define:  $ f_k(x)\equal{}g(x)x^k$ where $ g$ is continuous on $ [0,1]$ and $ g(1)\equal{}0$.  Show that $ f_k\rightarrow 0$ uniformly on $ [0,1]$.

[hide="Hint"]
First show that $ f_k(x)$ is uniformly convergent in $ [0,1)$ because $ x^k$ is convergent to $ 0$ on the interval and $ g$ is continuous so that it is bounded.
Then show that $ f_k(1)$ is also convergent to $ 0$ because $ 1^k$ is convergent to $ 1$ and $ g(1)\equal{}0$ so $ f_k(1)\equal{}0$ when $ k$ goes to infinity. I am sorry if I am wrong.
[/hide]


[quote="mmrproblem"]Let ABC be any right triangle and it's hypotenuse is divided two line segments by incircle that lengths 4 and 5.Find the area of ABC triangle.[/quote]

If r be the radius of the incircle, the arms of the right angle are r+4 and r+5

So Ar = (1/2)*(r+4)(r+5)..(1)

Again semiperimeter s = Area of ABC / r

=> (1/2)*(r+4 + r+5 + 9) = Ar(ABC)/r

=>  r(r+9) = Ar(ABC)...(2)

eQUATING (1) AND (2) and observing that Ar(ABC) = r(r+9), we get Area = 20 sq. units

Assume the angle $\measuredangle C$ is right. Let $R, S, T$ be the tangency points of the sides $a = BC$, $b = CA$, $c = AB$ with the incircle and $r$ its radius. The external tangent lengths from the vertices $A, B, C$ to the incircle are equal, $AT = AS$, $BT = BR$, $CR = CS$. Since the $\measuredangle C$ is right, $CR = CS = r$. The triangle sides are:

$c = AT + BT = 4 + 5 = 9$
$a = BS + CS = BT + r = r + 5$
$b = CR + AR = r + AT = r + 4$

$a^2 + b^2 = c^2$
$81 = (r + 5)^2 + (r + 4)^2 = r^2 + 10r + 25 + r^2 + 8r + 16$
$r^2 + 9r - 20 = 0$

$r = \frac{-9 \pm \sqrt{81 + 4 \cdot 20}}{2} = \frac{-9 \pm \sqrt{161}}{2}$

Since $r > 0$, only the positive root with the plus sign is acceptable.

$a = r + 5 = \frac{-9 + \sqrt{161}}{2} + 5 = \frac{ 1 + \sqrt{161}}{2}$
$b = r + 4 = \frac{-9 + \sqrt{161}}{2} + 4 = \frac{-1 + \sqrt{161}}{2}$

Area of the triangle $\triangle ABC$ is

$\frac{ab}{2} = \frac{(1 + \sqrt{161}) (-1 + \sqrt{161})}{8} = \frac{161 - 1}{8} = 20$

[quote]Let ABC be any right triangle and its hypotenuse is divided by the incircle
into two line segments of lengths 4 and 5.
Find the area of ABC triangle.[/quote]
I came up with my own solution,
but it is a combination of the solutions of santosh_blr and yetti.[hide]
From santosh_blr, we have: $\text{ Area} = \frac{1}{2}(r^2 + 9r + 20)\;\;[1]$

From yetti, we have: $r^2 + 9r - 20 = 0\;\hdots$ Hence: $r^2 + 9r = 20\;\;[2]$

Substitute [2] into [1]: $\text{Area} \;= \;\frac{1}{2}(20 + 20) \;= \;20$[/hide]


A circle is inscribed in a square. A second square is inscribed in the circle. A second circle is inscribed in the smaller square. What is the ratio of the area of the smaller circle to the larger circle? Express your answer as a common fraction.
[asy]draw(unitsquare);
draw((.5,0)..(1,.5)..(.5,1)..(0,.5)..cycle);
draw((.5,0)--(1,.5)--(.5,1)--(0,.5)--cycle);
draw((.25,.25)..(.25,.75)..(.75,.75)..(.75,.25)..cycle);[/asy]

[quote="cgyao15"]The squares these circles are enscribed in are in the ratio of $ \ 1: 2$ so the diameters of the circles are $ \ 1: 2$ and so are the radius.
So we divide each area by pi and notice that the ratio will always be 1/4[/quote]

Hi,

It is actually 1/2 because that's what it said on FTW answers.

-nbute

Hello, the answer is not 1/2 because "that's what is said on FTW answers", but rather:
[hide="Solution:"] Let the side length of the large square be $2x$. The area of the larger circle is then just $x^2\pi$.
The side length of the smaller square is $\frac{2x}{\sqrt{2}}=\sqrt{2}\cdot x$ The area of the smaller circle is then $\frac{1}{2}x^2\pi$
Thus the ratio is $\boxed{\frac{1}{2}}$ [/hide]


integration
calculus
calculus computations

(i) f(1)=0
(ii) f'(1)=1
(iii) d/dx[f(2x)]=f'(x), for all x>0

a) Find f'(2).
b) Suppose f' is differentiable. Prove that there is a number c, 2<c<4, such that f''(c) = -1/8.
c) Prove that f(2x)=f(2)+f(x), for all x>0.

[b]Given:[/b]
$ (i)~~$ $ f(1)\equal{}0$
$ (ii)~$ $ f'(1)\equal{}1$
$ (iii)$ $ \frac{d}{dx}[f(2x)]\equal{}f'(x) ~\forall ~x>0$

$ a)$ Find $ f'(2)$.
$ b)$ Suppose $ f'$ is differentiable. Prove that there is a number $ c$, $ 2<c<4$, such that $ f''(c) \equal{} \frac{\minus{}1}{8}$.
$ c)$ Prove that $ f(2x)\equal{}f(2)\plus{}f(x) ~\forall~ x>0$

[b]Solution:[/b]
$ a)$ We'll use parts $ (ii)$ and $ (iii)$ in this problem.

[hide]$ \frac{d}{dx}[f(2x)]\equal{}f'(x)$
$ \implies 2 \cdot f'(2x) \equal{} f'(x)$
$ \implies 2 \cdot f'(2) \equal{} f'(1)$
$ \implies \boxed{f'(2) \equal{} \frac{1}{2}}$[/hide]

$ b)$ We'll have to use parts $ a)$, $ (ii)$, and $ (iii)$ on this part.

[hide]$ \frac{d}{dx}[f(2x)]\equal{}f'(x)$
$ \implies 2 \cdot f'(2x) \equal{} f'(x)$
$ \implies 2 \cdot f'(2\cdot 2) \equal{} f'(2)$
$ \implies 2 \cdot f'(4) \equal{} \frac{1}{2}$
$ \implies f'(4) \equal{} \frac{1}{4}$

Thus by the Mean Value theorem $ \exists ~ c \in (2,4)$ such that:

$ f''(c)\equal{}\frac{f'(4)\minus{}f'(2)}{4\minus{}2}$
$ \implies f'(c)\equal{}\frac{\frac{1}{4}\minus{}\frac{1}{2}}{2}$
$ \implies f'(c)\equal{}\frac{\frac{\minus{}1}{4}}{2}$
$ \implies \boxed{f'(c)\equal{}\frac{\minus{}1}{8} \text{ for some } c, \text{ such that } 2<c<4}$[/hide]

$ c)$ We'll use parts $ (i)$ and $ (iii)$ on this part.

[hide]$ \frac{d}{dx}[f(2x)]\equal{}f'(x)$
$ \implies \int \frac{d}{dx}[f(2x)] ~dx\equal{}\int f'(x)~dx$
$ \implies f(2x) \equal{} f(x) \plus{} C$

Now we'll use part $ (i)$:

$ f(2 \cdot 1)\equal{}f(1)\plus{}C$
$ f(2) \equal{} 0 \plus{} C$
$ C\equal{}f(2)$

Therefore $ \boxed{f(2x) \equal{} f(x) \plus{} f(2) ~\forall ~x>0}$[/hide]


Is the diagram fully describable from your description?

[quote="ZhangPeijin"][quote="Milkyway989558"]The diagram shows a trapezium ABCD.
E is a point on BC such that BE = 1 cm and EC = 4 cm.
AE divides ABCD into two parts.
The areas of the two parts are in the ratio 1 : 6.
Find the ratio of the lengths AB : DC.

hint: AD is on the left side and BC is on the right side[/quote]
What's a trapezium?
Is it the british way to say trapezoid?[/quote]

Yes, it's the British way.

[quote="Milkyway989558"]Here is the diagram[/quote]

I still don't see a diagram...Do you?

sorry im awfull at making diagrams but what ever i think youll get the idea.

[hide="answer"]

Answer: 2:5 or 2\5

Solution

So, I first just found the area of the triangle which if we say that the hight of the "trapezium"(im assuming its a trepeziod) is h, we can assign a random number to h because they are asking for ratis. then the area of the traingle=1/2(h=4) and  the area of the triangle is then 2.  so the area of the trapezoid the trapezoid is 7 times that. so 14=1/2(5+$ b_2$)4  so $ b_2 \equal{} 2$ therefore the ratio of the side lengths is 2:5 or $ \boxed{\frac{2}{5}}$.

Because we assigned a hight we need to check that assigning a different hight doesnt affect the outcome.  We can do the same thing with the hight of 2.  so the area of the of the triangle is 1 and the area of the trapeziod is 7.  so we have 1/2(5 + $ b_2$)2 so we have $ b_2 \equal{} 2$ and therefore AB:DC = 2:5 or $ \boxed{\frac{2}{5}}$[/hide]

That diagram isn't correct.
And sanghnlee, tip #142: Sarcasm does not travel over the internet.

whats wrong with it.  If its the fact that the triangle formed by ADE is not a right triangle, im sorry im bad at drawing diagrams! but if it is something else i would love to know.  And despite my incorrect diagram, is my solution incorrect as well?

and yes i agree that sarcasm does not get recieved over the internet, at first i thought that you thought my awful drawing was actually good!

Im not sure is this correct... just did some rough sketch
AB/DC=5/2

[quote="ZhangPeijin"]Ok 1, you didn't label where E is and two, E is on BC, not AB.
So yeah, all and all, failed diagram.
And to the british guy, you know when you post a reply, on the top it says Message Body and Add an Attachment, and the Message Body is bold, click on Add an Attachment and attack your attachment.[/quote]

the labeling of E is not necessary.  as long as you can identify it as that point, it's ok.  It's just [u]better[/u] to have it labeled.  But E isn't AB either. :wink:

[hide="quick solution?"]
Basically, if you draw EF parallel to CD, you get another trapezoid similar to the first one.  Do some stuff letting $ AB\equal{}x$, $ CD\equal{}y$, and the height of the trapezoid be $ h$, you get $ \frac{x}{y}\equal{}\frac{1}{6}$.
[/hide]

[quote="Nightmare124"]i dont see a diagram -.-[/quote]
Before posting, read ALL the other posts.


