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Difference between Reflexive and Symmetric in Discrete Maths?
This is what I understand:
Reflexive $\to <!--\; \to \; --><a,a=a>,<b,b=b>$ uses = to describe.
Symmetric $\to <!--\; \to \; --><a,b>,<b,a>$ uses $\le <!--\; \le \; -->,\ge <!--\; \ge \; -->,=$ to describe.
Transitive $\to <!--\; \to \; --><a,b>,<b,c>,<a,c>$ uses <, >, = to describe.
But why is this is a symmetric $<1,1>,<2,2>,<3,3>,<4,4>,<5,5>?$

Discrete Math Question: Induction
Define a sequence recursively as follows. $x_1=1$ and for $n\in N,x_{n+1}=\sqrt{(x_n{)}^2+1/(x_n{)}^2}$
Prove using mathematical induction that for all $n\in <!--\; \in \; -->N$, $1\le x_n\le \sqrt{n}$

Direct Proof Discrete Math.
Show that if n is an odd integer, then $n^2$ is odd.
Proof : Assume that n is an odd integer. This implies that there is some integer k such that $n=2k+1$. Then $n^2=(2k+1{)}^2=4k^2+4k+1=2(2k^2+2k)+1$. Thus, $n^2$ is odd.
Why does the solution assume n to be $2k+1$?
How do you know $n^2$ is odd based on $2(2k^2+2k)+1$?
I don't see how $2k+1$ for n and 2(2k2+2k)+1 for $n^2$ means an odd integer. Some clarification would be helpful.

Represent the statement that: “A car is either moving or stationary; if a car is stationary then its brakes are applied; the car does not have its brakes applied therefore the car is moving.”
I believe i have to use formal logic and connectives while solving this question and i have attempted it but not sure if it is correct. My answer: $(P\vee <!--\; \vee \; -->Q)\wedge <!--\; \wedge \; -->(Q\to <!--\; \to \; -->R)\wedge <!--\; \wedge \; -->(\mathrm{\neg <!--\; \neg \; -->}R\to <!--\; \to \; -->P)$
If this is wrong please correct me

There are three computers A, B, and C. Computer A has 10 tasks, Computer B has 15 tasks, and Computer C has 20 tasks. Each computer must complete its own tasks in order. After, each computer sends its output to a shared fourth computer. How many different orders can the outputs arrive at the fourth computer.

How many $x\in <!--\; \in \; -->\mathbb{N}$ such that $3\cdot <!--\; \cdot \; -->{10}^{13}<x<{10}^{14}$?
and one of the numbers 1, 2,. . . , 9 exactly once and no zeros next to each other?

Let p and q be two given propositions. Use equivalence laws shown in Table 6 and example 3 of section 1.3 to simplify compound propostion $\mathrm{\neg <!--\; \neg \; -->}[p\wedge <!--\; \wedge \; -->(p\to <!--\; \to \; -->q)]$. Indicate what law(s) you are using in each step.

