Can someone help me understand how to do this problem?
Let F:N -> {0,1} be defined by F(n)= 0 if 3|n or 1 if 3|n
a)Show F is NOT one-to-one.
b)Show F is onto.
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Can someone help me understand how to do this problem?
Let F:N -> {0,1} be defined by F(n)= 0 if 3|n or 1 if 3|n
a)Show F is NOT one-to-one.
b)Show F is onto.
Quote:
Originally Posted by matthayzon89
Look carefully at what you wrote after "..be defined by..." : it makes no sense.
Tonio
Why does it make no sense? it does make sense....Quote:
Originally Posted by tonio
read it like this:
Let the funtion that MAPS Natural numbers to 0,1 be defined by F(n)= 0 if 3|n or 1 if 3|n
It still makes no sense.Quote:
Originally Posted by matthayzon89
0 if 3|n or 1 if 3|n.
That reads 0 if three divides n or 1 if three divides n.
That is nonsense.
Sorry, but what are you using | to mean?
Quote:
Originally Posted by Plato
SORRY!!
This is what it was suppose to say, I found out that there was a TYPO on the assignment itself...
It is suppose to read 0 if 3(does NOT)|n or 1 if 3|n
Well then:Quote:
Originally Posted by matthayzon89
a) ifit is not one-to-one.
b). So...?
So b) is ONTO.Quote:
Originally Posted by PlatoWell then:
a) if
b)
p.S
I need help trying to understand the question and knowing what to look for.... I don't really need help solving it. Hints don't help much either.
Can someone please explain how to approach such a problem and what I need to look for in order to prove something is onto or one-to-one?
Any function is a set of ordered pairs.Quote:
Originally Posted by matthayzon89
A function is one-to-one is no two pairs have the same second term.
A function is onto if the range is the same as the co-domain (the final set).
So I know the answer b/c I went over it with my prof. but I still dont really understand it, can someone please explain?
Here it is:
F:N -> {0,1}
F(n)= {0 if 3 does not |n, 1 if 3|n)
a) is NOT one-to-one <- I know this is due to having more then one value mapped to the same second value, but I dont understand how this conclusion is figured out given the information of the problem.
Proof:
Consider n(base1)= 1 and n(base2)= 2
Then F(n(base1)=F(n(base2))=0 but n(base1)DOES NOT EQUAL 2.
Also,
Is F onto? YES
Proof:
Let y E {0,1}
if y=0,choose n=1 then, n E N and F(n)=y since F(1)=0
If y=1, choose n-3 then, n E N and F(n)=y since F(3)=1.
Can someone please explain how this proof was reached and the logic behind whats going on in detail so i understand it once and for all?
Thank You in advance,
Matt H