[quote="Franz Joseph"]Given $f(x)=x^{3}-x-1$ and $\alpha, \beta, \gamma$ are the roots of $f(x)$. Define $S_{n}= \alpha^{n}+\beta^{n}+\gamma^{n}$

Find $S_{4}, S_{8}, S_{12}, S_{16}$.

[hide="hint"] Find a $f(y)$ such that its roots are $\alpha^{2}, \beta^{2}, \gamma^{2}$[/hide][/quote]

[hide="Give-away hint"]
Newton Sums...
[/hide]

[quote="Franz Joseph"]Given $f(x)=x^{3}-x-1$ and $\alpha, \beta, \gamma$ are the roots of $f(x)$. Define $S_{n}= \alpha^{n}+\beta^{n}+\gamma^{n}$

Find $S_{4}, S_{8}, S_{12}, S_{16}$.

[hide="hint"] Find a $f(y)$ such that its roots are $\alpha^{2}, \beta^{2}, \gamma^{2}$[/hide][/quote]

[hide]Hint

Show that $S_{n}=S_{n-2}+3$, where $n\ge3$[/hide]

Newton's Sum? Interesting, I think I might try that method.

[quote="Elemennop"]isn't newton's sum not the overwhelmingly OBVIOUS method????[/quote]
if you know it, probably yea

[url]http://www.artofproblemsolving.com/Wiki/index.php/Newton_sums[/url]

[quote="Elemennop"]isn't newton's sum not the overwhelmingly OBVIOUS method????[/quote]

If you don't know it, then no.  :|


Find the ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle to the perimeter of an equilateral triangle inscribed in the same circle.

[hide]
$\frac {2\sqrt3}{3\sqrt3} =$

$\frac 23$

[/hide]


[quote="rrusczyk"][quote="Karth"]I constantly get around 120 - 130 on the AMC 12 and about 3 or 4 on the AIME. I really want to boost my scores, so which should I take? Also, I saw that the AIME Problem Series does mainly problems only... what about the AMC 12 Problem Series?[/quote]

The AMC 12 Problem Series focuses on the harder problems on the AMC 12, and the AIME Problem Series aims at the mid-hard problems on the AIME.  If you are working steadily through the summer, you'll probably be in place for the AIME class for the fall.[/quote]

I was actually considering taking the class this summer since I have somewhat of a heavy courseload in the fall. But for now, should I be focusing more on the AMC 12 or the AIME?

If you're scoring in the 120-130 range, I'd spend more time on the AIME.  That will help you with the last 5 problems on the AMC anyway.

[quote="Pakman2012"]I took AMC 12 Problem Series, I felt like I gained a greater confidence in my abilities for AMC.  But if you're in the 120-130 range, I do suggest AIME Problem Series, because I imagine it has a similar effect.[/quote]

*Taking mike from Saiem*  :D

I was in class with him on AMC 12 Problem Series, and just like any other AoPS class, this is an excellent class. On my first AMC-10 as freshman, I got 103. After taking AMC 12 Problem Series, I got 111.5 on AMC-12 as a sophomore and got 3 on AIME. I could've got a lot more (I got #1,3, and 9 ---> obviously, since I got 9, it means that I do have skills to do well) but I was so nervous that I couldn't think right (or draw). Choosing class really depends what you want.. Do you want to have high AMC score? Or do you want to have two digit AIME score? It's possible that you can achieve both but it's pretty tough to do both. AMC concentrates on speed while AIME is really on skills. If you're a fast problem solver and want to get around 140-150ish, take AMC 12 Problem Series and while taking the class, see how fast you can see the solutions to the problems. I recognize most of the problems but it was still worthwhile to review them. If you're a slow problem solver like myself, pay attention to AIME Problem Series and take that one. It'll be MUCH better for you that way because AIME Problem Series, as Mr. Rusczyk remarked, covers mid to hard AIME problems. It's not just "easy" AIME problems so yeah.

Think about it.  :)


Let $ S$ be the set of points $ (a,b)$ in the coordinate plane, where each of $ a$ and $ b$ may be $ \minus{} 1$, $ 0$, or $ 1$. How many distinct lines pass through at least two members of $ S$?

$ \textbf{(A)}\  8 \qquad \textbf{(B)}\  20 \qquad \textbf{(C)}\  24 \qquad \textbf{(D)}\  27\qquad \textbf{(E)}\ 36$

[hide="Solution"]I got E.

First, we know that nine points may be formed. Thus, there are $\binom{9}{2}=36$ points going between them.[/hide]

But you didn't take into account that there are collinear points among those 9. Hence, not all of the 36 lines are distinct.

[hide]If I count correctly, there are three with the slope 0; three with the slope $\infty$; three with the slope 1; three with the slope -1; two with the slope 2; two with the slope 1/2; two with the slope -2; two with the slope -1/2 - for the total of 20.[/hide]

[hide]Initially we count ${9\choose2}$, and we run into the problem, like Farenhajt said, about overcounting.

Consider the set of collinear points (1,0), (0,0), and (-1,0). By doing ${9\choose2}$, we three lines. However, these three lines can be combined into one line connecting all three dots. Every time there are three collinear points, we subtract 2 from ${9\choose2}$. There are 8 of these lines, so we have 36-16=20 lines. [/hide]

[hide]B. Simpler method is just counting =p
Of course, that would be impractical if their were many more points than these.[/hide]


I wasn't sure whether to post this here or in the AMC forum, but:

In the Resources section, the 2000 AMC-10 test (Olympiad Resources --> USA --> AMC 10 --> 2000) is incorrect; questions #6 and #9 are identical. The correct statement of question 6 is:

The Fibonacci sequence 1,1,2,3,5,8,13,21... starts with two 1s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?

(a) 0
(b) 4
(c) 6
(d) 7
(e) 9


Prove that the sine of any rational multiple of $ \pi$ is never transcendental.

Also, tell me if this problem is too easy for this forum, as I get the impression that it is.

[url=http://www.artofproblemsolving.com/Forum/weblog_entry.php?t=246555]How to overkill this problem[/url].


Why does showing that a subset of a vector space is nonempty and closed sufficient to show that it is a subspace?

Also, why is the zero vector in every subspace of a vector space?

Look at the axioms for a vector space. If we take a subset of any vector space, these axioms are automatically satisfied as long as the operations keep us inside the subset. If $ x\plus{}y\equal{}y\plus{}x$ for all $ x,y\in V$, then $ x\plus{}y\equal{}y\plus{}x$ for all $ x,y\in U\subset V$. The only apparent exception is the existence of a zero element- but we can always just multiply by zero to get there.


Another nice problem:

If $ b^2 \equal{} \frac{a}{4}$ and $ \frac{a}{2} \equal{} 4b$ and $ b \neq 0$, then what does $ a$ equal?

[hide]Divide the second equation by $ 2$.  It then becomes $ \frac {a}{4} \equal{} 2b$.  By substitution, we can rewrite this as $ b^2 \equal{} 2b$.  Subtract $ 2b$ from both sides to get $ b^2 \minus{} 2b \equal{} 0$.  Factoring the left side, we get $ b(b \minus{} 2) \equal{} 0$.  This means that either $ b \equal{} 0$ or $ b \minus{} 2 \equal{} 0$.  Since the problem states that $ b$ does not equal $ 0$, $ b$ must equal $ 2$.  Plugging this value back into the first equation, we get $ 2^2 \equal{} \frac {a}{4}$.  Multiplying both sides by $ 4$, we find $ a$ to be $ 16$.  Using the second equation to verify, we see that $ \frac {16}{2}$ does indeed equal $ 4\cdot2$.  Thus, the answer is $ 16$.[/hide]


This is a question from AMC 12.

How many perfect squares are divisors of the product 1! x 2! x 3! ... x 9! ?

I know it is 1^9 x 2^8 x 3^7 ... x 9^1

But how do I find the number of perfect squares?

I understand everything except for the part where you did 16 x 7 x 3 x 3.

it is like the divisor formula, there are 16 possibilities for the power of 4, ($4^0$,$4^1$,...)
and there are 7 possibilities for the powers of 9, similarily for the other ones, and since they are independant, we can multiply them to find the total number

Sorry, I still don't get it. Wouldn't it be 16 + 7+ 3 + 3?

However, the answer is 672.

http://www.unl.edu/amc/b-registration/b1-archive/2003-2004/04bro1012/04bro1012html/04bro1012-que.html
Question #23

you want all the possibilities of the squares, we are saying if you mulitplied them, you would get unique perfect squares, i.e. 4^0*9^0, 4^0*9^1...4^0*9^6, then you would repeat that with 4^1,4^2,etc, you should review the fundamental counting principle if you still dont get how this gives all the possibilities

Ahhh, ok, thanks Altheman. I didn't realize the product of 2 squares is always a square.

However, if this is the case, why is the answer according to AMC 12, 672?

Perhaps 1 us not counted as a solution?

I will try:

$1^9 \cdot 2^8 \cdot 3^7 \cdot 4^6 \cdot 5^5 \cdot 6^4 \cdot 7^3 \cdot 8^2 \cdot 9^1$

$2^8 \cdot 2^{12} \cdot 2^4 \cdot 2^6 \cdot 3^7 \cdot 3^4 \cdot 3^1 \cdot 5^5 \cdot 7^3$

$2^{30} \cdot 3^{12} \cdot 5^5 \cdot 7^3$

now making them powers of perfect squares.

$4^{15} \cdot 9^6 \cdot 25^2 \cdot 49^1 \cdot 3 \cdot 5 \cdot 7$

So using the divisor formula:

$16 \cdot 7 \cdot 3 \cdot 2 = \boxed{672}$

I get the right answer....

The previous solver's error was that he changed $7^3$ into $49^2 \cdot 7$ which is $7^5$


inequalities
LaTeX
inequalities proposed

1) Given $n$ real numbers $x_1,x_2,\ldots,x_n$ such that $0\leq x_i\leq 1$, $\forall i=\overline{1,n}$, prove that
\[ \frac{3n-n^2}2\leq x_1+x_2+\cdots+x_n-x_1x_2-x_1x_3-\cdots-x_{n-1}x_n\leq1.  \]

2) Given $n$ real numbers $x_1,x_2,\ldots,x_n$ such that $0\leq x_i\leq 1$, $\forall i=\overline{1,n}$, prove that
\[ x_1+x_2+\cdots+x_n-x_1x_2-x_2x_3-\cdots-x_nx_1\leq \left [\frac n2\right].  \]

3) Given $n$ real numbers $x_1,x_2,\ldots,x_n$ such that $0\leq x_i\leq 1$, $\forall i=\overline{1,n}$, prove that
\[ \frac1n\sum_{i=1}^nx_i^2-\left(\frac1n\sum_{i=1}^nx_i\right)^2\leq \begin{cases} \frac14 \;\; n \;\;\text{is odd} \\ \frac14 -\frac1{4n^2}\;\; n \;\;\text{is even} \end{cases}.  \]
4) (Han Ngoc Duc) Let $n\geq 2$, and $x_1,x_2,\ldots, x_n$ be positive real numbers in the interval $[a,b]$, $a>0$, find the maximum value of
\[ Q=\sum_{1\leq i<j\leq n}(x_i-x_j)^2. \]

the second:

denote the left side of the inequality by S, it is lineair in each  of his variables, it's enough then to check the inequality when x_i belongs to {1,0}.

