Explore Discrete Math Examples

Without loss of generality, what may we validly assume?
So I came across the question where it was asked: Given that a,b,c are positive integers, in a proof to the theorem $a^3+b^3+c^3\ge <!--\; \ge \; -->a^{2}b+b^{2}c+c^{2}a$, we may assume without loss of generality that:
1. $a\ge <!--\; \ge \; -->c,b\ge <!--\; \ge \; -->c$
2. $a\ge <!--\; \ge \; -->b,a\ge <!--\; \ge \; -->c$
3. $a\ge <!--\; \ge \; -->b,b\ge <!--\; \ge \; -->c$
4. $a\ge <!--\; \ge \; -->c,c\ge <!--\; \ge \; -->b$
I deduced that the correct answer to the question would be (3) since those are the possible cases and if that's not true, we could reorder a,b,c around. Is my idea of it correct?

What is the semantic meaning of in?
I have this assignment homework in an intro to discrete mathematics class, and I'm talking a class mate about a statement I wrote, and how he interprets such a statement.
$$11x+3=10\in <!--\; \in \; -->{\mathbb{Z}}_{18}$$
My friend thinks that this means that the literal untouched equation of the left of $\in <!--\; \in \; -->$ is inside the set, as in $\{11x+3=10,0,1,2,\cdots <!--\; \cdots \; -->\}$
I think that equations means $\mathrm{\exists <!--\; \exists \; -->}x\in <!--\; \in \; -->{\mathbb{Z}}_{18}$ such that $11x+3=10$
The original homework question is:
Solve the equation $11x+3=10\in <!--\; \in \; -->Z_{18}$
Which is why I used the previous form, does the statement change because I've dropped the word "solve" or replaced "in" with "∈" (he is very adamant that these are not synonyms).
How should I read my original equation? Should I avoid this form?

Let $G=(V,E)$ be a tree. Prove that at most one vertex can have degree at least $|V|/2+1$.
I tried to solve this by using a proof by contradiction. I assumed that at least two vertices can have a degree of $|V|/2+1$. But It didn't help that much. Any ideas?

Self-Study in Discrete Mathematics: Linear Recurrence Relations
Through self-study of mathematics, I accidentally stumbled into an area called Linear Recurrence Relations. After poking around in a few different websites I think I'm starting to grasp the idea. I'm just looking to clarify something I read to see if I'm understanding something correctly. On one of the of the websites I was looking at it, it gave the following example:
$x_1=3,x_n=3x_{n-<!--\; -\; -->1}$
$x_2=3x_1=9$
$x_3=3x_2=27$
$x_4=3x_3=81$
Having "solved" the equation for multiple instances, it is noticed that the answers are powers of 3. Thereby making the formula $x_n=3^n$.
I have a few questions.
1. Is it valid to re-write the starting formula as $x_{n+1}=3x_n$ or does this fundamentally change something with the base formula?
2. Is it necessary to solve for several iterations of the base equation to derive the formula or is there a method to have inferred the $x_n=3^n$ from the original formula?
3. Are the solvable iterations necessary for proof?
I will probably have more questions. As this area of mathematics is new to me, and has been self-study. Also, I apologize for any improper use of LaTeX. I'm just now learning of it and how to write with it. So if something doesn't look right, it's probably due to my inexperience with it.

