Explore Discrete Math Examples

Finding the first forward difference of $f(x)=x-<!--\; -\; -->\frac{q(1-<!--\; -\; -->q^x)}{p}+q^{x}c$
So I am trying to unravel the derivation of some literature equations and one equation I'm struggling at, is to find the first forward difference of the following equation:
$f(x)=x-<!--\; -\; -->\frac{q(1-<!--\; -\; -->q^x)}{p}+q^{x}c$
So, actually, how to find a simplification of
$f(x+1)-<!--\; -\; -->f(x)=((x+1)-<!--\; -\; -->\frac{q(1-<!--\; -\; -->q^{x+1})}{p}+q^{x+1}c)-<!--\; -\; -->(x-<!--\; -\; -->\frac{q(1-<!--\; -\; -->q^x)}{p}+q^{x}c))$
Now I think that the difference between first and third term of the left and right part is equal to
$1-<!--\; -\; -->c{q}^x$
But I don't seem to get that second part involved, especially if I look at the desired result, which is:
$f(x+1)-<!--\; -\; -->f(x)=1-<!--\; -\; -->q^x(pc+q)$

Is $K_6$ a minor of $K_{2,2,2,2,2}$.
I am trying to find out whether the $K_6$ graph is a minor of $K_{2,2,2,2,2}$, but I am struggling. The $K_{2,2,2,2,2}$ graph is pretty complicated to draw. Logically, I can see that $K_5$ is a minor of $K_{2,2,2,2,2}$ by contracting each 2-partite set into one vertex. However, this does not yet give me information on whether $K_6$ is a minor of $K_{2,2,2,2,2}$.

matrix representation of relations over $A=\{1,2,3,\dots <!--\; \dots \; -->,1000\}$
I have an exercise in discrete math about matrix representation of relations that I want to be sure I have solved properly:
How many non-zero entries does the matrix representing the following relations on $A=\{1,2,3,\dots <!--\; \dots \; -->,1000\}$ consisting of the first 1000 positive integers have if R is:
a) $R=\{(a,b)\mid <!--\; \mid \; -->a+b=1000\}$
b) $R=\{(a,b)\mid <!--\; \mid \; -->a+b\le <!--\; \le \; -->1001\}$
c) $R=\{(a,b)\mid <!--\; \mid \; -->a\ne <!--\; \ne \; -->0\}$
In the c case, one million since every entry will be 1.
In the a case, every entry, from the first one to the 999-th one, excluding the 1000th row, will have one non-zero entry, so 999
In the b case the whole first row satisfies the condition, the whole second row satisfies the condition and in the third row everything but the entry (3,999) satisfies the condition. In the fourth, everything but (4,999) and (4,998) satisfies the condition.
So we will have $1000+1000+999+998+\mathrm{997...}+1$ non-zero entries in the matrix representing the relation. $999+998+\mathrm{997...}+1=99(100)/2$
So we have $499500+2000=501500$ non zero-entries.
Is this the right way to solve the exercise?

Total number of subsets in a set
I was reading about subsets, in that, the article suggests the total number of subsets in a set is $2^n$, where n is the number of elements in the set. For example -$\{1,2,3,4,5\}$ the total number of subsets is 32 because n is 5 and $2^5$ is 32 by multiplicative principle.
But the multiplicative principle is that if m events can happen in n ways then the possible outcomes are $m\times <!--\; \times \; -->n$. So in the subsets problem if every element has 2 possibilities of it being in set or not being in set why is it not $2\times <!--\; \times \; -->5$ and $2^5$? I know that the $2^5$ is correct but not able to visualize it.

Suppose we have a system of linear equations:
$x^2-<!--\; -\; -->x+y^2-<!--\; -\; -->y=n$
$x^2+y^2-<!--\; -\; -->x-<!--\; -\; -->y+2xy=2n$ where $n\in <!--\; \in \; -->{\mathbb{Z}}^{\mathbb{+}}$

How to use structural induction on lists?
I don't really understand structural induction and how to use it. The question is use structural induction on lists to prove that $\mathrm{rev}<!--\; \; -->(\mathrm{rev}<!--\; \; -->(L))=L$. You may use the lemma that $\mathrm{rev}<!--\; \; -->(\mathrm{app}<!--\; \; -->(L,M))=\mathrm{app}<!--\; \; -->(\mathrm{rev}<!--\; \; -->(M),\mathrm{rev}<!--\; \; -->(L)).$.
So far I believe the base cases are:
$L=[]$
But after that I am little confused as to how to go through the proof. Any help would be appreciated!

