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- Mar 29th 2009, 04:26 PM #1

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## 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!

- Mar 29th 2009, 04:39 PM #2

- Mar 29th 2009, 06:46 PM #3

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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

- Mar 29th 2009, 11:01 PM #4

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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 toof the even numbers.**all**

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.

- Mar 30th 2009, 03:56 PM #5

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Update:

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

for a)

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)

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 when the odd number is squared and one is added will always yield an even number

I'm still working on c) and d)

- Apr 2nd 2009, 10:15 AM #6

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- Apr 2nd 2009, 10:47 AM #7

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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