If exactly k of the x_i are equal to 0,and the others are equal to 1,then S=<n-k, and since the sum x_1x_2+.....x_nx_1 is

at least n-2k, S is less than or equal to n-k-(n-2k)=k. So the maximum of S is min (k,n-k) wich is at most [n/2].

it follows than for n even, equality holds when (x_1,...x_n)=(0,1,0,1....0) or (1,0,1,0,.....1).For n odd ,equality holds when
all pairs (x_i,x_i+1) consist of a zero and a one ,exept for one pair which consists of two ones.

the first is wrong when n=2 and (x_1,x_2)=(0,0) :D

For these ones, we need two lemmas:

Lemma 1. For $f(x)=ax+b$, we have for $x\in [\alpha,\beta]$
\[ \min\{f(\alpha),f(\beta)\}\leq f(x)\leq \max\{f(\alpha),f(\beta).\}  \]

Lemma 2. For $f(x)=ax^2+bx+c$, $a\geq 0$, we have
\[ f(x)\leq \max\{f(\alpha), f(\beta)\},  \]
with $x\in [\alpha,\beta]$.

Altusi, please learn to use LATEX so that your post is more readable.


Why does $ 1\minus{}\sin{2x}\equal{}(\sin x\minus{}\cos x)^2$?

$ 1\minus{}\sin{2x}\equal{}1\minus{}2\sin{x}\cos{x}\equal{}\cos^2{x}\plus{}\sin^2{x}\minus{}2\sin{x}\cos{x}\equal{}(\sin{x}\minus{}\cos{x})^2$.

Try expanding $ (\sin{x}\minus{}\cos{x})^2$ and know your trig identities ($ \sin{2x}\equal{}2\sin{x}\cos{x}$, $ \sin^2{x}\plus{}\cos^2{x}\equal{}1$).


trigonometry
trig identities
Law of Cosines

I there any simple solution ?

I couldn't  find.

[quote="CircleSquared"][hide="I would"]
trig bash it with a bunch of variables, law of Cosines twice, and an equation... but that's just me.  I would explain in further detail but I need more labeled points.
[/hide][/quote]

Or coordinates... or stewart's/angle bisector theorem :wink:

It's pretty useless to post a ugly solution without actually giving it.

[quote="CircleSquared"]I'm just giving him an idea of how to start...  :ninja:[/quote]

I could'n understand.

Any link about?

Try labeling the segments further (namely split up the bottom segment and label the other trisector y), and use Law of Cosines to figure out the angles.  Then use Law of Cosines to figure out the bottom segment in terms of x and y... that's all I can say.


quadratics
trigonometry
algebra
Trigonometric Equations
IMO
IMO 1959

Let $a,b,c$ be real numbers. Consider the quadratic equation in $\cos{x}$ \[ a \cos^2{x}+b \cos{x}+c=0.  \] Using the numbers $a,b,c$ form a quadratic equation in $\cos{2x}$ whose roots are the same as those of the original equation. Compare the equation in $\cos{x}$ and $\cos{2x}$ for $a=4$, $b=2$, $c=-1$.

In the given equation $a \cos{x}^2+b \cos{x}+c=0$, we substitute
\[ \cos{x}=\pm \sqrt{\dfrac{1+\cos{2x}}{2} }\]
obtaining
\[ a \left(\dfrac{1+\cos{2x}}{2}\right) \pm b \sqrt{\dfrac{1+\cos{2x}}{2}+c=0 }\]
which simplifies to
\[ (1) a^2 \cos^2{2x}+(2a^2+4ac-2b^2)\cos{2x}+a^2+4ac+4c^2-2b^2=0 \]
For $a=4, b=2, c=-1$, the given equation is
\[ 4\cos^2{x}+2\cos{x}-1=0 \]
and the left member of $(1)$ is $4(4\cos^2{2x}+2\cos{2x}-1)$, so that the quadratic equation for $\cos{2x}$ has coefficients proportional to those of the given equation.

hello, multiplying the equation by $ 4(a\cos(x)^2\minus{}b\cos(x)\plus{}c)$ we obtain $ 4a^2\cos(x)^4\plus{}2(4ac\minus{}2b^2)\cos(x)^2\plus{}4c^2\equal{}0$. Plugging in this term $ 2\cos(x)^2\equal{}1\plus{}\cos(2x)$ we get after some algebraic manipulation
$ a^2\cos(2x)^2\plus{}(2a^2\plus{}4ac\minus{}2b^2)\cos(2x)\plus{}a^2\plus{}4ac\minus{}2b^2\plus{}4c^2\equal{}0$
For $ a\equal{}4,b\equal{}2$ and $ c\equal{}\minus{}1$ we have $ 4\cos(x)^2\plus{}2\cos(x)\minus{}1\equal{}0$ and $ 16\cos(2x)^2\plus{}8\cos(2x)\minus{}4\equal{}0$ which implies $ 4\cos(2x)^2\plus{}2\cos(2x)\minus{}1\equal{}0$.
Sonnhard.

I don't get the simplification part, could someone do it step by step?

[quote=qntty]By $ \cos{x^2}$, do they mean $ \cos^2{x}$?[/quote]

($ \cos{x}$)$^2$=$ \cos^2{x}$

It's important to notice:
$ \cos({x^2})$ does NOT equal  $ \cos^2{x}$, but it is a common typing mistake, like in this case.

[color=#f00]The official IMO website has the problem transcribed as $\color{red}\cos^2 x$ as you suggest, so I've edited it.  Thanks! ~dj[/color]

Clearly, $\cos x = \sqrt{ \frac{\cos(2x)+1}{2}}$. Therefore, $$a\left(\frac{\cos (2x) +1}{2} \right) + b\sqrt{\frac{\cos (2x) + 1}{2}} + c = 0 \implies b^2 \left(\frac{\cos (2x) + 1}{2}\right) = \left( c + a\left( \frac{\cos (2x) + 1}{2}\right) \right)^2 \implies 0 = \cos ^2 (2x) \cdot a + \cos(2x)\left( -2b^2+2a^2+4ac\right) + 4ac + a^2 + 4c^2 - 2b^2$$Plugging in $a=4,b=2, $ and $c=-1$, we have that $\cos x = \frac{-2 \pm \sqrt{4-4(4)(-1)}}{8} = \frac{-1 \pm \sqrt{5}}{4}$. Plugging this into our equation for $\cos (2x)$, we have that $\cos(2x) = \frac{-1 - \sqrt{5}}{4}$.

Let $\cos x=r,s$ be the roots of $a\cos^2x+b\cos x+c$. Since $\cos2x=2\cos^2x-1$ by the double-angle formula, we seek the quadratic with roots $2r^2-1$ and $2s^2-1$. Furthermore, because $r+s=-\tfrac{b}{a}$ and $rs=\tfrac{c}{a}$ by Vieta's Formulas, $r^2+s^2=(r+s)^2-2rs=\tfrac{b^2}{a^2}-\tfrac{2c}{a}=\tfrac{b^2-2ac}{a}$. Hence,
$$(2r^2-1)+(2s^2-1)=2(r^2+s^2)-2=\frac{2b^2-4ac}{a^2}-2=\tfrac{2b^2-4ac-a^2}{2}$$
and
$$(2r^2-1)(2s^2-1)=4r^2s^2-2(r^2+s^2)+1=\frac{4c^2}{a^2}-\frac{2b^2-4ac}{a^2}+1=\frac{a^2-2b^2+4c^2+4ac}{a^2},$$
so $2r^2-1$ and $2s^2-1$ are the roots of the quadratic
$$t^2+\frac{2a^2-2b^2+4ac}{a^2}t+\frac{a^2-2b^2+4c^2-4ac}{a^2}=0,$$
or, after replacing $t$ with $\cos2x$ and scaling by $a^2$,
$$\boxed{a^2\cos^22x+(2a^2-2b^2+4ac)\cos2x+(a^2-2b^2+4c^2+4ac)=0}.$$
When $(a,b,c)=(4,2,-1)$, we have
$$a\cos^2x+b\cos x+c=4\cos^2x+2\cos x-1=0$$
and
$$4^2\cos^22x+(2\cdot4^2-2\cdot2^2+4^2)\cos2x+(4^2-2\cdot2^2+4\cdot1^2-4^2)=16\cos^22x+8\cos2x-4=0,$$
which do indeed have the same roots in $\cos x$ and $\cos2x$, respectively, as each is a scalar multiple of the other. $\blacksquare$

To be honest, I am slightly confused about the problem:
In general, the quadratic equation $a\cos(x)^2+b\cos(x)+c=0$ will have two solutions for $\cos(x)$ and hence two solutions $x \in [0,\pi]$.
However, the corresponding quadratic equation in $\cos(2x)$ will be a degree-4 equation in $\cos(x)$ and hence will in general have four solutions for $\cos(x)$ and hence four solutions $x \in [0,\pi]$.
So I fail to see how it could be even possible to produce such an equation for general $a,b,c$ "whose roots are [b]the same[/b] as those of the original equation" and I doubt that the solutions presented in this tread as well as the official one work.  :maybe:
(What they produce, in my opinion, is a equation whose solutions [b]include [/b]those of the original equation.)
Can anyone clear/confirm my doubt?


MATHCOUNTS
AMC
USA(J)MO
USAMO
summer program
Mathcamp

I saw that Massachusetts placed 4th and Xiaoyu placed 11th but I know nothing else. If the MA team or Mr. Frost could fill in the scores, team score, and ranking, that would be helpful. Also, how was the test? Hard? Easy? And also, how was Denver and the trip so far?

EDIT: Oh yeah, what calculators did you guys get?

I know the top score was a 43, this is from a person i talked to that attended nats that I forgot who.

Well if the high score was 43, the test was probably relatively easy. According to Mr. Frost, rumor is that Alan came in 13th.

Lyndon Ji let me down.  :(

Well, let me rephrase that, I think the top score is a 43. What Alan was 13th...... :( and  :) he would have pwned in countdown(at least fun to watch). To add with you, so did Nathan Salo.

