# Discrete math textbook problem

1. Jan 7, 2009

1. The problem statement, all variables and given/known data
find the domain and image of f such that
f(x) = {(x,y) $$\in R \times R \vert x = \sqrt{y+3}$$
and domain and image of g such that
g = {$$(\alpha,\beta) \vert \alpha is a person, \beta is a person, \alpha is the father of \beta$$

2. Relevant equations
the domain and image are what would be expected of domain and image. No unstandard definitions.

3. The attempt at a solution
I think the answers in the back of the book are wrong. I will state my answers, the ones in the book have the opposite for domain and range (although it is a copyright 2009 book). It is also possible that I am wrong.

okay for f, the function can be rewritten by substituing $$\sqrt{y+3}$$ for x.Then ,
f = {$$(\sqrt{y+3},y) \vert y \in R$$. This means that the domain will consist of values that are bigger or equal to -3. So, $$dom f = \{x \vert x \in R , x \geq -3 \}$$. And following that, the image of f is the set of values that are greater than or equal to zero. $$im f = \{y \vert y \in R, y \geq 0 \}$$.

For the second one. The domain is all people who are fathers, and the image is all people.(assuming that any person has a father where father denotes only a biological relationship).

Any suggestions?

2. Jan 7, 2009

### Staff: Mentor

For the first problem, and assuming that by domain the text's author means the set of x values, x has to be >= 0. The range is {y >= -3}.

3. Jan 7, 2009

Ah yes, because $$x = \sqrt{y+3}$$, it would be impossible to have an x that is negative. Although, in case y is as small as possible, x can be zero. Thus, x >= 0. Now on the case of y, flip the radical and get that y = x^2 - 3. x >= 0, thus y >= -3. I see, I guess I fell into the trap the author wanted me to fall into.
g = $$\{ (\alpha,\beta) \vert \alpha,\beta \in P, R(\alpha,\beta) \}$$