Advanced Calculus Help

**algebrapro18** · September 11th 2008, 07:05 AM

A) For each part, find a function f: R -> R that has the desired properties: neither onto nor one-to-one

B)Under what conditions does A\(A\B) = B?

C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n element of N(natural numbers).

D) Given A = {1,2,3,4,5}, B = {2,3,4,5,6,7} and C = {a,b,c,d,e} state an example of f: A ->B, g: B-> C, such that g(f)(the composition of g onto f) is 1-1 but g is not 1-1.

**bkarpuz** · September 11th 2008, 07:14 AM

Originally Posted by algebrapro18

B)Under what conditions does A\(A\B) = B?

See #2 of http://www.mathhelpforum.com/math-he...ts-proofs.html by kathrynmath.

**algebrapro18** · September 11th 2008, 09:29 AM

I read that thread and still can't see the answer.

**bkarpuz** · September 11th 2008, 09:47 AM

As we have msntioned previously, $A\backslash B=A\cap B^{c}$ , where $B^{c}$ is the compliment set of $B$ .
Noting that $(B^{c})^{c}=B$ , and simplifying the given expression, we get
$A\backslash(A\backslash B)=A\backslash(A\cap B^{c})$
.............. $=A\cap(A\cap B^{c})^{c}$
.............. $=A\cap(A^{c}\cup B)$
.............. $=\underset{\emptyset}{\underbrace{(A\cap A^{c})}}\cup(A\cap B)$
.............. $=A\cap B.\qquad(*)$
Since (*) is equal to $B$ , we infer that $B\subset A$ is true.

**algebrapro18** · September 11th 2008, 09:57 AM

Thanks, that helps me with part B now all I need help with is C because I got A and D done by my self.

**bkarpuz** · September 11th 2008, 10:11 AM

Originally Posted by algebrapro18

Thanks, that helps me with part B now all I need help with is C because I got A and D done by my self.

Can you please explain C a little bit more?
What is J or should we find what it is?
Also do we still want the function f not to be onto and into again?

**algebrapro18** · September 11th 2008, 10:14 AM

C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n which is an element of N.

I need to use what was given(the line above is all that is given) to find: the image of f, if f is 1-1, if f is onto, and if f is onto I need to find its inverse, the domain of the inverse, and the range of the inverse.

here is what I have so far:

The Im(f) has to be all odd numbers because that is what you get when you plug numbers into f(n)=2n-1.

from there I get stumped.

**bkarpuz** · September 11th 2008, 10:37 AM

Originally Posted by algebrapro18

C)Define f:J -> N(natural numbers) where f(n) = 2n-1 for each n element of N(natural numbers).

Although the following remark is not applicable for this exercise, I would like to tell it.

Remark. Let $f:A\to B$ be a function and $A,B$ be finite sets if f is onto, then it is one-to-one; vice versa, if f is one-to-one, then it is onto.

if $J\not\subset K:=\Big\{1,\frac{3}{2},2,\frac{5}{2},\ldots\Big\}$ , then $f$ can not be a function with an image which is a subset of $\mathbb{N}$ (pick an element which is not in the set $K$ , and see that it is mapped into the set $\mathbb{R}\backslash\mathbb{N}$ ).
Therefore, we must have $J\subset K$ .
If $J=K$ , then $f$ is one-to-one and onto.
If $J\neq K$ , then $f$ is only one-to-one.

Just, try to figure it out by yourself by letting $f(n)=2n-1=m$ , where $n\in K$ and $m\in\mathbb{N}$ . Note that $f$ is strictly increasing, which indicates that it is one-to-one.
Then obtain the set $K$ by isolating $n$ ...

**algebrapro18** · September 11th 2008, 11:00 AM

In english please...I understood about 1 persent of what you said...

**bkarpuz** · September 11th 2008, 11:50 AM

Originally Posted by algebrapro18

In english please...I understood about 1 persent of what you said...

If $f$ is a function from $J$ to $\mathbb{N}$ defined by $f(n)=2n-1$ , we see that $2n-1\in\mathbb{N}$ for all $n\in J$ .
This means that for every $n\in J$ , there exists $m\in\mathbb{N}$ such that $2n-1=m$ holds.
Note that for every $m\in\mathbb{N}$ , we may not find $n\in J$ .
Therefore, we see that
$<br /> n=\frac{m+1}{2}\text{ for }m\in\mathbb{N}<br />$
holds.
Since $m\in\mathbb{N}$ , we this indicates that
$<br /> n\in K:=\Big\{1,\frac{3}{2},2,\frac{5}{2},3,\frac{7}{2} ,\ldots\Big\}.<br />$
Hence, the domain of $f$ can be picked to be any subset of $K$ , i.e. $J\subset K$ .
Now consider the following possible cases.
If $J=K$ , then $f$ is one-to-one and onto.
If $J\neq K$ , then $f$ is only one-to-one.

Hint. $f$ is strictly icreasing, and it is hence one-to-one.

I guess it is more clear now?

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