# Thread: MATH1081 Discrete Maths

1. ## MATH1081 Discrete Maths

Don't mind me...just setting up some threads for more of my stupidity this upcoming semester.

UNSW course outline: https://www.maths.unsw.edu.au/sites/...81-s2_2016.pdf

2. ## Re: Discrete Maths Sem 2 2016

$Let A=\left\{a_1,a_2, \dots, a_n \right\}\\ and let B=\left\{A, b_1,b_2,\dots, b_n \right\}$

$Are both of the following true or is the former the only one true?\\ A \in B\\ A \subseteq B$

$Also, would you have A \notin P(B) but \{A\} \in P(B)?$

3. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
$Let A=\left\{a_1,a_2, \dots, a_n \right\}\\ and let B=\left\{A, b_1,b_2,\dots, b_n \right\}$

$Are both of the following true or is the former the only one true?\\ A \in B\\ A \subseteq B$

$Also, would you have A \notin P(B) but \{A\} \in P(B)?$
A is an element of B, but not a subset (assuming the b's aren't coincidentally just equal to the a's).

Since A isn't a subset of B, it's not in the power set of B. But {A} is in P(B), since {A} is a subset of B, since A is an element of B.

4. ## Re: Discrete Maths Sem 2 2016

Originally Posted by InteGrand
A is an element of B, but not a subset (assuming the b's aren't coincidentally just equal to the a's).

Since A isn't a subset of B, it's not in the power set of B. But {A} is in P(B), since {A} is a subset of B, since A is an element of B.
Alright this last bit was what I needed to see. I need to have my foundations but just making sure:

$A \in B\iff \{A\} \subseteq B$

5. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
Alright this last bit was what I needed to see. I need to have my foundations but just making sure:

$A \in B\iff \{A\} \subseteq B$
That's correct (by definition of subset essentially) .

6. ## Re: Discrete Maths Sem 2 2016

ahahahaha goodluck with that subject

7. ## Re: Discrete Maths Sem 2 2016

Can I please have my proof checked?

$\\Let A = \{x \in \mathbb R | \cos x = 1 \}\\ and B = \{ x \in \mathbb R | \sin x = 0\}\\ RTP: A \subset B$

The video solution was clever in how it used a Pythagorean identity here to match up A and B, however I did it by solving. Just want to check on its validity

\begin{align*}A&=\{x \in \mathbb R | \cos x = 1 \}\\ &= \{x \in \mathbb R |k \in \mathbb Z |x = 2k \pi \}\\ & \subseteq \{x \in \mathbb R |k \in \mathbb Z |x = k \pi \}\\ &=\{x \in \mathbb R | \sin x = 0 \}\\ &= B \end{align*}

$So A \subseteq B however \pi \in B yet \pi \notin A. \therefore A \subset B$

8. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
Can I please have my proof checked?

$\\Let A = \{x \in \mathbb R | \cos x = 1 \}\\ and B = \{ x \in \mathbb R | \sin x = 0\}\\ RTP: A \subset B$

The video solution was clever in how it used a Pythagorean identity here to match up A and B, however I did it by solving. Just want to check on its validity

\begin{align*}A&=\{x \in \mathbb R | \cos x = 1 \}\\ &= \{x \in \mathbb R |k \in \mathbb Z |x = 2k \pi \}\\ & \subseteq \{x \in \mathbb R |k \in \mathbb Z |x = k \pi \}\\ &=\{x \in \mathbb R | \sin x = 0 \}\\ &= B \end{align*}

$So A \subseteq B however \pi \in B yet \pi \notin A. \therefore A \subset B$
A is a proper subset of B, since there are elements in B that are not in A. You are claiming that A is the set B itself.

Recall that sinx = 0 does not imply cosx = 1

The solutions to sinx = 0 can be divided into the solutions to cosx = 1 and cosx = -1

Edit: sorry did not read the final line

Yes it looks good.

9. ## Re: Discrete Maths Sem 2 2016

Originally Posted by Paradoxica
A is a proper subset of B, since there are elements in B that are not in A. You are claiming that A is the set B itself.

Recall that sinx = 0 does not imply cosx = 1

The solutions to sinx = 0 can be divided into the solutions to cosx = 1 and cosx = -1

Edit: sorry did not read the final line

Yes it looks good.
Lol. Yeah the question first asked to prove it was just a subset before claiming it was a proper subset. So I put x=π on the end to contradict they're the same set.

