MATH1081 Discrete Maths (2 Viewers)

leehuan

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Re: Discrete Maths Sem 2 2016

if i remember its easier to prove by Venn Diagrams
Sounds right. Jim Franklin seems to prefer to drill us into using slightly more formalised proofs though.

I just wasn't sure how to manipulate my x in. And yet again I forgot that you had to prove they are a mutual subset of each other to be equal sets.
 

RenegadeMx

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Re: Discrete Maths Sem 2 2016

Sounds right. Jim Franklin seems to prefer to drill us into using slightly more formalised proofs though.

I just wasn't sure how to manipulate my x in. And yet again I forgot that you had to prove they are a mutual subset of each other to be equal sets.
havent heard of him before, but had a look at your course outline you lucky fucks will have peter brown
 

leehuan

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Re: Discrete Maths Sem 2 2016



I can see why it is but how would you prove it
 

leehuan

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Re: Discrete Maths Sem 2 2016



I feel like my proof is inconclusive and achieves nothing because I don't effectively use both criteria.





 
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InteGrand

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Re: Discrete Maths Sem 2 2016

To show A is a subset of B, you should start off by letting x be an element of A (you did A∩C), and then show it is an element of B. (You may want to consider cases along the way to help use the given info.)

I think what you essentially did is show A∩C is a subset of B.
 
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seanieg89

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Re: Discrete Maths Sem 2 2016

Another way of doing it that doesn't involve taking elements is:



Where the step follows from applying the question assumptions to replace with the supersets respectively.
 

leehuan

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Re: Discrete Maths Sem 2 2016

Another way of doing it that doesn't involve taking elements is:



Where the step follows from applying the question assumptions to replace with the supersets respectively.
Nice that makes sense. I'm just used to element wise because the lecturers said so lol.
___________________________

The answers don't even tell me if it's true or false so I assumed it was true (because I could not find any counterexample) but then I got stuck proving the latter part.



Basically I got stuck proving the RHS is a subset of the LHS
 
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InteGrand

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Re: Discrete Maths Sem 2 2016

Nice that makes sense. I'm just used to element wise because the lecturers said so lol.
___________________________

The answers don't even tell me if it's true or false so I assumed it was true (because I could not find any counterexample) but then I got stuck proving the latter part.



Basically I got stuck proving the RHS is a subset of the LHS
A way to show RHS is a subset of LHS.

Note the RHS is a set of ordered pairs. Let (x,y) be an element of RHS.

Then (x,y) is in A x B or in A x C.

If it's in A x B, then x is in A and y is in B ==> x is in A and y is in B union C. (As B is a subset of B union C.)

The other case you can similarly show that x is in A and y is in B union C.

So in either case, (x,y) is in A x (B union C).

So RHS is a subset of LHS.
 

leehuan

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Re: Discrete Maths Sem 2 2016



It means the symmetric difference in this context.

 

seanieg89

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Re: Discrete Maths Sem 2 2016

Have you learned that the symmetric difference is associative? If not, then prove this as an exercise as it is easy and has this claim is a trivial consequence.

Indeed, We have A*C=B*C => (A*C)*C=(B*C)*C => A*(C*C)=B*(C*C) => A*e=B*e => A=B where e denotes the empty set and * denotes the symmetric difference operation.

(In fact we are proving a property of abstract groups here. An algebra of sets endowed with the symmetric difference is an abelian group.)
 

leehuan

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Re: Discrete Maths Sem 2 2016

I will work on it later. It doesn't look hard if I give it a bit of thought.

We weren't actually taught what symmetric difference is really. It's just a homework question regarding it.
 

seanieg89

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Re: Discrete Maths Sem 2 2016

Yep, it really isn't.

As it happens (A*B)*C=A*(B*C)=The set of things that are elements of exactly one of the three sets.
 

leehuan

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Re: Discrete Maths Sem 2 2016

I think I'm forgetting something.

I understand how a function f is surjective iff the codomain of f is the range. However what's the problem if the codomain was a proper subset of f?

Also, are these supposed to be two notations for the same thing
 
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seanieg89

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Re: Discrete Maths Sem 2 2016

"Codomain was a proper subset of f"? What do you mean by this?

Yes to your second question. Read that kind of arrow as "maps to".
 

leehuan

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Re: Discrete Maths Sem 2 2016

Alright cool.

My bad on the first qn, meant a subset of the range. So something like



Edit: Oh crap, is it because when x=-1 (which is in R), there is no output value?

With that maps-to notation though I have another question. How would you write something like f:R->R, f(x)=x?
 

seanieg89

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Re: Discrete Maths Sem 2 2016

That doesn't make any sense, when you talk about a function, you have to specify a domain and a codomain for that function. Loosely speaking a function is a pair of sets X,Y and a rule that assigns to every element of X an element of Y.

The range is the set of elements of Y that are actually hit, ie the range is always a subset of the codomain.
 

leehuan

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Re: Discrete Maths Sem 2 2016

Wait what? Now I'm confused.

I resolved my question mainly based off something like this:


I was wondering why something like this 'function' is not surjective:


But then I realised that it's not a function because when x<0, there are no output values.

(Edit: Or maybe I do actually. I think I might've abused the term "range")
 

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