Explore Discrete Math Examples

I'm studying for my upcoming discrete math test and I'm having trouble understanding some equivalences I found in a book on the subject. I guess I'm not really familiar with these rules and I would like someone to walk me through the steps if they don't mind.
I know the elementary laws, De-Morgan's, absorption, distribution, associativity, symmetry, and idempotent laws. But I don't recognize how this person transforms the predicates. Could someone point out the name of the law I need to study?
The transformations are as follows:
$(notPandnotQ)or(notQandnotR)$
$\Leftarrow <!--\; \Leftarrow \; -->\Rightarrow <!--\; \Rightarrow \; -->(notPor(notQornotR))and(notQor(notQornotR))$
$\Leftarrow <!--\; \Leftarrow \; -->\Rightarrow <!--\; \Rightarrow \; -->(notPornotQornotR)and(notQornotQornotR)$
$\Leftarrow <!--\; \Leftarrow \; -->\Rightarrow <!--\; \Rightarrow \; -->(notPornotQornotR)and(notQornotR)$

The combination of a certain combination lock consists of three numbers from 1 through 20 in a sequence with no two consecutive numbers the same. How many different combinations are possible?
This is very difficult and I couldn't figure out the answer. Can anyone help me?
Also I had a question in mind, I am a computer science major and I'm currently taking discrete math. Discrete math consists of all logic questions. This question is based on logic, though I couldn't figure it out. Computer science is all about logic, does a typical "good" computer scientist able to figure this problem out logically without any help? In other words, you need to have a good way of thinking/ logic to become a computer scientist. You should be able to solve complicated problems if you want to become a "good" programmer. Is this an easy question that a computer science major student should be able to answer in top of his head? Idk if that made any sense but your answer would be appreciated.

What does a relation to the power of -1 mean? $R^{-<!--\; -\; -->1}$
Context: If relation R is antisymmetric, then $R^{-1}$ is antisymmetric. Thank you for your help!

Uncountable set of uncountable equivalence classes
I am trying to find $A/\sim <!--\; \sim \; -->$, a set A with an equivalence relation $\sim <!--\; \sim \; -->$ such that the set of equivalence classes is uncountable and the equivalence classes contain an uncountable amount of elements.
I already tried with $A=\mathbb{R}$ and $x\sim <!--\; \sim \; -->y:=x-<!--\; -\; -->y\in <!--\; \in \; -->\mathbb{Z}$. The equivalence classes are the sets of reals with the same fractional part. You can show that $A/\sim <!--\; \sim \; -->=[0,1)$. Which is easy to show is uncountable with an injection $f:\{0,1{\}}^{\mathrm{\infty <!--\; \infty \; -->}}\to <!--\; \to \; -->[0,1)$ (this is the standard method I use to show a set is uncountable).
However, I think all elements in my equivalence classes are countable, as they are all natural numbers plus a specific fractional part. Am I right to think like this, does it make my example false ? In case it is false, is there a way to "fix" it?

Creating equivalent expressions by changing the domains and predicates
I'm having trouble finding a third way to make a logical expression.
Translate this statement into a logical expression in 3 different ways by varying the domain and by using predicates with one and with two variables.
Someone in your school has visited Mexico
I've made 2 translations:
Domain = person in your school
$C(x)=x$ has visited Mexico
$\exists xC(x)$
Domain = people in the world
$S(x)=x$ in your school
$\exists x(S(x)\wedge <!--\; \wedge \; -->C(x))$
Yet I don't know what to change to create a third translation.