Should be ez

11. ## Re: Discrete Maths Sem 2 2016

$Simplify \left(\left[(A \cup B)^C \cup C\right]\cup B^C \right)^C$

They don't have an answer so I am suspecting that my answer is wrong lol. I started from the outside in my working.

\begin{align*}\left(\left[(A \cup B)^C \cup C\right]\cup B^C \right)^C &= \left[(A \cup B)^C \cup C\right]^C\cap B \\ &= \left[(A \cup B) \cap C^C\right]\cap B \\ &= \left[ (A \cap C^C) \cup (B \cap C^C) \right] \cap B\\ &= (A \cap C^C \cap B)\cup (B \cap C^C \cap B)\\ &= (A \cap (C^C \cap B))\cup((C^C \cap B))\\ &= A \cap B \cap C^C\end{align*}

Edit: After line 2 one of my friends used associativity like this

$= [(A \cup B) \cap B] \cap C^C$

But ends up with a final answer of just $B \cap C^C$ instead. Is this justified?

12. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
Can I please have my proof checked?

$\\Let A = \{x \in \mathbb R | \cos x = 1 \}\\ and B = \{ x \in \mathbb R | \sin x = 0\}\\ RTP: A \subset B$

The video solution was clever in how it used a Pythagorean identity here to match up A and B, however I did it by solving. Just want to check on its validity

\begin{align*}A&=\{x \in \mathbb R | \cos x = 1 \}\\ &= \{x \in \mathbb R |k \in \mathbb Z |x = 2k \pi \}\\ & \subseteq \{x \in \mathbb R |k \in \mathbb Z |x = k \pi \}\\ &=\{x \in \mathbb R | \sin x = 0 \}\\ &= B \end{align*}

$So A \subseteq B however \pi \in B yet \pi \notin A. \therefore A \subset B$
just fyi on the second line of the A it reads "Let A be the set of x in the real numbers such that k is in the integers such that x = 2k*pi" which makes zero sense

try

$\{ x \in \mathbb{R} \mid x=2k\pi, k \in \mathbb{Z} \}$

13. ## Re: Discrete Maths Sem 2 2016

Originally Posted by Shadowdude
just fyi on the second line of the A it reads "Let A be the set of x in the real numbers such that k is in the integers such that x = 2k*pi" which makes zero sense

try

$\{ x \in \mathbb{R} \mid x=2k\pi, k \in \mathbb{Z} \}$
That was a part of why I put that question up. What's the difference between using | and ,

Edit, ok my prediction is | means such that whereas , means where. In that case, if the second | was replaced with , would that still be nonsensical?

14. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
$Simplify \left(\left[(A \cup B)^C \cup C\right]\cup B^C \right)^C$

They don't have an answer so I am suspecting that my answer is wrong lol. I started from the outside in my working.

\begin{align*}\left(\left[(A \cup B)^C \cup C\right]\cup B^C \right)^C &= \left[(A \cup B)^C \cup C\right]^C\cap B \\ &= \left[(A \cup B) \cap C^C\right]\cap B \\ &= \left[ (A \cap C^C) \cup (B \cap C^C) \right] \cap B\\ &= (A \cap C^C \cap B)\cup (B \cap C^C \cap B)\\ &= (A \cap (C^C \cap B))\cup((C^C \cap B))\\ &= A \cap B \cap C^C\end{align*}

Edit: After line 2 one of my friends used associativity like this

$= [(A \cup B) \cap B] \cap C^C$

But ends up with a final answer of just $B \cap C^C$ instead. Is this justified?
To get the final answer your friend got (which looks correct), use an absorption law at the second last line of your proof. (Unfortunately your simplification in your last line isn't valid. But if we just apply the absorption law there we'll get the answer. )

(See: https://proofwiki.org/wiki/Absorptio...h_Intersection.)