I'm taking a discrete math course, and were on Bézout Coefficients right now. I kind of understand the algorithm, the generalization. However the example in the book is throwing me off.
The steps in the Euclidean algorithm to find gcd(101,4620) are:
$\begin{array}{rl}4620& =45\cdot <!--\; \cdot \; -->101+75\\ 101& =1\cdot <!--\; \cdot \; -->75+26\\ 75& =2\cdot <!--\; \cdot \; -->26+23\\ 26& =1\cdot <!--\; \cdot \; -->23+3\\ 23& =7\cdot <!--\; \cdot \; -->3+2\\ 3& =1\cdot <!--\; \cdot \; -->2+1\\ 2& =2\cdot <!--\; \cdot \; -->1\end{array}$
This I understand. Now to find the Bézout coefficients they follow these steps.
$\begin{array}{rll}1& =3-<!--\; -\; -->1\cdot <!--\; \cdot \; -->2& \\ & =3-<!--\; -\; -->1\cdot <!--\; \cdot \; -->(23-<!--\; -\; -->7\cdot <!--\; \cdot \; -->3)& =-<!--\; -\; -->1\cdot <!--\; \cdot \; -->23+8\cdot <!--\; \cdot \; -->3\\ & =-<!--\; -\; -->1\cdot <!--\; \cdot \; -->23+8\cdot <!--\; \cdot \; -->(25-<!--\; -\; -->1\cdot <!--\; \cdot \; -->23)& =8\cdot <!--\; \cdot \; -->26-<!--\; -\; -->9\cdot <!--\; \cdot \; -->23\\ & =8\cdot <!--\; \cdot \; -->26-<!--\; -\; -->9\cdot <!--\; \cdot \; -->(75-<!--\; -\; -->2\cdot <!--\; \cdot \; -->26)& =-<!--\; -\; -->0\cdot <!--\; \cdot \; -->75+26\cdot <!--\; \cdot \; -->26\\ & =-<!--\; -\; -->0\cdot <!--\; \cdot \; -->75+26\cdot <!--\; \cdot \; -->(101-<!--\; -\; -->1\cdot <!--\; \cdot \; -->75)& =26\cdot <!--\; \cdot \; -->101-<!--\; -\; -->35\cdot <!--\; \cdot \; -->75\\ & =26\cdot <!--\; \cdot \; -->101-<!--\; -\; -->35\cdot <!--\; \cdot \; -->(4620-<!--\; -\; -->45\cdot <!--\; \cdot \; -->101)& =-<!--\; -\; -->35\cdot <!--\; \cdot \; -->4620+1601\cdot <!--\; \cdot \; -->101\end{array}$
My problem is with the second line, where are they getting this +8 from? I've tried just about every algebraic trick I know, but I can't seem to find what they are actually doing.
I think I'm just missing some really simple algebra logic, but maybe I'm not understanding the steps to get Bézout coefficients?

Discrete Math logically equivalent?
Show that $(p\wedge <!--\; \wedge \; -->q)\vee <!--\; \vee \; -->(\mathrm{\neg <!--\; \neg \; -->}p\wedge <!--\; \wedge \; -->\mathrm{\neg <!--\; \neg \; -->}q)\equiv <!--\; \equiv \; -->p\leftrightarrow <!--\; \leftrightarrow \; -->q$
How would I go about doing this?
Do I use a truth table or a more "algebraic" process?

Discrete Math Proof Method
1. Give a direct proof of the fact that $a^2-<!--\; -\; -->5a+6$ is even for any integer a.
2. Suppose a and b are integers and $a^2-<!--\; -\; -->5b$ is even. Prove that $b^2-<!--\; -\; -->5a$ is even.

All books are assigned a 10 digit ISBN ($d_{10}{d}_9{d}_8...d_2{d}_1$) which has the following property:
$\underset{i=1}{\overset{10}{\sum <!--\; \sum \; -->}}i{d}_i\equiv <!--\; \equiv \; -->0(\mathrm{mod}(11))$
Prove that if you swap two adjacnt digits in an ISBN#, it is no longer a valid ISBN#.

Discrete Math Compositions
I am having trouble with these compositions.
$T=\{(a,a),(a,b),(b,c),(b,d),(c,d),(d,a),(d,b)\}$
$U=\{(a,a),(a,d),(b,c),(b,d),(c,a),(d,d)\}$
I need to find $T\circ <!--\; \circ \; -->T$, $U\circ <!--\; \circ \; -->T$, and $T\circ <!--\; \circ \; -->U$.
My problem is when I get down to, for example $U\circ <!--\; \circ \; -->T$ where (d,a) corresponds with both (a,a) and (a,c). This seems to happen for everyone of these problems. Is it even possible to take the composition of these?

1. There is a function that is both $O(n^2)$ and $\mathrm{\Omega <!--\; \Omega \; -->}(n^3)$.
2. Given two functions f(n) and g(n), if $g(n)=O(f(n))$ and $f(n)=O(n^2)$, then $g(n)=O(n^5)$.
3. Given two functions f(n) and g(n), if $g(n)=O(f(n))$ and $f(n)=O(n^2)$, then $g(n)=\mathrm{\Omega <!--\; \Omega \; -->}(n)$.
I think the first one is false, but not sure about other two. Can anyone help me?

