X^3 and x belong to n is finite
To clarify, it seems like you're asking about the set of numbers where x^3 and x both belong to the set of natural numbers (n), and whether this set is finite.
Let's break this down:
The natural numbers, denoted by N, are the set of counting numbers: 1, 2, 3, 4, and so on.
Now, if we consider the equation x^3, it represents all the possible values of x raised to the power of 3. For example, if x = 2, then 2^3 = 2 * 2 * 2 = 8. So, x^3 would produce the set of values with all possible cubes.
So, the question is whether there exists a finite set of numbers that satisfies both x^3 and x being natural numbers.
In this case, the answer is yes, and the set is finite. We can determine this by considering the values of x that would satisfy the equation x^3 = x:
1. If x = 0, then 0^3 = 0, which satisfies the equation.
2. If x = 1, then 1^3 = 1, which also satisfies the equation.
These are the only two solutions. Any other natural number x would not satisfy the equation x^3 = x. For example, if x = 2, then 2^3 = 8, which is not equal to 2.
Hence, the set of numbers where x^3 and x both belong to the set of natural numbers is finite, as it contains only the values x = 0 and x = 1.