Explore Discrete Math Examples

Discrete Math Proofs Involving Real Numbers
1 . Prove that for every three positive real numbers a, b, and c that
$(a+b+c)\ast <!--\; \ast \; -->(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\ge <!--\; \ge \; -->9$
2. Prove that for every three positive real numbers a, b, and c that $a^2+b^2+c^2\ge <!--\; \ge \; -->ab+bc+ac$.
I have tried direct proof and have not gotten anywhere significant. I won't put the work on there since it is way too long and I don't think it will help. There must be some sort of trick involved, but for the life of me, I cannot figure it out.

Prove that at least one student solved all the problems using PHP
In a physics exam, 5 problems were given to a class of N students. Suppose every two of these problems were solved by more than $\frac{3N}{5}$ students.Prove that at least one student solved all the problems
I have used PHP here. We know if at least $(k.n+1)$ objects are distributed among n boxes, then one of the boxes must contain at least $(k+1)$ objects. So according to this problem $(k.n+1)$ = 2 |where $$n=\frac{3N}{5}$$. But unable to prove at least one student solved all the problems. I have to prove basically $(k+1)$ $$=5$$. It will be helpful for me if someone helps me with this.

Discrete Math - Calculating $\underset{k}{\overset{n}{\sum <!--\; \sum \; -->}}$
I have a question regarding on how to calculate expressions like (I need to write it without $\sum <!--\; \sum \; -->$ and without k):
$$\underset{k=2}{\overset{25}{\sum <!--\; \sum \; -->}}(\genfrac{}{}{0ex}{}{25}{k})(\genfrac{}{}{0ex}{}{k}{2})$$
And also like:
$$\underset{k=1}{\overset{10}{\sum <!--\; \sum \; -->}}k(\genfrac{}{}{0ex}{}{10}{k})(\genfrac{}{}{0ex}{}{20}{10-<!--\; -\; -->k})$$
Unfortunately I didn't learn how to do it, but this is a part of the material for my exam today.

Discrete Math Binomial Theorem Formula.
Binomial Theorem Formula:
- Sigma Notation: $\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}$ $(\genfrac{}{}{0ex}{}{n}{i})$ $x^i$ $y^{(n-<!--\; -\; -->i)}$
Note: $(n-<!--\; -\; -->i)$ is the exponent of y.
Logical Question:
Can the place of exponents for x and y be switched ?
Ex. $\underset{i=0}{\overset{n}{\sum <!--\; \sum \; -->}}$ $(\genfrac{}{}{0ex}{}{n}{i})$ $x^{(n-<!--\; -\; -->i)}$
Note: Now $(n-<!--\; -\; -->i)$ is the exponent of x.

Show that the relation R consisting of all pairs (x, y) such that x and y are bit strings of length three or more that agree in their first three bits is an equivalence relation on the set of all bit strings of length three or more.
I am new to this kind of math, and what I am stuck at is how do I convert (x,y) into strings of bits?
Thus far I have made the sets: $R=\{(x,x),(x,y),(y,x),(y,y)\}$ but I need to make them into strings of length three or more. Do I just pick random 1's and 0's???

Interpretation of supercript and subscript on sigma symbol
Given $\underset{1}{\sum <!--\; \sum \; -->}=\{a,b\},and\underset{2}{\sum <!--\; \sum \; -->}=\{1,2,3\}$
What does $\underset{1}{\overset{3}{\sum <!--\; \sum \; -->}}$ or $\underset{2}{\overset{3}{\sum <!--\; \sum \; -->}}$ mean? Note that the 1 and 2 are meant to appear as subscripts of $\sum <!--\; \sum \; -->$, I wasn't sure how to do that here on stack exchange.