15. ## Re: Discrete Maths Sem 2 2016

Originally Posted by InteGrand
To get the final answer your friend got (which looks correct), use an absorption law at the second last line of your proof. (Unfortunately your simplification in your last line isn't valid. But if we just apply the absorption law there we'll get the answer. )

(See: https://proofwiki.org/wiki/Absorptio...h_Intersection.)
Ahh I absorbed wrong

16. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
$Simplify \left(\left[(A \cup B)^C \cup C\right]\cup B^C \right)^C$

They don't have an answer so I am suspecting that my answer is wrong lol. I started from the outside in my working.

\begin{align*}\left(\left[(A \cup B)^C \cup C\right]\cup B^C \right)^C &= \left[(A \cup B)^C \cup C\right]^C\cap B \\ &= \left[(A \cup B) \cap C^C\right]\cap B \\ &= \left[ (A \cap C^C) \cup (B \cap C^C) \right] \cap B\\ &= (A \cap C^C \cap B)\cup (B \cap C^C \cap B)\\ &= (A \cap (C^C \cap B))\cup((C^C \cap B))\\ &= A \cap B \cap C^C\end{align*}

Edit: After line 2 one of my friends used associativity like this

$= [(A \cup B) \cap B] \cap C^C$

But ends up with a final answer of just $B \cap C^C$ instead. Is this justified?
Yes what your friend did is valid. Since intersection is both associative and commutative, we can do intersections in any order (like, (X cap Y) cap Z = X cap (Y cap Z) = X cap (Z cap Y) = (X cap Z) cap Y, using associativity and commutativity. I used 'cap' to mean intersection symbol.). Then using absorption law finishes it.

17. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
That was a part of why I put that question up. What's the difference between using | and ,

Edit, ok my prediction is | means such that whereas , means where. In that case, if the second | was replaced with , would that still be nonsensical?
Don't view "," and "|" in the same way (btw ":" is a common alternative for "|" that is my personal preference). The former is basically informal formatting here, whilst the latter is part of the formal set builder notation syntax.

{x in A: Mathematical statement P(x) about x}

is the general way of denoting the collection of x in A such that P(x) is true. Comma is just formatting of that mathematical statement in this case, to be translated as "for some". Although this certainly isn't unambiguous notation, its intended meaning should be pretty obvious from context. We are often slightly lazy in writing mathematical statements because writing things formally with quantifiers in each line would be needlessly tedious in long proofs.

18. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
That was a part of why I put that question up. What's the difference between using | and ,

Edit, ok my prediction is | means such that whereas , means where. In that case, if the second | was replaced with , would that still be nonsensical?
no, but then you're saying the set consists of x and k, when you really want to just have the x's that satisfy the condition

the thing i wrote reads "x in the real numbers such that x = 2k*pi where k is any integer"

19. ## Re: Discrete Maths Sem 2 2016

This bugger.

$\\Prove the uniqueness of complement: If A \cup B = U and A \cap B = \emptyset, then B = A^C$

20. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
This bugger.

$\\Prove the uniqueness of complement: If A \cup B = U and A \cap B = \emptyset, then B = A^C$
if i remember its easier to prove by Venn Diagrams

21. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
This bugger.

$\\Prove the uniqueness of complement: If A \cup B = U and A \cap B = \emptyset, then B = A^C$
$\noindent Well A^c is the set of all things in U (the universal set) that are not in A. Let A \cup B = U and A \cap B = \emptyset. Suppose x\in B \subseteq U. Then since A\cap B = \emptyset (i.e. A and B are disjoint), x is not in A, or in other words, x is in A^{c}. So B\subseteq A^{c}.$

$\noindent Conversely, suppose x\in A^{c}\subseteq U. Then x\notin A. Since x\in U, we have x\in A\cup B (as U = A\cup B). So x\in A or x\in B. But we already said x\notin A, so x must be in B. Therefore, A^{c}\subseteq B.$

$\noindent It follows that B=A^{c}.$

22. ## Re: Discrete Maths Sem 2 2016

Originally Posted by RenegadeMx
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.

23. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
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

24. ## Re: Discrete Maths Sem 2 2016

$\text{Is it true that if }P(A)=P(B)\text{ for two sets }A,B\text{ then }A=B?$

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

25. ## Re: Discrete Maths Sem 2 2016

Originally Posted by leehuan
$\text{Is it true that if }P(A)=P(B)\text{ for two sets }A,B\text{ then }A=B?$

I can see why it is but how would you prove it
$\noindent Note for sets S and T we have T \in P\left(S\right) iff T \subseteq S. Also for any set S, S is in its power set.$

$\noindent Now, suppose A and B are sets with P\left(A\right) = P(B). Since A \in P(A), we have A \in P(B) (as P(A)=P(B)), so A\subseteq B. Reversing the roles of A and B shows B\subseteq A. So A=B.$

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