An operation bar v on the set of real nvmbarsis defined by a bar b=3a+b-1 all over5 for all a,bER.
a. Determine whether or not the set R is closed vnder bar v
Response not known
since R is closed under addition (and thus by subtraction) and under multiplication and division where the divisor is not zero, what do you think?
To determine if the set of real numbers (R) is closed under the operation bar v, we need to check if any two real numbers, a and b, when operated with bar v, produce another real number.
Let's take any two real numbers a and b from the set R and perform the operation bar v on them.
The operation bar v is defined as b = (3a + b - 1) / 5.
Let's substitute the two real numbers a and b into the equation and see if the result is also a real number.
Using the given equation, we have:
b = (3a + b - 1) / 5
Let's simplify the equation:
5b = 3a + b - 1 (multiply both sides by 5 to eliminate the fraction)
4b = 3a - 1 (subtract b from both sides)
4b + 1 = 3a (add 1 to both sides)
a = (4b + 1) / 3 (divide both sides by 3)
Now, we have an expression for a in terms of b. To determine if the set of real numbers R is closed under bar v, we need to check if for any real number b, the corresponding value of a is also a real number.
Since b is an arbitrary real number, the expression (4b + 1) / 3 represents a linear function with a slope of 4/3 and a y-intercept of 1/3.
Since a linear function is defined for all real numbers, we can conclude that for any real number b, the corresponding value of a will also be a real number.
Therefore, the set of real numbers R is closed under the operation bar v.