Are you saying Nathan Salo let me down? I didn't really care about Nathan Salo...

Lol,
My predictions:
MA would get 5th
Xiaoyu would be in the top 12 places
Alan would not make the top 12
Winner of the written round is Bobby Shen
Winner of the competition is Bobby Shen
Results:
MA got 4th
Check
Check
Check
Bobby got overconfident after that 3rd question and lost.
:|
BTW guys, what was the white shirts about?
And how did xiaoyu change into a red shirt that quickly?
:huh:

Hey Guiys,

We won't know specifics until tomorrow, but Xiaoyu is pretty sure he got 36 and Alan got 35 so the cutoffs were about the same as last year.  Xiaoyu(24/12) Alan (21/7) is what we are thinking with Jason and Yongyi in the low 30s.  Team round was a lot harder this year.  Only one 10 that we know of and three 9s(MA being one of them and again due to a misread :( oh well).

Last year MA missed 2nd by a team round, I bet tomorrow I am reporting we were even closer this year.

The white shirts have print front and back.  All kids in countdown had a red shirt thrown over their heads as they walk on stage

There were lots of 7th graders in top 12 and a 6th grader won it, so Peijin and Hao you guys better get to work :lol:

Great job to all the MA kids!!

Frost

PS the calculators are the new TI Nspire which are pretty cool actually

Guess the calculators caught everyone by surprise. Everyone thought it would be the TI-84 Silver Edition. :rotfl:

What the heck...I think there is something up with MA. Since 2003 I believe, one of two things has happened:

a) Carl has messed up the team thus putting them in 6th;
b) The team has come in 4th;

both of which are mutually exclusive.

Official news flash : Alan placed 13th
MA was .25 behind 2nd and 3rd(tied) place and held the tie breaker over 3rd.

Great job everyone.

Alan Chou has officially surpassed Noah as being the king of missing important things (or scoring appropriately to miss important things) while in middle school.

(edit) for the second year in a row, the highest placing person who gets no trophy is from clarke. there is something up with this. but this year was closer, as one single question would have given alan two trophies, while i needed two separate questions.

That is so bad for Alan. First, states, he misses cutoff for nats, then, at nats alan misses cutoff for cd. Wow, I say that 2008 is a bad year for alan. Hmmmm...didn't they predict that gas prices will raise to like a lot in 2008?

like didn't i JUST SAY THAT gosh hopefully my comments don't need elaboration...but if they do he also missed the USAMO by half of an "index" point. yes, "index," not index.

also we all know that gas prices are rising, wtf

like seriously i see no point in the above post.

Hao was trying to prove that 2008 is an unlucky year.
:|
Yay, got accepted to MP...
3500 for 4 weeks?
thats like 1k for a week.
So like 130 or 140 per day...
Let's see...
Say meals are 10bucks.
30bucks for meals...
~100 for classes and dorms...
That's not horrible...

eh what why is it only 3500 now

mathcamp apparently has lots of apps so it's taking longer this year...

bleargh i probably got rejected

(edit) you guys better be partying or you'll go down as probably the least social mathcounts team from MA ever...i have to admit that having no white kids puts you in not-so-good shape to start.

What is this? Is Lexington invading Mathcamp?

Does anyone here know how much MA missed 1st place?

Bobby said that Texas was 3.75 ahead of 2nd, and frost13 says that MA was .25 behind 2nd, so that would make 4 on the nose.

Oh nvm, MP is 4500...
1125 per week
~160 per day..
So 130 for beds and classes...
Still not bad.
How did Yongyi and Jason do?

According to Jason's signature he came in 43rd place with a 31

target was easy, messed up #8 though

[quote="CatalystofNostalgia"]Alan Chou has officially surpassed Noah as being the king of missing important things (or scoring appropriately to miss important things) while in middle school.[/quote]

hmm, I figured.  who's next?

Yongyi... um, I think I'll let him tell you guys

it also seems I was lucky at states to get unlucky at nats :huh:

arg, maybe I'll start trying to do actual math now, except I suck too much to do that

It has never occured to me that the "Home » National & Local Communities » North America » Massachusetts" forum actually existed.

Official results indicate that I got somewhere between 27 and 27, but rumor has it that I got a 31. I'll say I got a 27 to prevent Alan from jumping on me (but I really got a 31).

Thus placing me... not in the top 40.


floor function
induction
function
algebra proposed
algebra

A sequence {$ a_n$) satisfies  $ a_1$ =1 and   $ a_{n+1}$ =$ \frac{a_n}_{n}$ +$ \frac{n}{a_n}$  ($ n$$ \geq$ 1)

Prove that $ \lfloor$ $ a_n^2$ $ \rfloor$ =$ n$when $ n$ $ \geq$ 4

You are wrong. Obviosly $ a_n\equal{}2\sqrt n(1 \plus{}o(1)), a_n^2\equal{}4n(1\plus{}o(1)).$

show that $ \sqrt{n}<\equal{}a_{n}<\equal{}\frac{n}{\sqrt{n\minus{}1}}$ for all $ n>\equal{}3$ by induction (use the function $ f(x)\equal{}x/n\plus{}n/x$)
and the rest is easy

$ a^2_{n \plus{} 1} \equal{} \frac {a^2_n}{n^2} \plus{} \frac {n^2}{a^2_n} \plus{} 2$ and prove by induction that $ n \plus{} \frac 2n < a^2_n < n \plus{} 1$ for all $ n \geq 4$
Could you prove that $ a_n^2\minus{}n$ is convergent and find its limit  :?:  :D
 :cool:

Let $ b_n \equal{} a_n^2$, then $ b_{n \plus{} 1} \equal{} \frac {n^2}{b_n} \plus{} 2 \plus{} \frac {b_n}{n^2}, b_4 \equal{} 5 \minus{} \frac {11}{36}$.
Obviosly $ f_n(x) \equal{} \frac {n^2}{x} \plus{} 2 \minus{} \frac {x}{n^2}$ decrease, when $ 0 < x < n^2$.
Let $ n < y_n < b_n < z_n < n \plus{} 1$, then if $ y_{n \plus{} 1} < f(z_n) < f(y_n) < z_{n \plus{} 1}$, we get $ y_n < b_n < z_n \ \forall n\ge 4.$
I chek, that $ y_n \equal{} n \plus{} \frac {3}{n \plus{} 1},z_n \equal{} n \plus{} 1 \minus{} \frac {1}{n}$ work for $ n\ge 4$.
For $ n \equal{} 4$ $ 4 \plus{} \frac {3}{5} < 5 \minus{} \frac {11}{36} < 5 \minus{} \frac {1}{4}$ and $ f(n \plus{} 1 \minus{} \frac {1}{n}) > y_{n \plus{} 1} \equal{} n \plus{} 1 \plus{} \frac {3}{n \plus{} 2}$ and
$ f(n \plus{} \frac {3}{n \plus{} 1}) < z_{n \plus{} 1} \equal{} n \plus{} 2 \minus{} \frac {1}{n \plus{} 2}.$

[quote="TTsphn"]$ x_n \equal{} a_n^2 \minus{} n$
Then:
$ x_{n \plus{} 1} \plus{} n \plus{} 1 \equal{} \frac {n^2}{x_n \plus{} n} \plus{} \frac {x_n \plus{} n}{n^2} \plus{} 2$
$ x_{n \plus{} 1} \equal{} \frac {n^2}{x_n \plus{} n} \plus{} \frac {x_n \plus{} n}{n^2} \plus{} 1 \minus{} n$
Easy to check $ x_{n \plus{} 1} < x_n$ and $ \{x_n\}$  is bound so it converge.[/quote]

How do you check it  :?:   $ x_4 \equal{} \frac {25}{36}\equal{}0.694444... , x_5 \equal{}0.70168...$  :!:
 :cool:


1. Show that the angle bisectors of a triangle will intersect at one point, that is they are concurrent.

2. Show that the medians of a triangle will intersect at one point, that is they are concurrent.

3. Show that the heights of a triangle will intersect at one point, that is they are concurrent.

[hide="1"]
Let the angle bisectors $AU$ and $BV$ meet at $I$, where $U$ lies on $BC$ and $V$ on $AC$.
Since $I$ lies on the bisector $AU$, the perpendiculars from $I$ to $AC$ (let this meet $AC$ at $Y$) and $AB$ (let this meet $AB$ at $Z$)  are the same length.
Similarly, since $I$ lies on the bisector $BV$, the perpendiculars from $I$ to $AB$ (this is $IZ$) and $BC$(let this meet $BC$ at $X$) are the same length.
So $IY = IX$.

Triangles $CIX$ and $CIY$ are congruent, since they each have a right angle, $IX=IY$ and $CI$ is a common length. Hence $CI$ is the bisector of angle $ACB$- and all three bisectors meet at $I$.[/hide]

2 can be proved using [url=http://www.artofproblemsolving.com/Wiki/index.php/Ceva%27s_Theorem]Ceva's Theorem[/url]

#3. First, show perpendicular bisectors are concurrent:

If you circumscribe the triangle (let's call it $\triangle ABC$),
the perpendicular bisectors of $AB, BC, CA$ all go through the center. (Converse of a radius always being a perp. bisector of a given chord).

Therefore, the perpendicular bisectors are concurrent.

Now, to prove it for altitudes, draw $\triangle DEF$ so that
$\triangle DEF \sim \triangle ABC$ and $A$ bisects $EF$, $B$ bisects $DF$, and $C$ bisects $DE$.

In other words, let $\triangle ABC$ be the medial triangle of
$\triangle DEF$.

Then, the altitudes of $\triangle ABC$ are the perpendicular bisectors of $\triangle DEF$, so they are concurrent.

If you don't follow, draw it out. I don't explain very well... :blush:

I think all of them can be proved by Ceva  :D .


Match the words above the line with the words below the line in a logical fashion and explain your reasoning.

bed, tooth, gold, bread, arm
------------------------------------
or, bras, lit, dent, pain

It's more about words than math, but then again, it's "puzzles and brainteasers"!! :D

[quote="MysticTerminator"]french![/quote]
 :o  I'm a Canadian and I didn't know this!!! Shame on me. :(

[hide="So It Becomes Obvious"]bed tooth gold bread arm matches up with lit dent or pain bras[/hide]

Interesting puzzle; the "bras" gave it away to me.

Hmm, I'm not seeing it. (what is the question even asking you to do?)

Match the words together. "Bras" is French for "arm," "lit" is French for "bed," "dent" is French for "tooth," "pain" is French for "bread," and so I deduced that "or" is French for "gold" (which I didn't know before).


