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1

In the $n$-dimensional vector space $F_5^n$, every non-zero vector generates a one-dimensional vector space with 5 elements. Each pair of such vector spaces intersect only at the zero vector. Therefore, $F_5^n$ can be divided into one-dimensional sub-spaces (all containing the zero vector). If the zero vector is removed, we have partitioned remaining ...


1

"Drawing a figure" is a bit tricky for higher dimensions, but one could consider expanding an "$n-$ cube" with "edge length" of 4 . To make it a bit easier to work with, we would extend each "edge" by $ \ \frac {1}{2} \ $ in both "directions", rather than simply adding 1 . For a line segment , we have $ \ 4 \ = \ 5 \ - \ 2 \ \cdot \ \frac{1}{2} \ $ , ...


4

Look at the following for $n = 2$: $$ \bullet \bullet \bullet \bullet \diamond \\ \bullet \bullet \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet \bullet \\ \bullet \bullet \bullet \bullet \bullet $$ The black dots visualize $5^2 - 1$. Now one easily sees a division of this in sets of $4$ and therefore ...


9

Using base-$5$, this is pretty obvious: $$10^n_5 - 1 = \underbrace{444\ldots4_5}_n = 4\times \underbrace{111\ldots 1_5}_n$$


9

Algebraic proof: Consider the field $K:=\mathbb{F}_{5^n}$. We shall show that the group of units $K^\times =K\setminus\left\{0\right\}$ has a subgroup $H$ of order $4$. By Lagrange's Theorem, $4$ must then divide the order of $K^\times$, which is $5^n-1$. Now, this group $H$ is given by the subset $\{1,2,3,4\}=\mathbb{F}_5\setminus\{0\}$. The result ...


15

Let there be $5$ different characters $\{\_\ , a,b,c,d\}$ to form a string of length $n$. There are $5^n$ of those strings. Apart from the string of all $\_$'s $"\_\ \_\ \_\ldots \_"$, all other strings can be grouped into exactly one of four groups in this way: Take the first character along each string that is not an $\_$, which can be one of $a,b,c$ or ...


3

Given $a x^2 + b xy + c y^2 + d x + e y + f = 0$:   [Complete the square to get rid of the $xy$ term.]   $( 2a x + b y )^2 - b^2 y^2 + 4a ( c y^2 + d x + e y + f ) = 0$.   Let $z = 2a x + b y$.   Then $z^2 + (4ac-b^2) y^2 + 2d z + (4ae-2bd) y + 4af = 0$.   [Complete the squares.]   $(4ac-b^2) (z+d)^2 + ( (4ac-b^2) y + ...


0

The "Synopsis of Pure and Applied Mathematics" is precisely what it claims to be. It is an abbreviated form of other nineteenth century texts for which it provides reference lists. You are not going to understand the contents of this synopsis properly unless you read these other texts as well, all of which, being 150+years old, are freely available online ...


0

Write $x=\lfloor x\rfloor+\{x\}$. Then we have to solve $$3\lfloor x\rfloor+2\{x\}=1\ .$$ There is no solution with $\lfloor x\rfloor<0$ or $\lfloor x\rfloor\geq1$. When $\lfloor x\rfloor=0$ we need $2\{x\}=1$, which implies $x={1\over2}$. It is easy to check that this is indeed a solution.


1

Note that for any integer $n$, $\;\lfloor x+n\rfloor=\lfloor x\rfloor+n $,and that $$\lfloor 2x\rfloor=\begin{cases}2\lfloor x\rfloor&\text{if}\enspace 0\le x-\lfloor x\rfloor<\dfrac12,\\2\lfloor x\rfloor+1&\text{if}\enspace \dfrac12\le x-\lfloor x\rfloor<1.\end{cases} $$ The given equation implies $\;1=\bigl\lfloor\lfloor x\rfloor+ ...


1

$x-1<\lfloor x\rfloor \leq x$, so $3x-1<\lfloor x\rfloor +2x=1\leq 3x$, this implies $x\in [1/3, 2/3)$, so $\lfloor x\rfloor=0$, then $2x=1$, $x=1/2$.


