Thread: functions question N/A

Recall that N is the set of natural numbers, i.e., N = {0, 1, 2, . . .}.
1. Consider the set W of words over the alphabet {a, b, c}, i.e., the set ofall finite non-empty sequences of letters chosen among a, b and c – forexample, aa, babb and bbc are all elements of W.
Let counta be the function from W to N that counts the number oftimes the letter a occurs, i.e., given w ∈ W, counta(w) is the numberof times a occurs in w.


(a) Calculate counta(aa) and counta(bbc).


(b) Is counta injective? Justify your answer.


(c) Is counta surjective? Justify your answer.


ANY HELP WOULD BE GREATLY APPRECIATED, I'M STRUGGLING.

Originally Posted by Libya

Recall that N is the set of natural numbers, i.e., N = {0, 1, 2, . . .}.
1. Consider the set W of words over the alphabet {a, b, c}, i.e., the set ofall finite non-empty sequences of letters chosen among a, b and c – forexample, aa, babb and bbc are all elements of W.
Let counta be the function from W to N that counts the number oftimes the letter a occurs, i.e., given w ∈ W, counta(w) is the numberof times a occurs in w.


(a) Calculate counta(aa) and counta(bbc).


(b) Is counta injective? Justify your answer.


(c) Is counta surjective? Justify your answer.


ANY HELP WOULD BE GREATLY APPRECIATED, I'M STRUGGLING.

surely you can do (a)

(b) injective means one to one. $counta(aaa)=3$, $counta(baaa)=3$, so clearly $counta()$ is not injective

(c) surjective here would mean that every $n \in \mathbb{N}$ has some word $w \in W$ such that $counta(w)= n$

For any given $n$ the word $w$ consisting of $n$ $a$'s is such that $counta(w)=n$ so $counta()$ is surjective

Originally Posted by romsek

surely you can do (a)

(b) injective means one to one. $counta(aaa)=3$, $counta(baaa)=3$, so clearly $counta()$ is not injective

(c) surjective here would mean that every $n \in \mathbb{N}$ has some word $w \in W$ such that $counta(w)= n$

For any given $n$ the word $w$ consisting of $n$ $a$'s is such that $counta(w)=n$ so $counta()$ is surjective

If that is a correct reading, then as I said this is a childish trivial question.

Originally Posted by Plato

If that is a correct reading, then as I said this is a childish trivial question.

as you said where?

Originally Posted by romsek

as you said where?

In the first reply;

Originally Posted by Plato

I confess that I am also confused. Either this is a totally trivial question or the worst worded.

Is this the way anyone else understands how the function works?
$\text{counta}(bbb)=0 $
$\text{counta}(babcc)= 1$
$\text{counta}(abcab)=2~????? $

If not, please correct me

Originally Posted by Plato

In the first reply;

odd, I'm not seeing that post.

Originally Posted by romsek

odd, I'm not seeing that post.

Thank you, I will tell Ted. We are having real trouble with a spam filter.
Do you agree that it is trivial under that reading?

Originally Posted by Plato

Thank you, I will tell Ted. We are having real trouble with a spam filter.
Do you agree that it is trivial under that reading?

I suppose it's a pretty remedial intro to injection and surjection so not entirely trivial but the function itself is pretty trivial

lmao you guys are saying this question is trivial this question represents my life, i suck at discrete maths. how are you guys so smart?

Originally Posted by Libya

lmao you guys are saying this question is trivial this question represents my life, i suck at discrete maths. how are you guys so smart?

being able to count the number of a's in a word does not make one so smart.

Originally Posted by Libya

lmao you guys are saying this question is trivial this question represents my life, i suck at discrete maths. how are you guys so smart?

Do you understand that this problem is asking you to count?

My experience is that people who are "not good at maths" simply do not grasp the definitions- often trying to read more into the definitions than they should.

"Calculate counta(aa)". Can you count to 2?