Yes they are real numbers
Some examples of sets of real numbers include: The set of positive integers: {1, 2, 3, 4, ...} The set of rational numbers: {1/2, -3/4, 5/6, ...} The set of whole numbers: {..., -2, -1, 0, 1, 2, ...} The set of natural numbers: {0, 1, 2, 3, 4, ...} The set of irrational numbers: {√2, π, e, ...}
1
Yes, -1 and 1 are real numbers. Real numbers consist of irrational numbers, rational numbers and integers.
There are no real numbers that will satisfy the requirements. In the complex field, the solution is 1+i and 1-i where i is the imaginary square root of (-1); that is, i^2 = -1
The range from 2 to -3 includes all the integers and real numbers between these two values. If considering only integers, the numbers are 2, 1, 0, -1, -2, and -3, totaling six integers. If including all real numbers, there are infinitely many numbers in that range.
Yes. :S real numbers are real numbers. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Some examples of sets of real numbers include: The set of positive integers: {1, 2, 3, 4, ...} The set of rational numbers: {1/2, -3/4, 5/6, ...} The set of whole numbers: {..., -2, -1, 0, 1, 2, ...} The set of natural numbers: {0, 1, 2, 3, 4, ...} The set of irrational numbers: {√2, π, e, ...}
1
a number that cant be written on a number line. (ex. √-1) (the check is a square root sign)
Yes, -1 and 1 are real numbers. Real numbers consist of irrational numbers, rational numbers and integers.
The set of natural numbers or counting numbers N is a subset of the set of real numbers R. N = {1, 2, 3, ...) R = {..., -2, -1, -0.5, 0, 1, √2, 2, 3, π, ...}
Rational
There are no real numbers that will satisfy the requirements. In the complex field, the solution is 1+i and 1-i where i is the imaginary square root of (-1); that is, i^2 = -1
Natural numbers, whole numbers, integers, rational numbers
0
The range from 2 to -3 includes all the integers and real numbers between these two values. If considering only integers, the numbers are 2, 1, 0, -1, -2, and -3, totaling six integers. If including all real numbers, there are infinitely many numbers in that range.
There are no real numbers that can do that.The two numbers that can are- 1/2 - 1/2 j sqrt(7)and - 1/2 + 1/2 j sqrt(7)