$\lfloor \frac{x+1}{2}\rfloor =\lfloor \frac{2x+1}{3}\rfloor$
I want to find the solutions in R of $\lfloor \frac{x+1}{2}\rfloor =\lfloor \frac{2x+1}{3}\rfloor$, where ⌊x⌋ is the unique integer such that $\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->\le <!--\; \le \; -->x<\lfloor <!--\; \lfloor \; -->x\rfloor <!--\; \rfloor \; -->+1$.
I solved it as follows. Let $\lfloor \frac{x+1}{2}\rfloor =n$. By the definition, $n\le <!--\; \le \; -->\frac{x+1}{2}<n+1$. Then $2n-<!--\; -\; -->1\le <!--\; \le \; -->x<2n+1$, $4n-<!--\; -\; -->2\le <!--\; \le \; -->2x<4n+2$, $4n-<!--\; -\; -->1\le <!--\; \le \; -->2x+1<4n+3$, and finally, $\frac{4n}{3}-<!--\; -\; -->\frac{1}{3}\le <!--\; \le \; -->\frac{2x+1}{3}<\frac{4n}{3}+1$. (1).
Now I analyzise n modulo 3.
If $n=3k$, then (1) becomes $4k-<!--\; -\; -->\frac{1}{3}\le <!--\; \le \; -->\frac{2x+1}{3}<4k+1$. Then either $\lfloor \frac{2x+1}{3}\rfloor =4k-<!--\; -\; -->1$, or $\frac{2x+1}{3}=4k$. From the equation itself, it follows that $3k=4k-<!--\; -\; -->1$, or $3k=4k$, so $k=0$ or $k=1$. It's easy to turn this into 2 intervals for x.
Similarly, I can continue for $3k+1$ and $3k+2$. This seems to lead to a solution, but is there perhaps a better way? Thank you for any ideas/verification of my solution.

Discrete Math (Combination with Repetitions)
Given $(x+y+z{)}^{50}$. How many terms does it have?

Discrete Math: Functions and Set Questions
1) Consider the function: $f:\mathbb{R}\to <!--\; \to \; -->\mathbb{R}$ (Real to Real Number), where $f(x)=2+x^2$, what would be all of the preimages of 3?
1) 11
2) 11, -11
3) 1, -1
4) 1
2) Let $D=\{a,b,c\}$ and let $E=\{2,4\}$, we will define the function f as $f:D\to <!--\; \to \; -->E$ with the following facts.
$f(a)=2$
$f(b)=2$
$f(c)=4$
Based on this information, what is accurate regarding the function of f?
1) f is a "bijection"
2) f is considered to be "one-to-one"
3) f is "onto" and "one-to-one"
4) f is "onto"

Discrete Math
$C=\{1,2\}$
$D=\{1,2,6,7\}$
Considering the sets above are the following statements correct?
$C\cup <!--\; \cup \; -->D=\{1,2,6,7\}$
$C\cap <!--\; \cap \; -->D$
$(C\times <!--\; \times \; -->D)\cap <!--\; \cap \; -->(D\times <!--\; \times \; -->C)=\{\{1,1\},\{1,2\}\}$

Discrete math, antisymmetric proof question
Show that if R is an antisymmetric relation on S then any relation $R^{\prime }\subseteq <!--\; \subseteq \; -->R$ is also antisymmetric. I think how you solve it is if (x,y) are in R, they must be in R' since R' is a subset of R, thus (x,y) are in R'. To be antisymmetric x has to be related to y and y has to be related to x such that $x=y$. And since R is antisymmetric and x,y are in R and R' is a subset then x,y in R' follow the same condition of R making them antisymmetric. That's one way I think it's solved but not 100% sure.

Is each graph a subdivision of itself? On proving reflexivity of the topological minor relation.
The minor relation and the topological-minor relation are partial orderings on the class of finite graphs, i.e. they are reflexive, antisymmetric and transitive
I proved the three properties for the minor relation but am confused when it comes to the topological-minor relation and proving that it is reflexive. Diestel does not really define if a graph G is counted as a subdividing itself, i.e. having an empty set of subdividing vertices and reading the definition on Proof Wiki sounds like one has to at least subdivide one edge resulting in a non-isomorphic graph G′ such that G′ is the smallest subdivision of G.
However, to prove reflexivity, i.e. G being a topological minor of G, G must be in the set of subdividing graphs TX. This is due to the fact that the topolocical minor relation is defined as follows: X is a topological minor of Y if there exists a subdivision of X, which is a subgraph of Y. Therefore, if $G\notin <!--\; \notin \; -->TX$ because G′ is the smallest subdivision of G. Then, no graph in TX can be a subgraph of G and hence, G can't be its own topological minor.
Am I misunderstanding how to prove the reflexivity of a topological minor or something in its definition?