Show that

$\lim_{n \to \infty} \left( \frac{ (n+1)^{n+1} }{n^n} - \frac{ n^n}{(n-1)^{n-1} } \right) = e$

[hide="Hint"]Write $a_{n}= \frac{ (n+1)^{n}}{n^n}$  [/hide]

Edit :I am not sure if this helps finally..... :?

Yes, it is easy. Consider the sequence
\[ x_n = \left(\frac{n+1}{n}\right)^n = \left(1+\frac{1}{n}\right)^n  \]
We know $\lim_{n \to \infty} x_n = e$ and $x_n - x_{n-1} < \frac{2}{n^2}$, hence
\[ \lim_{n \to \infty} \left( \frac{ (n+1)^{n+1} }{n^n} - \frac{ n^n}{(n-1)^{n-1} } \right) = \lim_{n \to \infty} \left((n+1) * x_n - n * x_{n-1}\right) \] \[ = \lim_{n \to \infty} \left(n* ( x_n - x_{n-1}) + x_n\right) = e  \]


Determine all natural numbers $n$ and $k>1$ such that $k$ divides each of the numbers $C(n,1), C(n,2),...,C(n,n-1)$.

I've solved (almost) the same  problem recently here is the link:
http://www.mathlinks.ro/Forum/viewtopic.php?t=133601


HCSSiM
geometry
geometric transformation
reflection
ARML
videos

What are some productive things an unproductive person can do during the summer? I didn't sign up for any academic camps and my school doesn't offer any credit courses, so what can I do? Or should be doing? This is because I pretty much wasted all of my last summer break.

Go study subjects on your own. That usually works.

Goof off that works too.

It might help if you tell us what your interests are, how far long the time line you are, etc.

Until then, generally, it never hurts to invest in your interests, and it never hurts to invest in your health.

Come on, isn't summer a time for fun? I don't see how summer should be viewed as an opportunity for more studying. You've put in the hard work over the school year, and now it's time to relax and pursue whatever you're interested in. If studying interests you, then do some, but if you feel like you've "wasted" a summer by not forcing yourself to study, you might want to take a step back rethink some things.

One of my teachers always says, "Be a kid! You're only a kid once. When you get older you have *a lot* of responsibility." Granted, I'm not in a position to back up this statement, but I really do think it makes sense. :)

Some math programs will still accept applications.  In particular, I think I applied to HCSSiM a few weeks before this time last year, and I had a great time there.  But if you want to do something like this, you should apply as soon as possible.  As in, waste no time.  Start now.  If you would like to go to such a program, that is.

Ok then for clarification, I am a rising sophomore. So they tell me I'm too young for research and that kind of stuff. And don't worry, I plan to have lots of fun in the summer. The thing is, I want to be productive in some way, you know, actually do or be doing something. I'm not a big fan of studying or working ahead, so it doesn't necessarily have to be that. Just something that can help me academically or be able to put on a resume. I know it's kind of vague

It still depends on your interests. Things to consider include:

► Science projects, if you're into this stuff. It might not be too early to get into research, but if you find that it is, maybe your science project will culminate in research

► Plan on starting a club. Starting a club takes planning, especially if you don't want it to die. If you think of a club you'd be interested in, and that other people would probably be interested in too, go for it. Two friends and I started a chess club in our sophomore years. And it looks really impressive to start up a successful, legitimate club as a sophomore -- evaluators know the emphasis that many schools have on bureaucracy and seniority

► Get a job, if you can. Jobs weigh in a lot. Many kids looking to build a resume overlook this. It can be anything for starters. Waged work is nice because it builds discipline and responsibility. Unofficial paid jobs, such as making websites for people for a lump-sum, etc, can also be very rewarding.

► Community service. Not just to "check it off," but it is actually worth the time, depending on what you do.

► Try a new sport, or get in shape if you're not already. Yes, you can put hobbies on your resume when you apply to college! People like to see balance. It's more impressive than just joining track to "check off the sports 'requirement' ".

► Make a personal website. Make it look sharp, presentable, and reasonably professional for a personal page. Let it reflect your interests. When you write your resume, put your webpage URL on it.

► Driving. I don't know how old you are, but start looking into drivers' education and driving. The earlier you get licensed, (1) the earlier you can drive, and (2) the less your insurance costs if you don't start driving right away. And it will indirectly help you build your resume in the future.

Yeah, I've always tried to make a personal website, but I always delete it in about 2 weeks 'cause it's always turns out really, really, ugly.

If you like the kind of stuff, you can try memorizing stuff (pi, periodic table  :lol: ).

Someone told you you're too young for research? Go find something you're interested in and push the bounds of knowledge just a little bit further. My best research so far (admittedly I just got done with my junior year) was done summer between 9th and 10th grades. (It was a new discovery of reduced-redundancy representation of improper fractional base integers, which is much less scary than it sounds.)

i usually go to camps that are centered around stuff that i'm interested in, but not just playing and fun and games and all that. i go to lots of tennis camps, and this year i'm planning on going to a violin camp also, and possibly a swimming camp.

One thing that you could do is learn to program in flash of basic. Making a full game can be very rewarding, and you can sell advertising space on the game page. So you can make money from it to.

I'd play tennis (or whatever sport of your choice) with my friends all summer and maintain that signature tennis tan (white feet, sleeve/short/collar lines, maybe even white bands around wrists, lol).

I also agree that you're never too young to start researching - I did my first project at school last year in eighth grade, in my 9th grade bio class. Then I decided to do another one the following summer with a professor on cancer gene therapy. Made me proud to have my name published.

But you really shouldn't work yourself that hard during the summer.. obviously would defeat the purpose of even having a summer break :)

I'd echo that you should do whatever you really want to do over the summer, i.e. don't do something just so it looks good.  Anyway, I remember being bored many a day during certain summers and thus it's a good idea to look at what other people have done that they enjoyed.  So here's what I've done over my summers:

$ \bullet \bullet$ High School $ \bullet \bullet$

$ \bullet$  Just about every summer since around middle school I've made it one of my top priorities to train.  Back in middle school it was for baseball.  I remember training for the state competitions in Florida since I was on the all star team.  Later in middle school I started to get much more into overall conditioning (i.e. lifting weights, running, etc).  Eventually this led to me realizing how much I love running and I ended up on the Cross Country and Track teams in high schools.  So in high school, each successive summer I saw myself logging more and more miles and increasing intensity.  I went to running camp 3 times (iirc) and it was a pretty awesome experience.

$ \bullet$ Math - It wasn't until late in my sophomore year that I really started hitting the books studying and practicing for competitions.  The summer after my sophomore year I went to ARML for the first time (got quite a bit lucky since I didn't actually make the team the standard way...gauss202 met me on the AoPS forums, the first of many great things that I got from being a part of the AoPS community =D).  After that, I had tons of inspiration and went through most of AoPS V 1/2 and took a few AoPS courses.

$ \bullet$ Research - After my junior year, I did some research at the University of South Carolina.  I opted for this instead of a math camp (I think both would provide you with a really awesome experience).  This was one of my best summers ever.  I got introduced to Electrical Engineering, continued training (woke up really early to get my runs in), got to explore the campus and met/lived with a bunch of really cool people.

$ \bullet$ Video games - yep, I remember gaming a bit in high school.

$ \bullet \bullet$ College $ \bullet \bullet$

$ \bullet$ Work - Since doing research my junior summer, I've learned that I really love the feeling of working each day and being able to come home at night feeling like I've really accomplished something.  Additionally, I can really appreciate the weekends.  My jobs so far have included AoPS as well as a computer programming position.  This summer I'll be doing a research internship, AoPS, and some tutoring.

$ \bullet$ Traveling - Last summer I went down to NYC once or twice.  I also went out to Lowell, MA randomly with a friend.  This summer, potentially I'll be going to Cape Cod, Vermont, NYC, possibly Montreal and maybe even Prague.

$ \bullet$ Trying new things - It's always fun and useful to try and learn/experience new things.  Every now and then I'll go out to a new restaurant in Boston.  Once I went sailing with my friends last summer.  I played soccer for the first time my freshman year and love it now.  I've also done stuff like learning how to fix certain parts of the house I live in, learning how to cook new dishes, watching good movies, exploring music, etc etc.

So basically I personally like to specialize and diversify.  I try to find one or two things to try and specialize in and then explore a lot of other things to see what else is out there and to get a decent understanding of the bigger picture of things.

Try getting a library card at your local university. Usually it costs some money but it'll give you access to all of their math books for a year (at least, where I live). And of course, there are plenty of math books to keep you busy.

Try emailing random professors about research opportunities. You'd be surprised how often it works.  :D


#8 2001 State Sprint

Sixty-four unit cubes are placed together to create a large cube. How many cubes with integer dimensions are in the 4 x 4 x 4 cube?

[hide]100[/hide]

note that the number of 1x1x1 cubes is 64.
2x2x2 = 27
3x3x3 = 8
4x4x4 = 1
Add these up and you get 100.

for problems like this it is interestsing to note that for a 4x4x4 cube, there are ((4+1)-1)^3 1x1x1 cubes
((4+1)-2)^3 2x2x2 cubes
((4+1)-3)^3 3x3x3 cubes, and
((4+1)-4)^3 4x4x4 cubes, so the total number of cubes is

$ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} 4^3$, which is equal to $ (1 \plus{} 2 \plus{} 3 \plus{} 4)^2$, which is $ 100$

sum of $ 1^3 \plus{} 2^3 \plus{} 3^3 \plus{} .... \plus{} n^3 \equal{} (1 \plus{} 2 \plus{} 3 \plus{} ...... \plus{} n)^2 \equal{} (n^2*(n\plus{}1)^2)/4$

yeah, in general its the n'th triangular number squared
like if its a 5*5*5 cube, it would be (1+2+3+4+5)^2 = 225


Two cards are drawn from a well shuffled pack of 52 cards. Find the probability that one of them is a red card & the other is a queen.

I think that is a simple problem:
The probability of getting a Queen while drawing a card is 1/52
So let's say P(Q) = 1/52
The probability of getting red card is 1/26
So let's say P(R) = 1/26

Then we have P(Q)*P(R) = 1 / 52*26 = 1 / 1352

I think that is the correct answer. If i'm mistaken feel free to correct me

[quote]Then we have P(Q)*P(R) = 1 / 52*26 = 1 / 1352 [/quote]

That does not take into account that 2 of the queens are red.

Anubis, both of your sentences that begin "the probability of getting" are false, and the computation you made with the resulting numbers has nothing to do with the question.

Probably it is useful to consider the following sub-questions:

1) How many pairs of cards are there in a deck?

2) How many red queens are there in a deck?

3) How many non-red queens are there in a deck?

4) How many red non-queens are there in a deck?

5) Which pairs of the categories red queens, non-red queens and red non-queens (possibly with duplication of a category) result in a pair of the desired form?

6) Using 2) - 5), how many good pairs are there?

