Online Course Discussion Forum

Math Challenge IC Handout 7

 
 
PengChristina的头像
Math Challenge IC Handout 7
PengChristina - 2018年04月23日 Monday 00:34
 

First, for questions 7,8, and 9 it seems to be the same question, just repeated. I'm guessing it's not intentional? I just repeated my answer.


Just to clarify in 7.21 part b, A^c means everything in omega other than A. So does that mean if the problem only mentioned A and B, A^c is just B without any A inter B?


For questions 7.28, 7.29, and 7.30, can you give me a few suggestions as to how to do it? For 30, I was doing a guess and check method with a 3 set venn diagram and it wasn't very effective.


Thankssss!!!!!



 
ReynosoDavid的头像
Re: Math Challenge IC Handout 7
ReynosoDavid - 2018年04月23日 Monday 13:02
 
Questions 7 and 8 have been corrected now, it was indeed not intentional. Thanks for letting us know! (you won't need to resubmit your homework in case you did already, but please take a look at the updated handout and give the question a try before the live homework discussion session!)


In 21 $\Omega$ is the set of whole numbers from $1$ to $100$ (inclusive).

For 28 try to think about what are the steps you would follow to come up with a function from $B$ to $A$ and use that to count them, then look at which of those would be and would not be surjections. 

In 29 you may want to do something similar as the example 9 from the video lecture. Recall $A\cap B\cap C = \varnothing$ means there is no element of $S$ that is in all three sets at the same time, but it is OK if some pairs have elements in common. Complementary counting will be quite useful as well. 

For 30 the PIE for $3$ sets will definitely help you out. A Venn diagram could help you set it up. 

PengChristina的头像
Re: Math Challenge IC Handout 7
PengChristina - 2018年04月24日 Tuesday 23:49
 

Thanks! Just curious, when doing a question like A {2,3,4} and B {1,2} and you need to come up with a function, why do we multiply out the choices with 2^3 instead of adding 2+2+2?

ReynosoDavid的头像
Re: Math Challenge IC Handout 7
ReynosoDavid - 2018年04月25日 Wednesday 10:31
 

If you want a function $A\rightarrow B$, you need to choose, for each of the elements of $A$, what element of $B$ to assign. For example $$2\mapsto 1\quad 3\mapsto 1 \quad 4\mapsto 2$$ or $$2\mapsto 2\quad 3\mapsto 1 \quad 4\mapsto 1.$$ So, in this particular case, for each of the $3$ elements of $A$ you have $2$ elements of $B$ to choose from. By the multiplication principle, in total there are $2\times 2 \times 2 = 2^3$ possible different functions from $A$ to $B$.

PengChristina的头像
Re: Math Challenge IC Handout 7
PengChristina - 2018年04月25日 Wednesday 15:50
 

Yes, everything makes sense except for why its multiplication. Writing all the functions out, I could at most get six when each element of A has two choices. I don't know why the answer is 2^3 instead of 2+2+2