What does the notation $\mathcal{F}\subset <!--\; \subset \; -->2^{[n]}$ mean?
What does the highlighted notation mean?
Theorem 1. Let $\mathcal{F}\subset <!--\; \subset \; -->2^{[n]}$ be such that |A| is odd for every $A\in <!--\; \in \; -->\mathcal{F}$ and $|A\cap <!--\; \cap \; -->B|$ is even for every distinct $A,B\in <!--\; \in \; -->\mathcal{F}$. Then $|\mathcal{F}|\le <!--\; \le \; -->n$.

Self-duality of two boolean functions
I'm stuck with one problem. I have two boolean functions:
1) $x\oplus <!--\; \oplus \; -->1$
2) $x\oplus <!--\; \oplus \; -->y\oplus <!--\; \oplus \; -->1$
The question is which of them is self-dual and which is not. I know the definition that one boolean function is self-dual iff it is equivalent to its dual function (i.e. if their outputs are respectively equal). As a result I get that 1) is not self-dual, and 2) is self-dual. When I checked the answer, it claims the absolute opposite - 1) is self-dual and 2) is not. I will show my truth table for case 1):
1) $x\oplus <!--\; \oplus \; -->1$
$$\begin{array}{|ccc|}\hline x& 1& x\oplus <!--\; \oplus \; -->1\\ 0& 1& 1\\ 1& 1& 0\\ \hline\end{array}$$
Dual of $x\oplus <!--\; \oplus \; -->1$: $\mathrm{\neg <!--\; \neg \; -->}(\mathrm{\neg <!--\; \neg \; -->}x\oplus <!--\; \oplus \; -->0)$
$$\begin{array}{|cccc|}\hline ⌝<!--\; ⌝\; -->x& 0& ⌝<!--\; ⌝\; -->x\oplus <!--\; \oplus \; -->1& ⌝<!--\; ⌝\; -->(⌝<!--\; ⌝\; -->x\oplus <!--\; \oplus \; -->0)\\ 1& 0& 1& 0\\ 0& 0& 0& 1\\ \hline\end{array}$$
The one mistake I've made could probably be putting the brackets - maybe negation should be applied before XOR, but then the second function will also be self-dual. I'm a little confused and will be grateful if anyone clarifies this to me. I've used the following definition for dual function:
The dual of f(x,y,z) is $\mathrm{\neg <!--\; \neg \; -->}f(\mathrm{\neg <!--\; \neg \; -->}x,\mathrm{\neg <!--\; \neg \; -->}y,\mathrm{\neg <!--\; \neg \; -->}z)$.

Orthogonality and Inverse of DFT Matrix
Since there are missing information about my question, I re-open the topic. I am trying to proof that the DFT matrix is orthogonal but i stucked. I just know to proof the orthogonality of two vectors, not a single matrix. How can we proof that the DFT matrix is orthogonal?
$$D_N=\left[\begin{array}{ccccc}1& 1& 1& \cdots <!--\; \cdots \; -->& 1\\ 1& W_N& W_N^2& \cdots <!--\; \cdots \; -->& W_N^{(N-<!--\; -\; -->1)}\\ 1& W_N^2& W_N^4& \cdots <!--\; \cdots \; -->& W_N^{2(N-<!--\; -\; -->1)}\\ ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->& \ddots <!--\; \ddots \; -->& ⋮<!--\; ⋮\; -->\\ 1& W_N^{N-<!--\; -\; -->1}& W_N^{2(N-<!--\; -\; -->1)}& \cdots <!--\; \cdots \; -->& W_N^{(N-<!--\; -\; -->1)(N-<!--\; -\; -->1)}\end{array}\right]$$

What are the integer solutions of the equation $2x^2-<!--\; -\; -->y^2-<!--\; -\; -->y=0$
Alice lives on a long street in which all houses are numbered consecutively (1....N). She noticed that the sum of all the house numbers above her house equals all the numbers above her house. Her house number has three digits. What is it? The general formula for the sum of integers $m....k,k>m$ consecutive integers is $(m+k)\cdot <!--\; \cdot \; -->(k-<!--\; -\; -->m+1)/2$. We form a sequence of consecutive integers $1.....j-<!--\; -\; -->1,j,j+\mathrm{1.....}k$ with j, the house number, and k the last number in the sequence. Then from the problem statement one can subtract the sum of the numbers above the house from the sum below the house and obtain the equation $k^2+k-<!--\; -\; -->2j^2=0$. Using the quadratic formula with j as a parameter one obtains $k=(1+-<!--\; -\; -->\sqrt{1+8\cdot <!--\; \cdot \; -->j})/2$, the “real solution”. We know j varies from 100 to 999. Using MATLAB with rounding I can iterate on j and find the correct answer j, that satisfies the real solution equation, and is the house number $=204$ and k the sequence end is 288. The iteration is very fast, and I can compute very large house numbers. My problem is that I know from Wolfram AlphaPro that there is a closed form “integer solution” that solves the equation for j and k as function n with n any integer.