I am stuck on 3 proofs for my discrete math class. Any help would be greatly appreciated.
Prove that if R is a partial order, then $R^{-<!--\; -\; -->1}$ is a partial order
Prove that if $R_1$ and $R_2$ are equivalence relations on the set A, then $R_1\cap <!--\; \cap \; -->R_2$ is an equivalence relation.

I need help understanding the concept of "at most one " using quantifers
$\forall x\forall y((\text{Cube}(x)\wedge \text{Cube}(y))\to x=y)$
To me, this statement means that for all cubes x and y, if you pick up a cube from the box, it will always be the same cube.
I understand this to mean that there is exactly one cube in the box.
I don't understand how the statement means that there is at most one cube in a box.

How would I convert this discrete math statement from logic/equation to English?
Given that
B(x) means "x is a bear",
F(x) means "x is a fish", and
E(x,y) means "x eats y",
what is the best English translation of
$\mathrm{\forall <!--\; \forall \; -->}x[F(x)\to <!--\; \to \; -->\mathrm{\forall <!--\; \forall \; -->}y(E(y,x)\to <!--\; \to \; -->B(y))]$?
How can I do solve this? I got "Every fish is eaten by some bear", but that is not the answer. I'm not entirely sure how to go about this since I am fairly new to Discrete Math.

Probability Discrete Math
{1,2,3,4,5,6,7,8,9} What is the probability that the sum of any of these three numbers is odd?

For the relation $R=(x,y):x+2y\le <!--\; \le \; -->3$, defined by $A=\{0,1,2,3\}$, determine if it is reflexive, symmetric, antisymmetric and transitive.
The 2y is throwing me a bit here. To determine if it is reflexive, I have done the following: $x+2x\le <!--\; \le \; -->3$, which is a no with counter example (1,2), as that would give me (1,4) and be greater than 3. Is every second number doubled? Ie. (x,2x) or (x,2y)?
I have concluded this relation is not symmetric, as it does not imply $y+2x\le <!--\; \le \; -->3$, on the basis that is $x=2andy=1$, this would result in $2+2(1)forx+2ybut1+2(2)$ for $y+2x$, which is greater than 3. I have no confidence in this answer however.
I'm floundering on this one. Every time it was touched on during the lecture, it was simple examples, like $x+y$ is even, or if there was an equality, there wasn't a defined set for it.
For determining transitive, what would I use as z?

How many ternary strings of length 10 that contain exactly 2 1s and 3 2s. Since it must contain exactly 2 1s and 3 2s the other 5 spaces must be 0s.But how many ways can we arrange this?

Passwords for a certain computer system are strings of uppercase letters. A valid password must contain an even number of X’s. Determine a recurrence relation for the number of valid passwords of length n. Note: 0 is an even number, so ABBC is a valid password.
Here’s a good way to think about it: to make a good password of length n you can either:
(a) add any non-X to the end of a good password of length $n-1$, or
(b) add an X to the end of a bad password of length $n-1$. For (b) you can use the Good = Total-Bad trick to count the number of bad passwords of length $n-1$.
*This is for an Introductory Discrete math course, so any sort of coding will not be helpful and I will not understand.
I also need to figure out the recursive solution for Xn, so if $X_n+1$ is shown, I have yet to understand how it applies to the recursive solution Xn. If you could PLEASE give me a STEP-FOR-STEP explanation on how this problem and the clues help answer this (throw in some examples as well), it would be so helpful to me. One thing for sure is I do not understand the n-1 part in the clues and how that applies to answering the question. Thus far, I have come up with $X_n=24(x(n-<!--\; -\; -->1))+{26}^{n-<!--\; -\; -->1}$, but when I set $n=1$, I end up with 1, which from using paper and pencil, is not correct (x1 should equal 25, not zero)*