1

Rewrite the given equation as $$\lfloor x \rfloor = 1-2x$$ Note that $1-2x$ must be an integer. So $x=\frac{k+1}{2}$. Now simplification is much easier by considering two cases (1)$k$ is even (2)$k$ is odd. OR alternatively you can do as follows for $x > 1/2$ the right side is negative but left side is non-negative so no equality. Likewise for $x ...


3

Note that the function $f(x) =\lfloor x \rfloor + 2x$ is strictly increasing. It is always between $3x-1$ and $3x$. Solving $3x-1=1$ and $3x=1$ gives that $x$ is between $\frac13$ and $\frac23$, because $f$ is strictly increasing. So $\lfloor x \rfloor = 0$, since $\frac13 < x < \frac23$. So we solve $2x=1$ so $x=\frac12$.


0

The number of famous mathematicians in history who were also great mental calculators are few and far between. Euler, who became blind in his latter years, is said to have been able to perform calculations in his head. One story is that Euler passed away while he was mentally calculating the orbit of the moon. Emilie du Chatelet is also another famous ...


0

There are two practice exams available in Amazon. There not by ETS but may be useful nonetheless. They were written as a part of a full prep course for the GRE math subject exam. http://www.amazon.com/s/ref=nb_sb_noss?url=search-alias%3Dstripbooks&field-keywords=subjectmath.com&rh=n%3A283155%2Ck%3Asubjectmath.com ...


0

$\text{Circumference}=2(x + x + 130)= 4x+260$ $\text{Area}=(x+130)\cdot (x)=x^2 + 130x = x^2 + 130x + 65^2 - 65^2 = (x+65+\sqrt{65})(x+65-\sqrt{65})$


0

There is a series of math children books in Russian by Владимир Артурович Лёвшин. To list some: Магистр рассеянных наук (translates roughly as Master (as in M.Sci) of the absent-minded sciences, though google translates it as Master scattered Sciences), Новые рассказы Рассеянного Магистра (New stories of the absent-minded Master), Путешествие по Карликании и ...


6

max{x,3x} + min{x,3x} = 4x max{x,3x} - min{x,3x} = |2x| ---------------------------- 2max{x,3x} = 4x + 2|x| max{x,3x} = 2x + |x| As a bonus, min{x,3x} = 2x - |x|


12

I will probably put more emphasis on those transformation properties of the $\max(\cdot)$ function which is reasonably "obvious" and easy to remember/use. For example, $\max(a+b,a+c) = a + \max(b,c)$. $\max(ab,ac) = a \max(b,c)$ when $a > 0$. $\max(a,-a) = |a|$. This may help a student to get familiar with such "tools" for attacking similar problems. ...


28

You can use the fact that, $\max\{x,y\}=\frac{x+y+|x-y|}{2}$. Therefore, $\max\{x,3x\}=\frac{x+3x+|x-3x|}{2} = \frac{4x+|-2x|}{2} =2x+|x|$.


2

Note that $f(-x) = -ax + b|x|+ c$, thus breaking into symmetrical and anti symmetrical combinations one obtains $$ f(x) - f(-x) = 2ax\\ f(x) + f(-x) = 2b|x| + 2c $$ Thus $$ a = \frac{f(1) - f(-1)}{2} = \frac{3 + 1}{2} = 2\\ c = f(0) = 0\\ b = \frac{f(1) + f(-1)}{2} - c = \frac{3 - 1}{2} = 1 $$ Second variation of the same idea: $$ f(-x) = \max(-x, -3x) = ...


4

I like the question! It is not immediately obvious to me, however, that it suffices to match your form to the given function at three points. Perhaps that is the case, but if so it requires a separate argument. Alternatively, having discovered the final form you could verify it directly. I would address the original problem this way: Your function, $f$, ...


8

How to overcome the temptation to read many books covering the same topics[?] Simply by working. I can't tell you how many things I am able to push out of my mind when thinking about the mathematics itself. I have no idea what you've been doing with respect to analysis, but over the past few months, you have asked several questions about analysis ...