Show that $T(n)=2^{n+1}-<!--\; -\; -->1$.
Show that $T(n)=2^{n+1}-<!--\; -\; -->1$, for the following,
$$T(n)=\{\begin{array}{ll}1& \text{ if, }n=0\\ T(n-<!--\; -\; -->1)+2^n& \text{ otherwise}\end{array}$$
I did it as follows,
$$\begin{array}{r}T(n)=T(n-<!--\; -\; -->2)+2^{n-<!--\; -\; -->1}+2^n\\ T(n)=T(n-<!--\; -\; -->3)+2^{n-<!--\; -\; -->2}+2^{n-<!--\; -\; -->1}+2^n\\ & ⋮<!--\; ⋮\; -->\\ T(n)=T(n-<!--\; -\; -->(n-<!--\; -\; -->1))+2^{n-<!--\; -\; -->n}+\cdots <!--\; \cdots \; -->+2^{n-<!--\; -\; -->1}+2^n\end{array}$$
From the last call, I can see that,
$$\begin{array}{r}T(0)=T(1)+2^0+\cdots <!--\; \cdots \; -->+2^{n-<!--\; -\; -->1}+2^n\\ 1=T(1)+2^0+\cdots <!--\; \cdots \; -->+2^{n-<!--\; -\; -->1}+2^n\\ 1=T(1)+\underset{i=1}{\overset{n}{\sum <!--\; \sum \; -->}}{2}^i\\ 1=T(1)+\frac{2^{n+1}-<!--\; -\; -->1}{2-<!--\; -\; -->1}\end{array}$$
How we can get rid of 1 and T(1) as they appeared in my solution please?

Fibonacci sequence is embedded in Pascal’s triangle by investigating “stretched diagonals”. While this is true, it is not obvious how the sequence is embedded. Redraw Pascal’s triangle to make it clear how the Fibonacci sequence is embedded in the triangle, and explain the calculation required to obtain the sequence. A formal proof is not required; an observation over 12 rows of the triangle will suffice. You may use the fact that the r-th number in the n-th row of the triangle is C(n; r) when explaining the calculation.

Help with Strong Induction proof
$$A_n=\{\begin{array}{ll}3,& n=1\\ 9,& n=2\\ 2A_{n-<!--\; -\; -->1},& n\ne <!--\; \ne \; -->1\text{ and }n\text{ odd}\\ 4A_{n-<!--\; -\; -->1},& n\text{ even}\end{array}$$
Need to prove $A_n\le <!--\; \le \; -->3^n$ $\forall n$ in $N\ge 1$ by strong induction. After case $n=1$ and $n=2$, i thought about going for something like $n=2k$ for n even and $n=2k+1$ for n odd but i got stuck and have yet no idea how to proceed after the two inicial cases.

Discrete Math: Induction and Recurrence
Show that the terms of the sequence that satisfy $a_n=3a_{n-<!--\; -\; -->1}-<!--\; -\; -->2$ and $a_1=4$ are given by the formula $a_n=3^n+1$ for all $n\ge <!--\; \ge \; -->1$.
I know this problem has to do with induction. In this case I think $n=n+1$ so I would have $3^n+1$ but i'm not sure what to do with the rest of the problem. Can someone help please?

Proof by Induction Help Discrete Math?
So I've gotten to a point in this induction problem where I really don't know where to go.
I've gotten the equation down to this:
$$\frac{k(k+1)(2k+3)}{2(2k+1)(2k+3)}+\frac{2(k+1{)}^2}{2(2k+1)(2k+3)}=\frac{(k+1)(k+2)}{2(2k+3)}$$
Original problem was this:
$$\frac{1^2}{1(3)}+\frac{2^2}{2(3)}+...+\frac{n^2}{(2n-<!--\; -\; -->1)(2n+1)}=\frac{n(n+1)}{2(2n+1)}$$

Basic Equivalence Class Discrete Math
I read through the textbook definition of the equivalence class, but still cannot clearly understand what an equivalence class is.
Does anyone have a good example with a definition that can hit me home?

Minimum requirements on a function to prove injectivity
I have one predicate and two finite sets A and B.
I(x): x is an injective function.
From these I construct the quantified statement:
$$\mathrm{\forall <!--\; \forall \; -->}\mathbf{x}\in <!--\; \in \; -->\mathbf{\text{U:[ I(x)]}}$$
$$\mathbf{\text{U = { f is a function | f: A}}\mapsto <!--\; \mapsto \; -->\text{B}}$$
What assumptions do I need to make on each of the sets to make each of the quantified statements true?
The answers in my book says the first statement is true when $|A|=0,|A|=1,or|B|=0$, though I'm not sure I follow the logic.
If $|A|=0$ then I can see that the statement is true in regards to the only function being the empty function which I've read is injective (though haven´t understood exactly why yet).
But if $|A|=1$, how can I assume f(x) will have an inverse without making any assumptions about B?
Also if $|B|=0$, is this statement just vacuously true with regards to injectivity?