Prove that $n^2-<!--\; -\; -->1$ is divisible by 8 for any odd integers n.
Below is my proof and I am confused about a few points. I am not sure the final lines are correct as I know that showing 2,4 are factors of $4k^2+4k$ is enough to prove that it is divisible by 8 and I have looked at some other examples. In an example of 30, both 3 and 6 are factors of 30 but 30 is not divisible by 18. But I am stuck on how to modify this proof to be complete.
To prove this statement, I intend to use direct proof. Since n has to be an odd integer as prescribed in the statement, by the definition of odd numbers, $(2k+1{)}^2-<!--\; -\; -->1$ must be divisible by 8 where $n=2k+1$ for some integer k. Next, we expand $(2k+1{)}^2-<!--\; -\; -->1$ to be $4k^2+4k+1-<!--\; -\; -->1$, which simplifies to $4k^2+4k$.
First, we use the distributive property to get the following $4(k^2+k)$. We let $m=k^2+k$ and therefore $4(k^2+k)$. Hence, we know that $4k^2+4k$ must be divisible by 4.
Then, we use the distributive property to factor 4k from the expression to get $4k(k+1)$. By the definition of even and odd number, if k is odd then $k+1$ must be even and if k is even then $k+1$ is odd. As an integer is Since 2 and 4 are common factors of $4k^2+4k$, which means 8 must also be a factor of $4k^2+4k$.
$n^2-<!--\; -\; -->1$ is therefore divisible by 8 where $n=2k+1$ for some integer k. Therefore, we have proven that $n^2-<!--\; -\; -->1$ is divisible by 8 for any odd integers n.

How to calculate this partial derivative?
given the function:
$v_h={[\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{r}_{h,i}\cdot <!--\; \cdot \; -->w_{h,i}-<!--\; -\; -->\frac{1}{T}\underset{t=1}{\overset{T}{\sum <!--\; \sum \; -->}}\underset{i=1}{\overset{N}{\sum <!--\; \sum \; -->}}{r}_{t,i}{w}_{t,i}]}^2$
I would like to compute the following:
$\frac{\mathrm{\partial <!--\; \partial \; -->}{v}_h}{\mathrm{\partial <!--\; \partial \; -->}{w}_h}=?$
$\frac{\mathrm{\partial <!--\; \partial \; -->}{v}_h}{\mathrm{\partial <!--\; \partial \; -->}{r}_h}=?$
$\frac{\mathrm{\partial <!--\; \partial \; -->}{v}_h}{\mathrm{\partial <!--\; \partial \; -->}{w}_h\mathrm{\partial <!--\; \partial \; -->}{r}_h}=?$
The problem for me is the double sum in the second term. Example of the function with $N=2$ and $T=2$ for $t=1$:
$v_1={[(r_{1,1}\cdot <!--\; \cdot \; -->w_{1,1}+r_{1,2}\cdot <!--\; \cdot \; -->w_{1,2})-<!--\; -\; -->\frac{1}{2}(r_{1,1}\cdot <!--\; \cdot \; -->w_{1,1}+r_{1,2}\cdot <!--\; \cdot \; -->w_{1,2}+r_{2,1}\cdot <!--\; \cdot \; -->w_{2,1}+r_{2,2}\cdot <!--\; \cdot \; -->w_{2,2})]}^2$

Proof that the set is countable - my idea
The set: $S:=\{T\subset <!--\; \subset \; -->\mathbb{N}:T$ is finite or $\mathbb{N}\mathrm{\setminus <!--\; \setminus \; -->}T$ is finite}
Since all finite subsets of $\mathbb{N}$ are countable, can i just prove that since $\mathbb{N}\mathrm{\setminus <!--\; \setminus \; -->}T$ would simply be $\{t;t\in <!--\; \in \; -->\mathbb{N}$ $t\notin <!--\; \notin \; -->T\}$... All though i do have the question how could $\mathbb{N}\mathrm{\setminus <!--\; \setminus \; -->}T$ be a finite set?

Let G be a regular graph. Prove that every bridge of G is in every perfect matching of G.
Struggling with a proof for the following:
Let G be a regular graph. Prove that every bridge of G is in every perfect matching of G.
I have ran into this whilst doing some revision work on graph theory.
Can anyone point me in the right direction. I am considering a proof by contradiction though I am unsure how to continue from there.