0

Yep. If you read one book over another you'll likely miss problems that are covered in only one of those books but not the other. However, if you read one whole book and have a solid understanding of the general theory, it won't be too hard to learn other topics that you missed if needed. How should I overcome such temptations and anxiety? Don't worry ...


0

Ask a professor or more advanced math student (whose opinion you trust) what book is best on the subject. Find the best book, understand it cover to cover (make sure you understand all pre-requisites first), then advance to a higher level book on that subject (if so desired).


0

As claimed here any child can be turned into a genius. The secret is early education. So, you should start to learn math at a young age. The younger you are, the less hard wired your brain is and that allows you to hard wire "mathematical aptitude" in a way that is impossible to do at a later age. It is very unfortunate that the school curriculum for math ...


6

I empathize with your username @iLearnSlow. The main reason I'm decent at trigonometry, and I daresay this goes for many people here, is because I've done many, many, many problems. Like many people, I really struggled at first, and I definitely didn't grok the material until the second or third time through. I first did problems at home with my father. ...


6

My suggestion is to find a book (your local bookstore, or university library is likely to have tons of books covering these subject areas) with a lot of worked examples. Start by reading through the examples, and see if you understand them. Next, you try to work the examples yourself, with the book closed. Once you have gotten a "feel" for the proofs, you ...


0

http://www.mathtube.org/ is very good. There are also videos on the fields institute webpage http://www.fields.utoronto.ca/video-archive In solidarity


5

Aptitude means little. Hard work means a lot.


0

A nice starting place is the collection of notes by William Chen. It covers most of the undergraduate curriculum (and somewhat before) in a really accessible way. The Trillia group has a collection of nice texts too.


2

Much of cryptography today works on grounds of abstract algebra (and number theory). Clearly to show that some encryption technique has a corresponding decryption one requires proof, that is math. But that is just a small part of cryptography. The recent most talked about problems with cryptographic systems have been programming errors, disregard of the ...


1

The $k-th$ percentile in a distribution I a value larger than that of $k$% of the population. As an example, the $40$% percentile is a value ( a salary in this case ) higher than the salaries of $40$% of the population. The information you gave is not enough to determine the $50 th$ and $75th$ percentile. As an example, if half the people make 30,000 ...


5

In a sense, both of your ideas (computer science and pure mathematics) are both correct for what might be suitable for cryptography. RSA encryption is based on abstract algebra concepts, and also the computer science belief that factoring integers is hard unless on a quantum computer. You can read about RSA online, there's plenty of references and if you ...


9

As a computer science student who took a graduate course (albeit an introductory one) in cryptography last semester, I found myself pulling from my knowledge of number theory immensely more than I did from my knowledge of computer science. The field today is a highly mathematical one, with current state-of-the-art systems reliant on number theoretical ...


1

1) logarithmic spirals do occur in Nature. I doubt that exact golden spirals occur otherwise than by accident (i.e. with probability $0$), as I can see no physical law that would explain them. 2) clearly, no.


2

All the proofs? Don't think you can make such a blanket statement. But I'll give my opinion. The proofs matter, but it really depends. Is this your first time reading through a book or paper? Then maybe most of the proofs should either be skipped or briefly skimmed over until the next reading. The technical ideas can get in the way of forming the big ...


0

OOP is a paradigm to organize code and data. This can be used e.g. to model physical and geometric problems. The large arsenal of applied mathematics is also used for modeling and solving such problems. That is about the commonality with OOP I can see. There is a relationship between the theories of computation and constructive mathematics, e.g. this ...


0

It sounds like you need an opportunity to evaluate your own strengths and weaknesses as a math learner that meets you where you are today. If you are in a technical field, and you want to weigh exploring traditional subject matter vs. starting with foundations, I would propose modifying your sequence to percolate down Linear Algebra out from "everything ...


3

A certain kind of Lacoste pullovers comes in five colors, three different neck forms, and in six sizes. Let $$\eqalign{C&:=\{{\rm blue,\ red,\ yellow,\ green,\ brown}\}\ ,\cr N&:=\{{\rm round\ neck, \ V{-} neck,\ polo{-}neck}\}\ ,\cr S&:=\{{\rm XS,\ S,\>M,\ L,\ XL,\ XXL}\}\cr}$$ be the sets of colors, neck forms, and sizes, respectively. Then ...