Maybe it's my misunderstanding the problem :blush:
I'll reconsider it and post a proper solution. I apologise to the original poster for confusing him!
@JBL: I do not understand point 5. Can you explain a bit?

(Just for reference (and to answer my own questions 2, 3, 4): a standard deck has 52 cards of which 26 are red and 4 are queens.  Two of the queens are red.)

Edit: fixed stupid error -- thanks jmerry.

@Anubis: there are 26 red cards, and 2 of these are queens.  That means that there are 24 red non-queens.

That's 26 red; 13 per suit, two red suits.

to succeed, your first card must be a queen or red

which happens 28 ways
2 of these ways in via a black queen, giving you 26/51 for your next pull
2 of these ways is via a red queen, giving you 27/51 ways to win on your next pull
the other 24 ways are red non queens, giving 4/51 ways to win on the next pull

$ P\equal{}\frac{2}{52}\cdot \frac{26}{51}\plus{}\frac{2}{52}\cdot \frac{27}{51} \plus{}\frac{24}{52}\cdot \frac{4}{51}\equal{}\frac{101}{1326}\equal{}.0762$


[quote="Rynalia"]Hello. Find all f : $R_{+}\rightarrow R$ such that

$f(\frac{1}{x}) = f(x+1)-f(x).$[/quote]
+) Let $x=\frac{\sqrt{5}-1}{2}\Rightarrow f(\frac{\sqrt{5}-1}{2})=0$

Is that the only solution though?  A polynomial which has a root at $\phi-1$?


Five different CD-ROM games, , , , , and , are offered as a promotion by cereal companies. One game is randomly included with each box of cereal. Determine the probability of getting all  games if  boxes are purchased.

Answer: [hide][/hide]

[quote="1=2"][hide="solution"]We are looking for the number of positive integral solutions to the equation $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 12$, divided by the number of non-negative integral solutions to the equation $ a \plus{} b \plus{} c \plus{} d \plus{} e \equal{} 12$. There are $ \binom{12 \minus{} 1}{5 \minus{} 1} \equal{} \binom{11}{4} \equal{} 330$ solutions to the first equation, and $ \binom{12 \plus{} 5 \minus{} 1}{5 \minus{} 1} \equal{} 1820$ solutions for the second. Thus there is a probability of $ \frac {33}{182}$ that we get all 5 cds.[/hide][/quote]

I understand where the 1820 comes from but where does the formula for equation 1 come from?

i understood his solution and what he did when i first looked at it, but something still bugs me. Why do you have to subtract 1 from n and k of $ \binom{n}{k}$ in both equations.

Each of a,b,c,d, and e must be positive, so you subtract 5 from (12+5-1) to just get (12-1).

@instantnoodles: This is a method called balls and urns. Search it. I think that there's also a name for the first one (with all positive), but I'm not completely sure what it is. Stars and bars?

thanks, both are explained http://en.wikipedia.org/wiki/Stars_and_bars_(probability) if anyone is interested


How many integers have an absolute value less than $ 10\pi$?

If we approximate $ \pi \equal{} 3.14$, $ 10(3.14) \equal{} 31.4$. Since only integers are involved, we wish the find the number of integers $ n$ such that $ |n| \le 31$ We know the integers from $ \minus{}31$ to $ 31$ work. So there are $ \boxed{63}$ integers that work ($ 0$ is an integer!)


abstract algebra
group theory
superior algebra
superior algebra unsolved

Show that a group of order $24$ with no element of order $6$ is isomorphic to $S_{4}$ (the permutation group on $4$ elements).

(I give my solution so maybe somebody could check it and tell me if it's right.)

Note $d$ the number of $3$-Sylows of $G$. Then $d$ divides $2^{3}$ and $d \equiv 1 [3]$, whence $d=1$ or $d=4$.
Assume $d=1$, then $G$ admits a unique $3$-Sylow $H$ which is normal in $G$. Also the action of $G$ on $H$ by
conjugation gives an homomorphism $\phi: G \rightarrow S(H)$ whose kernel $K$ has a cardinal divisible by $4$.
But then $K$ has an element of order $2$, which is absurd since $G$ has no element of order $6$. Whence
$d=4$.
Now consider the action of $G$ on the set $Syl_{3}$ of its $3$-Sylow subgroups. This gives raise to an
homomorphism $\psi: G \rightarrow S(Syl_{3})$ whose kernel $L$ is the intersection of the normalizers of the
$3$-Sylows. But the normalizer $N(S)$ of a $3$-Sylow $S$ has order $6$, hence $|L|$ divides $6$. But $L \neq 6$
since $\forall S \in Syl_{3}, S \subseteq N(S)$, and the intersection of any two $3$-Sylows is trivial; also
$|L| \neq 3$ because no $3$-Sylow of $G$ is normal, and $|L|\neq 2$ since no subgroup of $G$ with order $2$ can
be normal (otherwise taking $g$ of order $2$ in $L$, for any element $x$ with order $3$, $xgx^{-1}=g \Rightarrow o(xg)=6$, which is absurd). As a result $L$ is trivial
and $G \cong S_{4}$.


ceiling function
logarithms
floor function

Can anyone tell me how to find the number of digits in $ 2^x$?

Unless I'm mistaken, it's $ \lceil x \log 2 \rceil$

Although that calculation returns the correct answer, in general the number of digits in a number $ n$ is not $ \lceil \log n \rceil$ (this gives the wrong answer for powers of $ 10$).  Rather, it is $ \lfloor \log n \rfloor \plus{} 1$.

Oh yeah, thanks

Can anyone supply a proof of this?

[hide="proof"]

Any positive integer $ n$ can be rewritten into the form $ m(10)^{d\minus{}1}$ where $ 1\leq m<10$ and $ d$ is the number of digits. This is a variant of scientific notation.

$ n \equal{} m(10)^{d\minus{}1}$

solving for d

$ d \equal{} 1 \plus{} \log{\frac{n}{m}}$

$ d \equal{} 1 \plus{} \log{n} \minus{} \log{m}$

note that $ 0 < \log{m} < 1$ so $ \log{n} \minus{} \log{m} \equal{} \lfloor\log{n}\rfloor$ giving the desired formula

$ d \equal{} 1 \plus{} \lfloor\log{n}\rfloor$
[/hide]

Given an $ x$.
Can we find the number of $ 2$s in $ 2^x$ (the base$ 10$ expression)?

Oh...
I thought there was an easier formula that had to do with dividing the exponent by 3...
Thanks anyways.

That is essentially the same way

[hide="look at this"]

$ d \equal{} 1 \plus{} \lfloor \log n \rfloor$

in your problem $ n \equal{} 2^x$

$ d \equal{} 1 \plus{} \lfloor \log 2^x \rfloor$

$ d \equal{} 1 \plus{} \lfloor x\log 2 \rfloor$

the $ \log 2$ is approximately $ \frac{1}{3}$ so that is probably where the formula you speak of comes from
[/hide]


Does anybody know what AMC's policy is for non-American citizens/permanent residents (especially Canadians) who live in the U.S.?

I know it says on their website that you must be a citizen or permanent resident, but I'm a Canadian citizen who lives in the U.S. So I don't know if I could take the USAMO if I get a high enough score since I'm not a permanent resident...

It'll suck if I can't...cuz the AIME will be the end of the line for me (unless I fly to Canada for the CMO or something).

Canadians can take the USAMO, so I think you're fine.

The current policy is that

IF
(1) you are a student residing in the United States (or its territories and possessions) or Canada,
 (2) attending a school or program that we can identify as being
in the United States (or its territories or possessions) or Canada by virtue of its address (including APO/FPO addresses for U.S. citizens living abroad  and attending Department of Defense schools)
(3) and you have qualifying scores on the AMC{10|12} and/or AIME,

THEN we will extend an invitation to the USAMO.

I am not a lawyer, and I do not have the ability to interpret the intricacies of citizenship  and immigration law in multiple specific cases. Also, we do not have enough reliable information on any individual student in order to make a reliable detemination of citizenship in order to include or exclude them from participation in our contests.  In addition, it is probably not legal to gather that information in the first place, and in my examination of past cases, most students (particularly younger students) do not reliably understand their citizenship status to report it to us anyway

At the time of selection of the USA delegation for the IMO, we will use the rules of the IMO to help guide us in the naming of the team members.  I certainly expect these six individuals will be U.S. citizens or permanent residents of the USA as has been the case for the last several years, although the IMO rules are vague on citizenship requirements.  It is easier to inspect citizenship status on 6 people than it is on 260+, ot 12,000+ or 250,000.

Steve Dunbar
AMC Director

So basically, no APO AP or FPO AP students made AIME this year I'm guessing?  There are no qualifiers on the site.

[quote="AMCDirector"]
However, Canadians will not be invited to the MOSP, nor will they be part of the USA team for the IMO. [/quote]

So basically, all qualifying  non-American citizens/permenant residents who live in the U.S. can take the AMC, AIME, and USAMO. But they cannot go to MOP or USA IMO even if they qualify, right?


a solid ball of a light material floats on the surface of water in a container exposed to the atmosphere.if the container is now covered and the air is pumped out,then the ball will.....

[color=blue]First of all , hello fellow bharti !![/color]Where exactly do you live?trying for JEE?

Now to business.
I'm interpreting it as a ball kept in a container of height H (open from top) which (the container) is floating in water.Initially, forces are balanced on the ball:-

1)Normal reaction from base of box , gravitational force ( \[P_{atm}\ will\ not\ act\ on\ ball\ since\ ball\ is\ surrounded\ from\ all\ sides\ by\ air\)\]

2)also the box is in balance with the forces :- normal reaction of ball on floor , g, bouyant force , \[P_{atm} \] (this atmospheric pressure will cancel the same term ocuurin in the expression of pressure of water = \[P_{atm}+ \varrho gh \]

But , when then the air is pushed out of the box , consider the floor of the box :-

1)\[P_{atm} \] which was previously there is gone , rest all forces are present and the same instantaneously.There is imbalance since the extra \[P_{atm} \] part of water pressure is still there.So the box will pushed out and will perform damped oscillations and will finally come to rest at equilibrium position.

[color=red]Hope the answer is correct !![/color]

[quote="Rushil"]I'm interpreting it as a ball kept in a container of height H (open from top) which (the container) is floating in water.[/quote]
This obviously is not the correct interpretation, because the ball would not have to be of a light material and actually, the container would not have to contain any ball at all. The container, partially filled with water, sits on the ground. The solid ball of a light material floats on the watter. Then the air is pumped out of the container, say through a valve, and the valve is closed.

