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, ...}
Yes, -1 and 1 are real numbers. Real numbers consist of irrational numbers, rational numbers and integers.
1
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
Given a function f(x), the domain is all of the numbers you are allowed to put in it (in other words x), and the range is all of the numbers you can get from it (in other words f(x)).Here are a few examples:f(x) = 3x+2D: All real numbers (-inf,inf)R: All real numbers (-inf,inf)f(x) = |x| + 2 (where |x| is the absolute value of x)D: All real numbers (-inf,inf)R: All real numbers greater than or equal to 2 [2,inf)f(x) = 1/(4x^2-1)D: All real numbers except 1/2 and -1/2 (because if you plug in either of those you get 1/0 which is undefined) (-inf,-1/2) u (-1/2,1/2) u (1/2,inf)R: All real numbers greater than or equal to -1 except 0. [-1,0) u (0,inf)
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, ...}
Yes, -1 and 1 are real numbers. Real numbers consist of irrational numbers, rational numbers and integers.
1
a number that cant be written on a number line. (ex. √-1) (the check is a square root sign)
Rational
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, ...}
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
0
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)
Given a function f(x), the domain is all of the numbers you are allowed to put in it (in other words x), and the range is all of the numbers you can get from it (in other words f(x)).Here are a few examples:f(x) = 3x+2D: All real numbers (-inf,inf)R: All real numbers (-inf,inf)f(x) = |x| + 2 (where |x| is the absolute value of x)D: All real numbers (-inf,inf)R: All real numbers greater than or equal to 2 [2,inf)f(x) = 1/(4x^2-1)D: All real numbers except 1/2 and -1/2 (because if you plug in either of those you get 1/0 which is undefined) (-inf,-1/2) u (-1/2,1/2) u (1/2,inf)R: All real numbers greater than or equal to -1 except 0. [-1,0) u (0,inf)
The question has no answer in real numbers. The solution, in complex numbers, are 2+3i and 2-3i where i is the imaginary square root of -1.