Number theory Discrete Math
prove $gcd(n,n+2)=1$ or 2. What possible values can $gcd(n,n+3)$ have what about $gcd(n,n+4)$C. What about $gcd(n.n+k)$

No man is weak, unless their name is Bob.
From this expression I see that every man who is weak must be called Bob $(W\Rightarrow <!--\; \Rightarrow \; -->B)$, but the expression does not equate to all Bob's are weak $(B\Rightarrow <!--\; \Rightarrow \; -->W)$, as being named Bob is simply the criteria for allocating everyone who isn't called Bob to the set of "all men who are not weak", meaning that the set of "all weak men" is a subset within the set of "all men named Bob".
I just want to make sure that my reasoning for why it can't be $(B\Rightarrow <!--\; \Rightarrow \; -->W)$ is correct, I know that this might be obvious but I just found this confusing initially.

Trying to solve a simple-looking recurrence relation
I have this recurrence relation that I've been trying to solve,
$$g_m=g_{m-<!--\; -\; -->1}+a_m{g}_{m-<!--\; -\; -->2}$$
with $g_1=g_0=1$ and $m\ge <!--\; \ge \; -->0$. Here, $a_m$ is the m-th term of a known sequence. Any ideas?
If it helps, $a_m=\frac{1}{L^2}(m-<!--\; -\; -->1)(r-<!--\; -\; -->m+2)$ where $L^2\in <!--\; \in \; -->\mathbb{Z},L^2\ne <!--\; \ne \; -->0$, $r\in <!--\; \in \; -->{\mathbb{N}}_0$ and $r\ge <!--\; \ge \; -->|L|$, but other than that, L and r are free. So, if I were to be extremely precise, $g_m$ and $a_m$ would actually be $g_{mrL}$ and $a_{mrL}$.

Show that this propositional equivalence is true:
$\neg (𝑟\leftrightarrow <!--\; \leftrightarrow \; -->𝑠)\equiv <!--\; \equiv \; -->(\neg 𝑟\wedge 𝑠)\vee (𝑟\wedge \neg 𝑠)$
My try out was to compare by truth table, but it is not that the exercice is asking. I need the resolution using arguments as "the morgan" and other simplifications as "$\neg (𝑟\leftrightarrow <!--\; \leftrightarrow \; -->𝑠)\equiv <!--\; \equiv \; -->(\neg 𝑟\wedge 𝑠)\vee (𝑟\wedge \neg 𝑠)$."
When I tryed the simplification on the right side, I could reach something like that:
Changing the sides to be easier...
Matching those two sides, in truth table, it doesn't get the same result. The left side gets the result: F V F F; and the right side results in F V V F. Proving that my resolution is wrong. Can someone help me with this argumentation?

Proving Mathematical Induction
For which nonnegative integers n is $2n+3\le 2^n$? Prove your answer.
I know that we first have to inspect for the basis case and so we find that $n>3$, since $n=1,2,3$ do not hold but 4 plus does, so $n>3.$. But for the Induction step I don't understand how to start it and the textbook solution is unclear, could anyone explain to me what this means?
Textbook solution: For the inductive step assume that P(k) is true. Then, by the inductive hypothesis, $2(k+1)+3=(2k+3)+2<2^k+2.$. But because $k\ge 1,2^k+2\le 2^k+2^k=2^k+1$. This shows that $P(k+1)$ is true.

Discrete math
What is the coefficient of $x^{10}$ in the expansion of $(5x^2+3{)}^{14}$
Leave answer as an expression rather than number.
Do I have to expand using the pascal triangle 10 times or is there a formula cos they said leave it as a mathematical expression?