Let $0 < \rho_A < \rho_B < \rho_W$ be the densities of air at atmospheric pressure, the ball and watter, respectively. Let $V$ be the ball volume and $V_S$ the submerged volume of the ball. By Archimedes law of buoyancy,

$\rho_W V_S = \rho_B V,\ \ V_S = \frac{\rho_B}{\rho_W} V$ [color=white].[/color] at vacuum

$\rho_W V_S + \rho_A (V - V_S) = \rho_B V,\ \ V_S = \frac{\rho_B - \rho_A}{\rho_W - \rho_A} V$ [color=white].[/color] at atmospheric pressure

Since $0 < \rho_A < \rho_B < \rho_W$ by the problem condition,

$\rho_B \rho_A - \rho_B \rho_W < \rho_W \rho_A - \rho_W \rho_B$

$\rho_B (\rho_A - \rho_W) < \rho_W (\rho_A - \rho_B)$

$\rho_B (\rho_W - \rho_A) > \rho_W (\rho_B - \rho_A)$

$\frac{\rho_B}{\rho_W} > \frac{\rho_B - \rho_A}{\rho_W - \rho_A}$

$\frac{\rho_B}{\rho_W} V > \frac{\rho_B - \rho_A}{\rho_W - \rho_A} V$

Therefore, the submerged volume of the ball at vacuum is greater than the submerged volume of the ball at atmospheric pressure. Since the ball floats on watter at atmospheric pressure, we have $0 < \frac{\rho_B - \rho_A}{\rho_W - \rho_A} < 1$. It follows that $\frac{\rho_B}{\rho_W} < 1$ as well and the ball will still float, when air is pumped out of the container. Since $\frac{\rho_B - \rho_A}{\rho_W - \rho_A}\ V < \frac{\rho_B}{\rho_W}\ V$, the ball will float less, but not completely sink, when air is pumped out of the container.

[quote="shadysaysurspammed"]I'M AFRAID U GOT THE WHOLE INTERP. WRONG RUSHIL.YETTI GAVE THE RIGHT AND STRIAGHT FORWARD SLOUTION.
YES RUSHIL,IAM GOIN FOR JEE.I AM IN INDORE.WHERE R U?COULD YOU TELL ME SOME HANDY BOOKS FOR JEE?

i thought of another method for this question although i'm not too sure 'bout it. The atm pressure acting on the water wont be there in vaccum and hence the water level will rise and hence the body sinks a bit.is this right? :?:[/quote]

YA, I realise my mistake now. All I can say is : DAMN !!!. But if my interpretation were the question asked , is my explanation correct?That would at least tell me I can analyse a situation.

I'm in chandigarh and am prepariong for JEE as well as the US and would prefer the latter if I get the top university there.Regarding books , there a whole box full of them available in the market and there is no point my telling you in general.Tell me yours in specific subjects and I'll be able to discuss the books in specific subjects.Also have you just entered 11th or 12 th? I've just entered 12th.

regarding your solution , I believe you're not taking the complete picture involving the ball as well.You can't just analyse one part and come to the answer.You need to compare the effects of P atm on water and ball and then get the answer.That's probably why you are getting a contradicting answer.

Nice talking to you.Hoping to see more posts from you.


function
LaTeX
geometry
geometric transformation
parameterization
set theory

Given that the definition of a function f, from A to B (sets) is:

1) f is a subset of AxB...............................................................................   $ f\subset AxB$ ,and

2)for all xεΧ ,there exists a unique yεΥ SUCH that (x,y)εf .......................................... $ \forall x [x\in X\Longrightarrow\exists! y(y\in Y\wedge (x,y)\in f)]$

prove that the following  is a function from A to B,where

A = {1,2}

B = { a,b }

f = { (1,a),(2,b) }

A LaTeX suggestion: for the multiplication cross, use \times instead of x.

What part of this question is giving you trouble?  You have a universal quantifier over a finite set, which just requires you to check each element independently.

[quote="JBL"]

  , which just requires you to check each element independently.[/quote]

show me

1) Let (x,y)εf ====>

$ (x,y)\equal{}(1,a)\vee (x,y)\equal{}(2,b)\Longrightarrow (x,y)\equal{}(1,a)\vee (x,y)\equal{}(1,b)\vee (x,y)\equal{}(2,b)\vee (x,y)\equal{}(2,a)\Longrightarrow (x,y)\ A\times B\Longrightarrow f\subset A\times B$

2) Let xεΑ ====> χ=1v x=2.

For x=1

put $ y\equal{}a\Longrightarrow y\equal{}a\vee y\equal{}b\Longrightarrow y\in B$

Also $ (x,y) \equal{}(1,a)\Longrightarrow (x,y)\equal{}(1,a)\vee (x,y)\equal{}(2,b)\Longrightarrow (x,y)\in f$

Thus there exists y such that, yεΒ AND (x,y)εf

So much for the existence part.

Now for the uniqueness part.For that we must prove:

$ \forall y\forall z[( y\in B\wedge (x,y)\in f\wedge z\in B\wedge (x,z)\in f)\Longrightarrow y\equal{}z]$

So let :

$ (x,y)\in f\wedge(x,z)\in f\Longrightarrow[(x,y)\equal{}(1,a)\vee (x,y)\equal{}(2,b)]\wedge[(x,z)\equal{}(1,a)\vee(x,z)\equal{}(2,b)]$

BUT $ 1\neq 2$ (an assumption needed in the intial hypothesis for f to be a function),thus

$ 1\neq 2\Longrightarrow 1\neq 2\vee y\neq b\Longrightarrow (1,y)\neq (2,b)\Longrightarrow (x,y)\neq (2,b)$,since x=1.

In the same way $ (x,z)\neq (2,b)$.

Hence (x,y)=(1,a) and (x,z) =(1,a) and so y=z

So:

$ X\equal{}1\Longrightarrow\exists ! y( y\in B\wedge (x,y)\in f)$

inthe same way:

$ x\equal{}2\Longrightarrow\exists ! y( y\in B\wedge (x,y)\in f)$

Therefor............................$ \forall x[ x\in A\Longrightarrow \exists ! y( y\in B\wedge (x,y)\in f)]$

..................................! meaning unique....................................................................................

can anybody see any short cuts to the above proof??

[quote="JBL"]Well, aside from the obvious one that you don't accept of not writing out in formal notation what can be easily understood from a few words in natural language, no -- all of those steps need to be there.[/quote]

Thanks for checking!!

The only thing you do when you use english language instead of logical symbol-spam is that you're constantly implicitly applying lemma's relating the english language to the logical symbols. His proof is fine.

[quote="Nukular"][quote="τρικλινο"][quote="Alp"]The following one is much shorter and easier to understand.

[hide]$ A \times B \equal{} \{ (1, a), (1, b), (2, a), (2, b) \}$
Thus :
$ f \equal{} \{ (1, a), (2, b) \} \subset A \times B$

1 done.

2 :
1 ($ \in A$) has a single image through, a ($ \in B$).
2 ($ \in A$) has a single image, through b ($ \in B$).
Thus each element of A has a single image in B.

2 done.[/hide][/quote]

This a very good example .

How can one convince Alp that his proof is not actually a proof but just a summary of a proof??[/quote]

* Define a completely rigorous system of symbols that we can call a "proof language"
* Write a computer program to interpret this proof language.
* Run his proof through the program; the program will return an error.

That's why formal logic systems were invented, and that's pretty much all they are good for. So
why not use them that way?

I'm sure, for reference, that you can even get computers to come up with formal proofs
all on their own -- at least, for simple theorems like the ones you are talking about. Formal
proofs for more complex theorems are unlikely to be short enough to be human readable,
or computer generated in any reasonable amount of time. Though, you could probably
create a programming language that converts Alp's type of proof (the ones people tend to
write) into formal logic jargon, that way it could be read and checked by machine.

Although this is very much like insisting that programmers write all their programs in machine
code, instead of something like C or Python :P.

Add-on: Here's a proof that [url=http://www.cs.unm.edu/~mccune/mace4/]Prover9[/url] generated:
(here I've used x@y to denote (x,y) and A*B to denote AxB).

[code]
Proof of First Part

1 (all z all x all y (member(z,x * y) <-> (exists a exists b (z = a @ b & member(a,x) & member(b,y))))) # label(definition_of_cartesian_product) # label(non_clause).  [assumption].
3 (all x all y (subset(x,y) <-> (all z (member(z,x) -> member(z,y))))) # label(definition_of_subset) # label(non_clause).  [assumption].
4 (all x (member(x,A) <-> x = 1 | x = 2)) # label(construction_of_A) # label(non_clause).  [assumption].
5 (all y (member(y,B) <-> y = a | y = b)) # label(construction_of_B) # label(non_clause).  [assumption].
6 (all x (member(x,f) <-> x = 1 @ a | x = 2 @ b)) # label(construction_of_f) # label(non_clause).  [assumption].
8 subset(f,A * B) # label(non_clause) # label(goal).  [goal].
15 member(x,y * z) | u @ w != x | -member(u,y) | -member(w,z) # label(definition_of_cartesian_product).  [clausify(1)].
20 subset(x,y) | member(f3(x,y),x) # label(definition_of_subset).  [clausify(3)].
21 subset(x,y) | -member(f3(x,y),y) # label(definition_of_subset).  [clausify(3)].
23 member(x,A) | 1 != x # label(construction_of_A).  [clausify(4)].
24 member(x,A) | 2 != x # label(construction_of_A).  [clausify(4)].
26 member(x,B) | a != x # label(construction_of_B).  [clausify(5)].
27 member(x,B) | b != x # label(construction_of_B).  [clausify(5)].
30 -member(x,f) | 1 @ a = x | 2 @ b = x # label(construction_of_f).  [clausify(6)].
33 -subset(f,A * B).  [deny(8)].
39 member(1,A).  [xx_res(23,b)].
40 member(2,A).  [xx_res(24,b)].
41 member(a,B).  [xx_res(26,b)].
42 member(b,B).  [xx_res(27,b)].
47 member(f3(f,A * B),f).  [resolve(33,a,20,a)].
48 -member(f3(f,A * B),A * B).  [ur(21,a,33,a)].
65 f3(f,A * B) != 1 @ a.  [ur(15,a,48,a,c,39,a,d,41,a),flip(a)].
69 f3(f,A * B) != 2 @ b.  [ur(15,a,48,a,c,40,a,d,42,a),flip(a)].
77 $F.  [ur(30,b,65,a(flip),c,69,a(flip)),unit_del(a,47)].

