discrete math. sets problem. true or false?

Question

let T = {(x,y) ∈ R x R(real numbers) | y = √x / (√(x-3))

Let D = {y ∈ R, | (x,y) ∈ T for x ∈ R}.

for each one, decide if true or false

a) 1/4 ∈ D

b) 4 ∈ D


why?

Answer 1

[a].  FALSE    If 1/4 ∈ D, then there is a real number x such that 

                        1/4 = √x/√(x-3).  So, 1/16 = x/(x-3).

 

                         Cross multiply to get 16x = x - 3

                                                              15x = -3

                                                                  x = -1/5  But, √(-1/5) ∉ R

 

[b]. TRUE     If 4 ∈ D, then there is a real number x such that 

                       4 = √x/√(x-3).  So, 16 = x/(x-3)

 

                       Cross multiply to get 16x - 48 = x

                                                            15x = 48

                                                                x = 48/15 = 16/5

 

                       (16/5,4) ∈ T and 4 ∈ D

Answer 2

a) False; y^2>1 and 1/16 is not greater than 1.

 

b) True; 4^2=16=x/(x-3) implies that 16x-16*3=x or x=16/5.

Answer 3

a) if 1/4 belongs to D, then 1/4 is Real and the y coordinate of an ordered pair in T. So what value of x leads to a y =1/4?

   1/4 = sqrt x / sqrt(x-3)....square both sides, cross multiply and x=-1/5. But that is and extraneous root since you can't

    take the square root of a negative number. So there is no value of x that leads to a y of 1/4, so 1/4 is not in D.

b) If 4 belongs to D, then 4 is real and the y coordinate of an ordered pair in T. So what value of x leads to a y=4?

    4 = sqrt x / sqrt(x-3)......square bothe sides, cross multiply and x = 16/5. When you plug it in to check, This number

    works and leads to the answer 4. Therefore 4 does belong to D.

 

Hope this helped.