Is there an efficient algorithm to this problem?
Let $f_i$ be n strictly decreasing, continuous functions on the positive real numbers with $\underset{x\to <!--\; \to \; -->0}{lim}{f}_i=\mathrm{\infty <!--\; \infty \; -->}$.
Let I be a positive real number.
I think I can prove that there always exists a unique set of n non-negative xi that sum to I and that have: $f_i(x_i)=f_j(x_j)$ for all i, j in 1, ..., n.
I think the greedy algorithm gives a solution to this problem, but is there anything better? In particular, is there a closed solution?

Is this analysis problem or discrete math problem?
Suppose $x_n\in <!--\; \in \; -->\mathbb{R},x_1=1,2x_{n+1}=x_n+3/x_n$. Then show that limxn exists and find its value.
So is this problem (real) analysis problem or a discrete math one?

Discrete math functions
Let $A=P(\{1,2,3,4\})$. Let h be the following function.
$h:\mathbb{N}\to <!--\; \to \; -->A$
defined by $h(x)=\{2,3\}\cap <!--\; \cap \; -->\{x\}.$
Write down h(1)
I'm a bit puzzled by this question. Does this just mean that $x=1$, therefore, there is no intersection or is there an actual answer to this question?

Discrete math evaluate
Evaluate $\underset{k=0}{\overset{n}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{2k}{k})(\genfrac{}{}{0ex}{}{2n-<!--\; -\; -->2k}{n-<!--\; -\; -->k})$

"Some computer science majors take discrete math"
S is the domain of all college students C(x) means "x is a computer science major" D(x) means "x takes discrete math"
Can someone please explain why the following statement is wrong according to the TA?
There exists x in S such that C(x) implies D(x)
I don't understand
EDIT: I can see how it could be: There exists x in S such that C(x) AND D(x), but I don't see why an implication is wrong for SOME x in S

When to use "$\to <!--\; \to \; -->$" vs "$\Rightarrow <!--\; \Rightarrow \; -->$" in discrete math?

All 15 players on the Tall U. basketball team are capable of playing any position.
(a) How many ways can the coach at Tall U. fill the five starting positions in a game?
(b) What is the answer if the center must be one of two players?

$R\circ <!--\; \circ \; -->S$I have only seen this circle operator with function compositions, so is this "Set Composition"? If so, then how does it work?
The question is
"Suppose R and S are relations on a set A. If R and S are reflexiverelations, then $R\circ <!--\; \circ \; -->S$ is reflexive" select true or false.

For each of the following cases, is
$\forall k\in <!--\; \in \; -->\mathbb{N}(P(k)⟹<!--\; ⟹\; -->P(2k))$
true, false or dependent on the value of P(k)?
a) $\forall n\in <!--\; \in \; -->\mathbb{N}P(n),$
b) $P(0)\wedge P(1),$
c) $\forall n\in <!--\; \in \; -->\mathbb{N}P(2n).$

1) How do I prove the following: Let $A=\{6a+4b\in <!--\; \in \; -->Z:a,b\in <!--\; \in \; -->Z\}$ and $B=\{2a\in <!--\; \in \; -->Z:a\in <!--\; \in \; -->Z\}$. Show that $A=B$.

Let n be a positive integer. Describe using quantifiers:
1. $x\in <!--\; \in \; -->\underset{k=1}{\overset{n}{\bigcup <!--\; \bigcup \; -->}}{A}_k$
2. $x\in <!--\; \in \; -->\underset{k=1}{\overset{n}{\bigcap <!--\; \bigcap \; -->}}{A}_k$
My work: $i=\{1,2,3,\dots <!--\; \dots \; -->,n\}$
1. $(\mathrm{\exists <!--\; \exists \; -->}x),(x\in <!--\; \in \; -->A_i)$
2. $(\mathrm{\forall <!--\; \forall \; -->}x),(x\in <!--\; \in \; -->A_i)$
What I need help is explaining with words. Currently I have:
a) There exists i for every $x\in <!--\; \in \; -->A_i$
b) There always is i for every x in $A_i$