Help with 1-1, onto fuctions; and functions! (emergency)

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• Oct 17th 2007, 09:20 PM
anu
Help with 1-1, onto fuctions; and functions! (emergency)
1) Let F be the function from R to R defined by f(x)=x^2. Find:
a) f^(-1) ({1})
b) f^(-1) ({x | 0 < x < 1})
c) f^(-1) ({x | x > 4})

2) Give an example of a function from N to N that is
a) one-to-one but not onto.
b) onto but not one-to-one.
c) both onto and one-to-one (but different from the identity function).
d) neither one-to-one nor onto.

Thank You very much in advance.
Anu
• Oct 17th 2007, 09:44 PM
angel.white
Quote:

Originally Posted by anu
1) Let F be the function from R to R defined by f(x)=x^2. Find:
a) f^(-1) ({1})
b) f^(-1) ({x | 0 < x < 1})
c) f^(-1) ({x | x > 4})

Okay, so the equation is $f(x)=x^{2}$

a) $f^{-1}(\{1\})$ is saying what value of f(x) will result in an answer where the inverse has a domain of 1. This sounds more complex than it is, because the range of the function is the domain of the inverse, all they're saying is what value of x will give you a y value of 1. Obviously there are two values for x that when squared give you 1. They would be positive and negative 1.

b) $f^{-1}(\{x | 0 < x < 1\})$ Okay, so just reword this to be what x values for f(x) will give you y values greater than 0 and less than 1. Well $0^{2} = 0$ and $1^{2}=1$ So you know that x values between 0 and 1 will give you y values between 0 and 1. Now, because the sign is greater than, but not equal to, and less than but not equal to, you know that x can never actually be 0 or 1. So the answer would be (0, 1) in parentheses.

c) $f^{-1}(\{x | x > 4\})$ Right, so what x values when plugged into the equation $f(x)=x^{2}$ will give you y values that are greater than 4?
• Oct 17th 2007, 09:48 PM
Jhevon
Quote:

Originally Posted by anu
1) Let F be the function from R to R defined by f(x)=x^2. Find:
a) f^(-1) ({1})

to find $f^{-1}(x)$ for x in the range of a function, is to find the element in the domain that maps to it. that is, if $f^{-1}(y)$ exists and $f^{-1}(y) = x$ then $f(x) = y$

so here, we want the element in the domain that maps to the element 1. for $f(x) = x^2$, there are two elements that map to one. thus: $f^{-1}( \{1 \}) = \{ 1, -1 \}$

since $f(1) = f(-1) = 1$

Quote:

b) f^(-1) ({x | 0 < x < 1})
here we want all the elements that map to the set {x | 0 < x < 1}. these are the values in the interval (-1,1) in the domain. do you see why?

so $f^{-1}( \{ x | 0 < x < 1 \}) = \left \{ y | -1 < y < 1, y \in \mathbb{R} \right \}$

Quote:

c) f^(-1) ({x | x > 4})
try this one by yourself.

Quote:

2) Give an example of a function from N to N that is
a) one-to-one but not onto.
take the even integers that are in $\mathbb{N}$.

so $f(n) = 2n$ for $n \in \mathbb{N}$ works. can you come up with any more functions? why is this function 1-1 but not onto?

Quote:

b) onto but not one-to-one.
how about $f(n) = \left \{ \begin{array}{lr} 1 , & \mbox{ if } n = 1 \\ n - 1, & \mbox { if } n \ge 2 \end{array} \right.$ for $n \in \mathbb{N} = 1,2,3,...$

why is this function onto but not 1-1?

Quote:

c) both onto and one-to-one (but different from the identity function).
$f(n) = \left \{ \begin{array}{lr} 2 , & \mbox{ if } n = 1 \\ 1, & \mbox { if } n = 2 \\ n, & \mbox { if } n \ge 3 \end{array} \right.$ for $n \in \mathbb{N} = 1,2,3,...$

Quote:

d) neither one-to-one nor onto.
you try this one

EDIT: Too slow, yet again
• Oct 17th 2007, 09:53 PM
anu
Quote:

Originally Posted by angel.white
c) $f^{-1}(\{x | x > 4\})$ Right, so what x values when plugged into the equation $f(x)=x^{2}$ will give you y values that are greater than 4?

x > 2, right?
• Oct 17th 2007, 09:58 PM
Jhevon
Quote:

Originally Posted by anu
x > 2, right?

what about, say, x = -5?
• Oct 17th 2007, 10:05 PM
anu
d) neither one-to-one nor onto.

f(n) = 1
then the function is neither 1-1 nor onto, right?

Quote:

Originally Posted by Jhevon
what about, say, x = -5?

Oh ya so x > 2 and x < -2
• Oct 17th 2007, 10:18 PM
angel.white
Righto ^_^

Of course, you need to put it in proper notation for your instructor, but I'm guessing you can handle that :P

And I'm impressed with how elegantly Jhevon answered the second question, I was sitting here laboring over it, only just finished question b when I realized it'd already been answered.

edit: I'm guessing you're using the book Discreet Mathematics and It's Applications by Kenneth H. Rosen? I'm actually taking that class right now myself :D this was in section 2.3, which will be on my test coming up this Wednesday (Sweating)
• Oct 17th 2007, 10:22 PM
Jhevon
Quote:

Originally Posted by anu
d) neither one-to-one nor onto.

f(n) = 1
then the function is neither 1-1 nor onto, right?

correct

Quote:

Oh ya so x > 2 and x < -2
again, correct. but it seems your questions demand notation that you are not employing here. use set notation to give the answer