Proving transitivity on relation: $aRb=7\mid <!--\; \mid \; -->|a-<!--\; -\; -->b|$
Proving transitivity on relation: $aRb=7|(|a-<!--\; -\; -->b|)$, so $aRb\wedge <!--\; \wedge \; -->bRc⟹<!--\; ⟹\; -->aRc$
What I tried:
$7k=|a-<!--\; -\; -->b|$
$7l=|b-<!--\; -\; -->c|$
$l,k\in <!--\; \in \; -->\mathbb{Z}$
Now I squared the two equations and subtracted the top one from the bottom one:
$49(l^2-<!--\; -\; -->k^2)=(b-<!--\; -\; -->c{)}^2-<!--\; -\; -->(a-<!--\; -\; -->b{)}^2=c^2-<!--\; -\; -->2b(a-<!--\; -\; -->c)-<!--\; -\; -->a^2\ne <!--\; \ne \; -->|a-<!--\; -\; -->c{|}^2$
I see that this approach does not work, since I can't get the square root of the distance between a and c, so that $\sqrt{49(l^2-<!--\; -\; -->k^2)}$ would be a rational number for all l and k (for instance we could have $l=3$ and $k=2,$, and we get $7\cdot <!--\; \cdot \; -->\sqrt{5}\notin <!--\; \notin \; -->\mathbb{Q}$, so my equation above does not imply that transitivity does not exist.
My question is what would be the best way to prove if it does or doesn't exist?

Approach to counting the number of sub-graphs of a given graph $G=(V,E)$
I have just currently started reading about Graph theory in computer science and while reading the concept of graphs I came across the concept of sub-graphs. As a sub-graph is essentially similar to a subset in set theory I became interested in the question
"How many sub-graphs exist for a given graph $G=(V,E)$"
While trying to count the number of sub-graphs I got a little confused as to what should be a generic approach in order to try figure out the number of sub-graphs quickly as it is highly dependent on the edge set the given graph possesses and this question confirmed my intuition : How many subgraphs of this simple graph?. But even this question doesn't answer the generic question as to what should be the general approach to figure out the number of sub-graphs for a given graph quickly. It will be great if someone could throw some light on the approach and explain it like they were explaining it to a 5 year old.

Set theory with empty sets confuses me. Why is $n-<!--\; -\; -->m=n-<!--\; -\; -->(n-<!--\; -\; -->5)=5$ false and why is $Ø\subset Ø,Ø$ true?
I get that $\mathrm{\varnothing <!--\; \varnothing \; -->}$ means the same as $\{\}$, it's still confusing to me. The above statements and their validity come from my University mathematics handbook on Discrete Mathematics.
The first statements makes it seem like $\{\mathrm{\varnothing <!--\; \varnothing \; -->}\}$ and $\{\{\mathrm{\varnothing <!--\; \varnothing \; -->}\}\}$ are the same thing. Then the second statement immediately dispels it by saying that $\{\{\mathrm{\varnothing <!--\; \varnothing \; -->}\}\}$ isn't the same thing as $\{\mathrm{\varnothing <!--\; \varnothing \; -->}\}$.
I want to mention that they make a distinction between $\subset <!--\; \subset \; -->$ and $\subseteq <!--\; \subseteq \; -->$.
- $A\subset A$ would be false
- $A\subset A,B$ would be correct
- $A\subseteq A$ would be true
- $A\subset A$ is correct too I think because A is a singleton and is distinct from just A.

The difference between "if" and "and" in symbolic logic
Let M(x,y) be “x has sent y an e-mail message” and T(x,y) be “x has telephoned y,” where the domain consists of all students in your class. Assume that all e-mail messages that were sent are received. Use quantifiers to express each of these statements.
1. There is a student in your class who has not received an e-mail message from anyone else in the class and who has not been called by any other student in the class.
2. Every student in the class has either received an email message or received a telephone call from another student in the class.
The answer in book for 1 is:
$\exists x\forall y(x\ne y\to (\neg M(y,x)\wedge \neg T(y,x)))$
and the answer for 2 is:
$\forall x(\exists y(x\ne y\wedge (M(y,x)\vee T(y,x)))).$
My problem is with parts $x\ne y\wedge$
and $x\ne y\to .$.
When should I use "→" or "∧" after $x\ne y?$ I don't understand the difference in their literature in the question. When I want to use "if" in any other question, the answer is "and" and vice versa.

How Would I Calculate this by Hand?
I know how to do summations and I know the equivalent equations for them. I am just lost on the second summation and how to proceed there. I have the wolfram alpha answer, I just need to know how do this by hand.
$\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}\underset{j=i}{\overset{n+4}{\sum <!--\; \sum \; -->}}\underset{k=3}{\overset{j+7}{\sum <!--\; \sum \; -->}}1$
I have the answer, but I want to know how to do it, or something similar in the future.