Marble Algorithms Using Discrete Math
During a job interview for a computer science job you are hypothetically given two identical marbles and then asked to figure out the following problem. Find an algorithm which minimizes the maximum number of steps needed to find the smallest floor j so that a marble breaks when dropped from floor j but not from floor $j-<!--\; -\; -->1$ in a skyscraper with 100 floors.
I think this problem has to do something with induction. I'm not sure how to start with this problem. Any help would be appreciated.

discrete math about Pigeonhole Principle
Prove that any set of 10 positive integers less than or equal to 100 will always contain two subsets with the same sum.
Can anyone help me with this problem?

Winning strategy at "Turning Turtles"
I'd like like to know if there is a winning strategy for this game.
From the written above I understand how to calculate the Nim-Value of it:
$H=1,T=0$
$HTTHTTHTHT=1001001010$, and Nim-Value of it is:
$(100)\oplus <!--\; \oplus \; -->(100)\oplus <!--\; \oplus \; -->(10)\oplus <!--\; \oplus \; -->(10)=0$ - this is loosing situation (if I understand right...).
So my question is:
How can I bring from loosing situation to a winning situation? (if it's possible)
And I don't understand something: lets say that this is the board $HTTHTTHTHT=1001001010$ - and it's my turn. I'm at loosing position (Nim-Val is 0), and I can do a move that make the next player at loosing position also:HTTHTTTTTT - and the next player is at Nim-Val of 0 (loosing position). Where is my mistake here?

Sets and Functions
Prove that If $𝐴0\subseteq <!--\; \subseteq \; -->𝑋$ and $𝐴1\subseteq <!--\; \subseteq \; -->𝑋$, then $𝑓(𝐴0)-𝑓(𝐴1)\subseteq <!--\; \subseteq \; -->𝑓(𝐴0-𝐴1)$
Prove that if $𝐴0\subseteq <!--\; \subseteq \; -->𝑋,𝐴1\subseteq <!--\; \subseteq \; -->𝑋$ and 𝑓 is one to one, then $𝑓(𝐴0)-𝑓(𝐴1)=𝑓(𝐴0-𝐴1)$

Number of different values that obtain from a given set
Suppose $S=\{1,2,\dots <!--\; \dots \; -->,7\}$, already we know that, S have $2^7$ subsets. But the problem is, when we multiply elements of each subset S, how many different value we can obtain from multiplication of elements of each subsets? In $2^7$ subsets, some of them are equal, for example, for subset of size 2, $3\times <!--\; \times \; -->4=6\times <!--\; \times \; -->2$. Are there a simple way that eliminate the duplicate values for any subsets?

Prove that $2^{n+1}>(n+2)\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(n)$ for all positive integers n.
Proof: If P(n) represents the given proposition.
(1) Basic Step: P(n) for $n=1$ is
$$2^{1+1}>(1+2)\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(1)$$
$$2^2>3\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(1)$$
$$4>3\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(1)$$
Since $\sin <!--\; \; -->(n)\le 1$ and $1<4/3$ and so $\sin <!--\; \; -->(n)<4/3$ by putting $n=1$, we obtain
$$\sin <!--\; \; -->(1)<4/3⟹<!--\; ⟹\; -->4>3\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(1)$$
Which is true.
(2) Induction Step: Suppose that P(n) is true for $n=k$, i.e. Let
$$2^{k+1}>(k+2)\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(k)$$
To prove that P(n) to be true for $n=k+1$
$$2^{(k+1)+1}=2^{k+1}2=2\cdot <!--\; \cdot \; -->2^{k+1}>2(k+2)\cdot <!--\; \cdot \; -->\sin <!--\; \; -->(k)$$
How to prove P(n) to be true for $n=k+1$. I got stuck here. Would appreciate for your assistance. Also review my proof if there is any mistake while writing.

State whether the following statements are true or false and justify your answers
(a) $\mathrm{\exists <!--\; \exists \; -->}x\in <!--\; \in \; -->Z\mathrm{\exists <!--\; \exists \; -->}y\in <!--\; \in \; -->Z(x+y=2)\wedge (x-<!--\; -\; -->y=3)$
(b) $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->R\mathrm{\forall <!--\; \forall \; -->}y\in <!--\; \in \; -->Rx>y-\to <!--\; \to \; -->x^2<y^2$
(c) $\mathrm{\forall <!--\; \forall \; -->}x\in <!--\; \in \; -->R\mathrm{\exists <!--\; \exists \; -->}y\in <!--\; \in \; -->R{x}^2+y^2>1$