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Question

Question: List all the elements of the following set: \[A=\left\\{ x:{{x}^{2}}\le 10,x\in Z \right\\}\]...

List all the elements of the following set:
A=\left\\{ x:{{x}^{2}}\le 10,x\in Z \right\\}

Explanation

Solution

In this question, we are given a set in set-builder form, and to solve that, we need to know that a set of set-builder forms like A = { f(x) : x follows a relation r, range}. Here, f (x) is the function that gives the elements of the set A, we will be always given a relation and the possible range.

Complete step-by-step answer:
In this question, we have to list all the elements of the set A=\left\\{ x:{{x}^{2}}\le 10,x\in Z \right\\}. Now, we have been given a relation that x210{{x}^{2}}\le 10. So, we can write 10x10-\sqrt{10}\le x\le \sqrt{10}. We have been also given that xZx\in Z, that is x belongs to integers. So, we can say that the value of x will be of the type \left\\{ 0,\pm 1,\pm 2,.... \right\\}. Now on using this relation 10x10-\sqrt{10}\le x\le \sqrt{10} and xZx\in Z we can say that the possible values of x can be {– 3, – 2, – 1, 0, 1, 2, 3}.
Now, we find the possible values of f (x) for different values of x. So, we will get the elements of A. We have been given f (x) = x. So,
For x = – 3, we get f(x) = – 3
For x = – 2, we get f(x) = – 2
For x = – 1, we get f(x) = – 1
For x = 0, we get f(x) = 0
For x = 1, we get f(x) = 1
For x = 2, we get f(x) = 2
For x = 3, we get f(x) = 3
Hence, we can say that the elements of set A are {– 3, – 2, – 1, 0, 1, 2, 3}.

Note: In this question, we have to be very careful while finding the value from the relationship because the relation is given for the higher power of x and we want the value of x. Also, we should remember that x2=(±x)2{{x}^{2}}={{\left( \pm x \right)}^{2}}, that is we get two values when we square root a number.