\{-\sqrt{2},\sqrt{2}\}$ which is non-empty.
