# MATH1081 Discrete Maths (1 Viewer)

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

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

#### InteGrand

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

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

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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:

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

#### InteGrand

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

##### Kosovo is Serbian
Re: Discrete Maths Sem 2 2016

ahahahaha goodluck with that subject

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

Can I please have my proof checked?

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

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

\bg_white \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*}

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

##### -insert title here-
Re: Discrete Maths Sem 2 2016

Can I please have my proof checked?

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

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

\bg_white \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*}

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

Last edited:

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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.

#### turntaker

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

Should be ez

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

\bg_white \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

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

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

Last edited:

##### Cult of Personality
Re: Discrete Maths Sem 2 2016

Can I please have my proof checked?

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

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

\bg_white \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*}

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

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

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

Last edited:

#### InteGrand

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

\bg_white \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

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

But ends up with a final answer of just $\bg_white B \cap C^C$ 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/Absorption_Laws_(Set_Theory)/Union_with_Intersection.)

Last edited:

#### InteGrand

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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

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

\bg_white \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

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

But ends up with a final answer of just $\bg_white B \cap C^C$ 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.

#### seanieg89

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

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.

##### Cult of Personality
Re: Discrete Maths Sem 2 2016

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"

#### leehuan

##### Well-Known Member
Re: Discrete Maths Sem 2 2016

This bugger.

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

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