Help with understanding this proof that all graphs contain an even number of odd vertices
Theorem: Every graph contains an even number of odd vertices.
Proof: Let G be a graph of size m. Since the sum of the degrees of the vertices of G is even, namely 2m, and the sum of the degrees of the even vertices of G is even, it follows that the sum of the degrees of the odd vertices of G is even. Therefore, G has an even number of odd vertices.
Question: Why does knowing the sum of the degrees of even vertices and odd vertices of G follow that G has an even number of odd vertices?

The range of a non-computable function that grows faster than computable functions is undecidable
Let f be a non-computable function that grows faster than every computable function. I have to prove that $R=\text{Range}(f)$ is non-decidable.
I'm trying to prove the statement by contradiction. Then the indicator function $1_R$ of R is computable. Now, to get a contradiction, I want to find a computable function g using $1_R$, s.t. f doesn't grow faster than g. Can you please give me some tips on the construction of g?
P.S. A precise definition of f is the following:
1. f is a non-computable totally defined $\mathbb{N}\to <!--\; \to \; -->\mathbb{N}$ function
2. Given a computable function h, there is $n_h\in <!--\; \in \; -->\mathbb{N}$, s.t. for all $m\ge <!--\; \ge \; -->n_h$ we have $f(m)>h(m)$ (if h(m) is defined)

How do I prove by using induction on k, that MergeSort uses $n({\log }_2<!--\; \; -->(n)+1)=2^k(k+1)$ comparisons?

How can I prove that a function is surjective?
I understood by scrolling through the old posts, that if i have a function like this:
$f:\mathbb{R}-<!--\; -\; -->\{2\}\to <!--\; \to \; -->\mathbb{R}-<!--\; -\; -->\{5\}\mid <!--\; \mid \; -->f(x)=\frac{5x+1}{x-<!--\; -\; -->2}$
If it is given to me something like this: $f:\mathbb{N}\times <!--\; \times \; -->\mathbb{N}\to <!--\; \to \; -->\mathbb{N}\mid <!--\; \mid \; -->f((n,m))=2^{n-<!--\; -\; -->1}(2m-<!--\; -\; -->1)$, how can i prove that is surjective? The fact that i have 2 variable is confusing me. Thanks, I hope the question is well asked.

Having trouble proving if two sets are equal or not
I am trying to figure out how to prove if two sets are equivalent or not by using the laws and rules of set theory but I am having a hard time with questions involving algebraic expressions.
$D=\{3r+1|r\in <!--\; \in \; -->\mathbb{Z}\}$
$E=\{3s+2|s\in <!--\; \in \; -->\mathbb{Z}\}$
Determine if this statement is true or false: $D=E$

Are the following sets of coordinates affinely equivalent?
Given the coordinates $\{(0,t^k),(t^k,0)\}$ for $k\in <!--\; \in \; -->\{1,...,n\}$ and for $t>1$, and $\{(\mathrm{\alpha <!--\; \alpha \; -->}{s}^{k-<!--\; -\; -->1},0),(0,\mathrm{\alpha <!--\; \alpha \; -->}{s}^{k-<!--\; -\; -->1})\}$ for $k\in <!--\; \in \; -->\{1,...,n\}$ and for $s>1$ and $\mathrm{\alpha <!--\; \alpha \; -->}>0$, are the two sets of coordinates affinely equivalent to one another? I know that affine equivalence must preserve the parallelity / collinearity of lines, which both sets of points seem to do with respect to the other set. Is there anything additional that I should also take into consideration?

$(p\wedge q)\leftrightarrow (p\vee q)=p\leftrightarrow q$ prove using logical statement
I am stuck on this question right now and it'd be great if someone can help me out. I managed to start it like this and it just stops there...
$(p\wedge <!--\; \wedge \; -->q)\leftrightarrow <!--\; \leftrightarrow \; -->(p\vee <!--\; \vee \; -->q)⟺<!--\; ⟺\; -->(p\wedge <!--\; \wedge \; -->q)\to <!--\; \to \; -->(p\vee <!--\; \vee \; -->q)\wedge <!--\; \wedge \; -->(p\vee <!--\; \vee \; -->q)\to <!--\; \to \; -->(p\wedge <!--\; \wedge \; -->q)$

Ok to mod exponents when solving modular expressions with integer exponents?
We have ${12}^{{12}^{12}}(\mathrm{mod}13)$. I tried solving it this way.
${12}^{{12}^{12}}\equiv <!--\; \equiv \; -->-<!--\; -\; -->1^{-<!--\; -\; -->1^{-<!--\; -\; -->1}}(\mathrm{mod}13)=-<!--\; -\; -->1^{\frac{1}{-<!--\; -\; -->1}}=\frac{1}{-<!--\; -\; -->1}=-<!--\; -\; -->1$. But the correct answer is 1, why?