Question

If $R = \{ (x,y):x,y \in Z,{x^2} + {y^2} \leqslant 4\} $ is a relation defined on Z, Write domain of R.

Answer 1

Hint: First, we will define domain and range.
Since the domain of the function is represented as $f(x)$ which is the set of all values with the function is defined And where the range of the function is like the set of all values that domain takes from the relation f. Co-domain is a function that gets the values from the domain elements or functions from the relation f.

Complete step-by-step solution:
From the given we have a function $R = \{ (x,y):x,y \in Z,{x^2} + {y^2} \leqslant 4\} $ where R is the real numbers and x, y are integers which is given as the expression of ${x^2} + {y^2} \leqslant 4$ is the relation on the set of all integers and we will need to find the domain of R.
From the definition of the domain, which is a function represented by $f(x)$ and maybe a polynomial or relation.
Since from the relation Z,${x^2} + {y^2} \leqslant 4$ there are only three possible outcomes $0, \pm 1, \pm 2$ (because the maximum output of the function is at most four, so the square of the values does not exist after two)
Put $x = 0$ in the given relation on the integers, we get ${x^2} + {y^2} \leqslant 4 \Rightarrow 0 + {y^2} \leqslant 4$.
Thus, we get ${y^2} \leqslant 4$(if we take the square root there is at most plus or minus two is the answer).
Because of less than equal function we get $y = 0, \pm 1, \pm 2$ (is ${y^2} = 4 \Rightarrow y = 2$).
Similarly Put $x = \pm 1$ in the given relation on the integers, we get ${x^2} + {y^2} \leqslant 4 \Rightarrow 1 + {y^2} \leqslant 4$.
Thus, we get ${y^2} \leqslant 3$(if we take the square root there is at most plus or minus one is the answer).
Because of less than equal function we get $y = 0, \pm 1$ (is ${y^2} = 3 \Rightarrow y = 1.73$).
Finally, for Put $x = \pm 2$ in the given relation on the integers, we get ${x^2} + {y^2} \leqslant 4 \Rightarrow 4 + {y^2} \leqslant 4$.
Thus, we get ${y^2} \leqslant 0$. Therefore$y = 0$.
Hence the range of the function R is$\{ 0, \pm 1, \pm 2\} ,\{ 0, \pm 1\} ,\{ 0\} $.
Therefore, the domain of R is $R = \{ (x,y):x,y \in R\} = \{ 0, - 1,1, - 2,2\} $ (which are the common terms from the overall set).

Note: R is the set of all real numbers, where Z is the set of all integers.
There is a result according to this; $Z \subseteq R$ the set of integers is contained in the set of real numbers.
Where real numbers are contained in the set of all complex numbers $R \subseteq C$.
Complex numbers are real and imaginary terms.