1. ## One-to-one and onto

Hey everyone, for homework we have to have one example of each under the condition that A = {odd numbers} and B = {even numbers}, thus:

give examples of functions from A to B which are:

a) One-to-one but not onto
b) Onto but not one-to-one
c) Neither one-to-one nor onto
d) One-to-one and onto

I've been working at this and I can't seem to get even the simplest of examples, ugh. Thanks for the help!

2. Originally Posted by teacast
Hey everyone, for homework we have to have one example of each under the condition that A = {odd numbers} and B = {even numbers}, thus: give examples of functions from A to B which are:
a) One-to-one but not onto
b) Onto but not one-to-one
c) Neither one-to-one nor onto
d) One-to-one and onto
Here is part a) $\displaystyle g(n)=4n$.
Reply by showing that the function works.

3. For a) g(n) = 4n,

Anything can be put into n, so when you apply g(n), then the number will always be even, thus always being in B? So 4 times any odd number will yield an even number, is that right?

As for b)

I would say that g(A) = B because there is at least one a in A such that g(a) = b

Let me know if/what I am doing or saying wrong, ugh 1-1 and onto escape me

4. 1-to-1 means no element in your domain A is simultaneously sent/mapped to two different images in your Range B. Nor are two different elements in your domain A mapped to the same image in the range B.

Onto means every element in your codomain B is 'hit'. That is, your map/function will map (some or all of) the odd numbers to all of the even numbers.

Your function needs to specify what it does to an element in A

so g(A) is ambiguous, as A is the whole set.

g(input) = output

your input is a member of A, your output is a member of B, g "maps" or "sends" your input to your output. You get to define the rule for this mapping.

5. Update:

So I looked it over, and read about onto and one to one and I think I get it.

for a)

$\displaystyle g(x) = 4x$ because whatever 4 is multiplied by, it will always yield an even number, thus satisfying what I want (input is odd, output is even).

for b)

$\displaystyle f(x) = x^2+1$ where x is in A

because Onto means that for every y in B, there is an x in A such that f(x)=y. So $\displaystyle x^2 + 1$ when the odd number is squared and one is added will always yield an even number

I'm still working on c) and d)

6. Originally Posted by teacast
Hey everyone, for homework we have to have one example of each under the condition that A = {odd numbers} and B = {even numbers}, thus:

give examples of functions from A to B which are:

a) One-to-one but not onto
b) Onto but not one-to-one
c) Neither one-to-one nor onto
d) One-to-one and onto

I've been working at this and I can't seem to get even the simplest of examples, ugh. Thanks for the help!
Hmmm...
the best notation to use would probably be the:

$\displaystyle f: \{ \begin{array}{l} \ A \rightarrow B\\ n \mapsto f(n) \end{array}$

c) might be:
$\displaystyle f: \{ \begin{array}{l} \ A \rightarrow B\\ n \mapsto f(0) \end{array}$
d) is probably pretty trivial
$\displaystyle f: \{ \begin{array}{l} \ A \rightarrow B\\ n \mapsto f(2n) \end{array}$

7. Originally Posted by teacast
For a) g(n) = 4n,

Anything can be put into n, so when you apply g(n), then the number will always be even, thus always being in B? So 4 times any odd number will yield an even number, is that right?
Yes, that is true. But now you need to show g is "one to one but not onto".

As for b)

I would say that g(A) = B because there is at least one a in A such that g(a) = b

Let me know if/what I am doing or saying wrong, ugh 1-1 and onto escape me
That makes no sense at all. What "g" are you talking about? Not g(n)= 4n, certainly, that was given to you as a function that is "one to one but not onto". You can't also say that it is "onto but not one to one"!