5

You have good historical reasons for interpreting $n:m$ as $n/m$. From http://jeff560.tripod.com/operation.html: The colon (:) was used in 1633 in a text entitled Johnson Arithmetik; In two Bookes (2nd ed.: London, 1633). However Johnson only used the symbol to indicate fractions (for example three-fourths was written 3:4); he did not use the symbol for ...


2

A reasonable definition would be $$a_0 : \cdots : a_{n-1} = \frac{1}{a_0+\cdots+a_{n-1}}(a_0,\ldots,a_{n-1})$$ For example, under this definition, we have: $$1:7 = (1/8,7/8)$$ Exercise. Show that $(a:b) = (a':b')$ iff $a/a' = b/b'.$ By the way, this can be used to take affine combinations. Given a $k$-tuple of real numbers $a$ and a $k$-tuple of ...


1

Arithmetically, 1:7 definitely means 1/7 May be, the moderator was comparing it with equating odds of 1:7 with a probability of 1/7 (as, alas, numerous people do, taking the two terms to be synonymous), which 0f course is incorrect and the probability is 1/8


1

If an amount of money is shared among A and B in the ratio $3:5$ then A gets ${3\over 8}$ of the total, but ${3\over5}$ as much as B. In my view an expression of the form $a:b$ is NAN (not a number) but a way of talking to be parsed in real time. Mathematically the pair $(a,b)$ can be considered as homogeneous coordinates of a point $p\in{\Bbb P}^1({\Bbb ...


5

If there are eight people and $1$ of them is tall and the rest short then the ratio of tall people to short people is $1:7$. However the ratio of tall people to all the people is $1:8$. I would say it makes more sense to relate $1:7$ with $\frac{1}{8}$ because it implies $\frac{1}{8}$ of the total satisfy the property. On the other hand one could say ...


1

As requested, here is my solution in answer form. if p(x) = $\sum {a_i}{x^i}$ is a polynomial then p(1) = $\sum {a_i}$ is the sum of the coefficients. similarly, p(-1) is the sum of the even numbered coefficients minus the sum of the odd numbered coefficients. To be clear: the constant term is the degree 0 term, an even numbered coefficient, so its sign ...


0

Not sure what you mean by "this is the $c$ term added up", but here are some examples: $P(x) = 3 x^4 - 2 x^3 + x^2 - 2$: $3 - 2 + 1 + 0 - 2 = 0$ so $1$ is a root. $3 + 2 + 1 - 0 - 2 = 4$ so $-1$ is not a root.


4

This 'trick' your professor taught you is actually what is called the remainder theorem. Basically what it says is that if you have a polynomial: $p(x)$ and you calculate the value of $p(d)$ then $p(d)$ is the remainder when $p(x)$ is divided by $(x-d)$. For example $p(x) = x^2 -2x+1$. Let $d=1$ and we get:$$p(d) = p(1) = 1^2-2(1)+1 = 1-2+1 = 0$$ Thus $p(x)$ ...


3

1. If $f$ is odd , then $|f|$ is even because $\left | f(-x) \right |= \left | -f(x) \right |= \left | f(x) \right |$. 2. If $f$ is odd and $g$ is even then $f\circ g$ is even. That is because if $h(x)=(f\circ g)(x)$ then: $$h(-x)=\left ( f\circ g \right )(-x)= f\left ( g(-x) \right )= f(g(x))=h(x)$$ 3. If $f$ is odd and $g$ is odd then $\varphi(x)=f(x) ...


0

Hint: Observe that $f$ is defined, as a real valued function, if and only if \begin{align*} &&3|x|-x-2&\ge 0\\ \iff && 3|x|&\ge x+2\\ \iff & &9x^2\ge x^2+4x+4 \quad&\text{or}\quad x+2\le 0\\ \iff&&2x^2-x-1\ge0\quad&\text{or}\quad x\le -2\\ \iff&&x\le-\frac{1}{2} \qquad\text{or}\qquad ...



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