1. ## Doubts in functions.

a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?

b) Prove that this relation need not hold, if f: A-->B and X ,Y are subsets of A.
f(X ^ Y) = f(X) ^ f(Y)
So is it enough if I just give a counter example or should I prove it?
and is this example right?

Say f:A---->B given by f(x) = a for all x in A,
Let X = {1,2,3}
Y = {4,5}
so here X ^ Y is null,
whereas f(X) =f(Y) = {1}
so f(X^Y) need not be equal to f(X)^ f(Y)

c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

I just wanted to know if my answer is correct.
# There are 2 functions which map A into B. That is I can say f(x) = 0 for all x in A. Or that f(x) = 1 for all x in B.

# From what I could see there are (2^n - 2) functions.
That is, i can map a element to one, and n-1 elements to the other - there are nC1 ways to do it.
I can may 2 elements to one, and n-2 elements to the other - there are nC2 ways to do it.
..
Similarly I can map n-1 elements to one, and 1 element to the other - there are nC(n-1) ways to do it.

So total number of ways to do it is nC1 + nC2 +.... nC(n-1) = 2^n-2

So are these right?
Have I gone wrong anywhere?

2. Originally Posted by poorna
c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

I just wanted to know if my answer is correct.
# There are 2 functions which map A into B. That is I can say f(x) = 0 for all x in A. Or that f(x) = 1 for all x in B.

# From what I could see there are (2^n - 2) functions.
That is, i can map a element to one, and n-1 elements to the other - there are nC1 ways to do it.
I can may 2 elements to one, and n-2 elements to the other - there are nC2 ways to do it.
..
Similarly I can map n-1 elements to one, and 1 element to the other - there are nC(n-1) ways to do it.

So total number of ways to do it is nC1 + nC2 +.... nC(n-1) = 2^n-2

So are these right?
Have I gone wrong anywhere?
The (c) part seems right.

4. Originally Posted by poorna

a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?

I think its just a whim...but look at the fallacy below.

Your question reminded me the following from my textbook:
$\sqrt{-1}.\sqrt{-1}=\sqrt{(-1)^2}=1$

5. ah, yes but that's (-1)^(1/2 + 1/2) = -1 right?

6. Originally Posted by poorna

a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?

b) Prove that this relation need not hold, if f: A-->B and X ,Y are subsets of A.
f(X ^ Y) = f(X) ^ f(Y)
So is it enough if I just give a counter example or should I prove it?
and is this example right?

Say f:A---->B given by f(x) = a for all x in A,
Let X = {1,2,3}
Y = {4,5}
so here X ^ Y is null,
whereas f(X) =f(Y) = {1}
so f(X^Y) need not be equal to f(X)^ f(Y)

c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

I just wanted to know if my answer is correct.
# There are 2 functions which map A into B. That is I can say f(x) = 0 for all x in A. Or that f(x) = 1 for all x in B.

# From what I could see there are (2^n - 2) functions.
That is, i can map a element to one, and n-1 elements to the other - there are nC1 ways to do it.
I can may 2 elements to one, and n-2 elements to the other - there are nC2 ways to do it.
..
Similarly I can map n-1 elements to one, and 1 element to the other - there are nC(n-1) ways to do it.

So total number of ways to do it is nC1 + nC2 +.... nC(n-1) = 2^n-2

So are these right?
Have I gone wrong anywhere?

For the (b) part ,since $X\cap Y = \emptyset$,you must prove 1st that :

$f(\emptyset) = \emptyset$ so you can show:

$f(X\cap Y)\neq f(X)\cap f(Y)$.

Also how do you prove that:

f({1,2,3}) = {1} ??

It would be easier if you put :

X ={1} and Y ={2}

FOR the (c) part,according to your formula for n=2 we have that:

$2^n - 2 = 2^2 -2 = 4 -2 =2$

Here is a counter example:

A = {1,2} , B={ a,b},then we have the following functions:

1) f(1) = a , f(2) =a

2) f(1) = b ,f(2 ) = b

3) f(1) = a ,f(2) = b

4) f(1) = b , f(2) = a

7. Originally Posted by xalk
FOR the (c) part,according to your formula for n=2 we have that:
$2^n - 2 = 2^2 -2 = 4 -2 =2$

Here is a counter example:
A = {1,2} , B={ a,b},then we have the following functions:
1) f(1) = a , f(2) =a
2) f(1) = b ,f(2 ) = b
3) f(1) = a ,f(2) = b
4) f(1) = b , f(2) = a
His/her part c) is correct.
The question is "how many onto (surjections) are there?"
There two in blue above.
The other two are not onto.

8. Originally Posted by Plato
His/her part c) is correct.
The question is "how many onto (surjections) are there?"
There two in blue above.
The other two are not onto.
her formula for n=2 ,gives two functions from A ={1,2} to B={a,b},while my counter example gives 4 functions

9. Originally Posted by poorna

a) In my book it says, if a > 0, the a^(x+y) = (a^x)(a^y)
Why have they mentioned it specifically for a>0. Is t just a whim or is there any reason behind it?
Hi poorna. The analytic definition of exponential with basis a is

a^x=e^(xlog a)

Hence, it is only well defined for a>0.

That is also the reason underlying because some mathematical programs only give the graphic representation of (x)^(1/3) for x positive, because it makes the calculation fot e^((1/3)(log x)). A lot of programs avoid the problem because this function can be easily defined for negative numbers
as -e^((1/3)log(-x)

10. Originally Posted by xalk
her formula for n=2 ,gives two functions from A ={1,2} to B={a,b},while my counter example gives 4 functions
zalk,
Please learn the vocabulary of set theory before correcting a posting.
The question asked for the number of onto functions (the number of surjections).
Two is the correct answer. I told you that above.

From the set $\left\{ {1,2,3, \cdots ,n} \right\}$ to the set $\left\{ {0,1} \right\}$ there $2^n -2$ surjecions.

Please be sure of the vocabulary before posting a correction to questions that you may not understand.

11. Originally Posted by poorna

c) If A = {1,2,....n} and B ={0,1}. How many functions map A into B? How many functions map A onto B?

# From what I could see there are (2^n - 2) functions.
HERE she asks for how many functions from A into B FIRSTLY AND then for the onto functions.