Proof of Second Part

2 (all x all y all z all w (x @ y = z @ w <-> x = z & y = w)) # label(ordered_pair_equality) # label(non_clause).  [assumption].
4 (all x (member(x,A) <-> x = 1 | x = 2)) # label(construction_of_A) # label(non_clause).  [assumption].
5 (all y (member(y,B) <-> y = a | y = b)) # label(construction_of_B) # label(non_clause).  [assumption].
6 (all x (member(x,f) <-> x = 1 @ a | x = 2 @ b)) # label(construction_of_f) # label(non_clause).  [assumption].
7 (all x all y all g (maps_under(x,y,g) <-> member(y,B) & member(x @ y,g))) # label(non_clause).  [assumption].
8 (all x (member(x,A) -> (exists y (maps_under(x,y,f) & (all z (member(x @ z,f) -> y = z)))))) # label(non_clause) # label(goal).  [goal].
16 x @ y != z @ u | x = z # label(ordered_pair_equality).  [clausify(2)].
17 x @ y != z @ u | u = y # label(ordered_pair_equality).  [clausify(2)].
18 x @ y = z @ u | x != z | u != y # label(ordered_pair_equality).  [clausify(2)].
19 -member(x,A) | 1 = x | 2 = x # label(construction_of_A).  [clausify(4)].
23 member(x,B) | a != x # label(construction_of_B).  [clausify(5)].
24 member(x,B) | b != x # label(construction_of_B).  [clausify(5)].
25 1 != 2 # label(1_and_2_distinct).  [assumption].
26 2 != 1.  [copy(25),flip(a)].
27 -member(x,f) | 1 @ a = x | 2 @ b = x # label(construction_of_f).  [clausify(6)].
28 member(x,f) | 1 @ a != x # label(construction_of_f).  [clausify(6)].
29 member(x,f) | 2 @ b != x # label(construction_of_f).  [clausify(6)].
32 maps_under(x,y,z) | -member(y,B) | -member(x @ y,z).  [clausify(7)].
33 member(c1,A).  [deny(8)].
34 -maps_under(c1,x,f) | member(c1 @ f4(x),f).  [deny(8)].
35 -maps_under(c1,x,f) | x != f4(x).  [deny(8)].
36 -maps_under(c1,x,f) | f4(x) != x.  [copy(35),flip(b)].
42 x @ y = z @ y | x != z.  [xx_res(18,c)].
45 member(a,B).  [xx_res(23,b)].
46 member(b,B).  [xx_res(24,b)].
48 2 @ x != 1 @ y.  [ur(16,b,26,a)].
49 member(1 @ a,f).  [xx_res(28,b)].
51 c1 = 1 | c1 = 2.  [resolve(33,a,19,a),flip(a),flip(b)].
96 maps_under(1,a,f).  [resolve(49,a,32,c),unit_del(b,45)].
105 c1 = 1 | c1 @ x = 2 @ x.  [resolve(51,b,42,b)].
646 c1 = 1 | member(c1 @ b,f).  [resolve(105,b,29,b(flip))].
768 c1 = 1 | maps_under(c1,b,f).  [resolve(646,b,32,c),unit_del(c,46)].
821 c1 = 1 | f4(b) != b.  [resolve(768,b,36,a)].
822 c1 = 1 | member(c1 @ f4(b),f).  [resolve(768,b,34,a)].
839 c1 = 1 | member(2 @ f4(b),f).  [para(51(b,1),822(b,1,1)),merge(b)].
853 c1 = 1 | 2 @ f4(b) = 2 @ b.  [resolve(839,b,27,a),flip(b),flip(c),unit_del(b,48)].
3898 c1 = 1 | f4(b) = b.  [resolve(853,b,17,a),flip(b)].
4129 c1 = 1.  [resolve(3898,b,821,b),merge(b)].
4130 -maps_under(1,x,f) | f4(x) != x.  [back_rewrite(36),rewrite([4129(1)])].
4131 -maps_under(1,x,f) | member(1 @ f4(x),f).  [back_rewrite(34),rewrite([4129(1),4129(4)])].
4132 member(1 @ f4(a),f).  [resolve(4131,a,96,a)].
4151 1 @ f4(a) = 1 @ a.  [resolve(4132,a,27,a),flip(a),flip(b),unit_del(b(flip),48)].
4152 f4(a) != a.  [resolve(4130,a,96,a)].
4153 x @ f4(a) != y @ a.  [ur(17,b,4152,a),flip(a)].
4154 $F.  [resolve(4153,a,4151,a)].
[/code]

Now [i]that's[/i] formal. And it took me barely any time whatsoever (beyond learning prover9).[/quote]

So what your computer tells you,that my proof,is a proof or Alp"s proof ,is a proof??

Because that was the question and not whether my proof was a formal proof or not.

Besides [b]practicaly any book in logic[/b] offers formal proofs .

But i suppose first of all we have to define what a formal proof is.

So go ahead define what a formal proof is and we take it from there

Since it's your question, you should probably have some definition that you are working from; but what you have
was not formal either. From what I recall, a formal proof would consist of a list of statements, each of which is
either a given, an axiom, or is the resolution of two previous steps through some rule of formal logic.

Certainly your proof is [i] not [/i] formal, since you do not break it up into a set of steps in some structured way,
nor do you justify these steps by citing a rule of logic. Chaining together statements, as you have done, is
a "summary" of a proof wherein the reader is forced to fill in which steps or axioms are used. Quite simply,
you have not written out all the steps. The output from the program did (although the reasoning for each
step may appear cryptic -- I would suggest checking the Prover9 documentation for details.)

One would also require a list of formal logic rules which are to be applied; since different texts can use different
rules, and many systems of logic are equivalent, we would not want to rely on intuition to decide which rules
are allowed.

[quote="Nukular"]Since it's your question, you should probably have some definition that you are working from; but what you have
was not formal either. From what I recall, a formal proof would consist of a list of statements, each of which is
either a given, an axiom, or is the resolution of two previous steps through some rule of formal logic.

[/quote].

So according to the above definition is the following a formal proof ??

Proof of First Part

1 (all z all x all y (member(z,x * y) <-> (exists a exists b (z = a @ b & member(a,x) & member(b,y))))) # label(definition_of_cartesian_product) # label(non_clause).  [assumption].
3 (all x all y (subset(x,y) <-> (all z (member(z,x) -> member(z,y))))) # label(definition_of_subset) # label(non_clause).  [assumption].
4 (all x (member(x,A) <-> x = 1 | x = 2)) # label(construction_of_A) # label(non_clause).  [assumption].
5 (all y (member(y,B) <-> y = a | y = b)) # label(construction_of_B) # label(non_clause).  [assumption].
6 (all x (member(x,f) <-> x = 1 @ a | x = 2 @ b)) # label(construction_of_f) # label(non_clause).  [assumption].
8 subset(f,A * B) # label(non_clause) # label(goal).  [goal].
15 member(x,y * z) | u @ w != x | -member(u,y) | -member(w,z) # label(definition_of_cartesian_product).  [clausify(1)].
20 subset(x,y) | member(f3(x,y),x) # label(definition_of_subset).  [clausify(3)].
21 subset(x,y) | -member(f3(x,y),y) # label(definition_of_subset).  [clausify(3)].
23 member(x,A) | 1 != x # label(construction_of_A).  [clausify(4)].
24 member(x,A) | 2 != x # label(construction_of_A).  [clausify(4)].
26 member(x,B) | a != x # label(construction_of_B).  [clausify(5)].
27 member(x,B) | b != x # label(construction_of_B).  [clausify(5)].
30 -member(x,f) | 1 @ a = x | 2 @ b = x # label(construction_of_f).  [clausify(6)].
33 -subset(f,A * B).  [deny(8)].
39 member(1,A).  [xx_res(23,b)].
40 member(2,A).  [xx_res(24,b)].
41 member(a,B).  [xx_res(26,b)].
42 member(b,B).  [xx_res(27,b)].
47 member(f3(f,A * B),f).  [resolve(33,a,20,a)].
48 -member(f3(f,A * B),A * B).  [ur(21,a,33,a)].
65 f3(f,A * B) != 1 @ a.  [ur(15,a,48,a,c,39,a,d,41,a),flip(a)].
69 f3(f,A * B) != 2 @ b.  [ur(15,a,48,a,c,40,a,d,42,a),flip(a)].
77 $ F.  [ur(30,b,65,a(flip),c,69,a(flip)),unit_del(a,47)].
Proof of Second Part
2 (all x all y all z all w (x @ y = z @ w <-> x = z \& y = w)) # label(ordered_pair_equality) # label(non_clause).  [assumption].
4 (all x (member(x,A) <-> x = 1 | x = 2)) # label(construction_of_A) # label(non_clause).  [assumption].
5 (all y (member(y,B) <-> y = a | y = b)) # label(construction_of_B) # label(non_clause).  [assumption].
6 (all x (member(x,f) <-> x = 1 @ a | x = 2 @ b)) # label(construction_of_f) # label(non_clause).  [assumption].
7 (all x all y all g (maps_under(x,y,g) <-> member(y,B) \& member(x @ y,g))) # label(non_clause).  [assumption].
8 (all x (member(x,A) -> (exists y (maps_under(x,y,f) \& (all z (member(x @ z,f) -> y = z)))))) # label(non_clause) # label(goal).  [goal].
16 x @ y != z @ u | x = z # label(ordered_pair_equality).  [clausify(2)].
17 x @ y != z @ u | u = y # label(ordered_pair_equality).  [clausify(2)].
18 x @ y = z @ u | x != z | u != y # label(ordered_pair_equality).  [clausify(2)].
19 -member(x,A) | 1 = x | 2 = x # label(construction_of_A).  [clausify(4)].
23 member(x,B) | a != x # label(construction_of_B).  [clausify(5)].
24 member(x,B) | b != x # label(construction_of_B).  [clausify(5)].
25 1 != 2 # label(1_and_2_distinct).  [assumption].
26 2 != 1.  [copy(25),flip(a)].
27 -member(x,f) | 1 @ a = x | 2 @ b = x # label(construction_of_f).  [clausify(6)].
28 member(x,f) | 1 @ a != x # label(construction_of_f).  [clausify(6)].
29 member(x,f) | 2 @ b != x # label(construction_of_f).  [clausify(6)].
32 maps_under(x,y,z) | -member(y,B) | -member(x @ y,z).  [clausify(7)].
33 member(c1,A).  [deny(8)].
34 -maps_under(c1,x,f) | member(c1 @ f4(x),f).  [deny(8)].
35 -maps_under(c1,x,f) | x != f4(x).  [deny(8)].
36 -maps_under(c1,x,f) | f4(x) != x.  [copy(35),flip(b)].
42 x @ y = z @ y | x != z.  [xx_res(18,c)].
45 member(a,B).  [xx_res(23,b)].
46 member(b,B).  [xx_res(24,b)].
48 2 @ x != 1 @ y.  [ur(16,b,26,a)].
49 member(1 @ a,f).  [xx_res(28,b)].
51 c1 = 1 | c1 = 2.  [resolve(33,a,19,a),flip(a),flip(b)].
96 maps_under(1,a,f).  [resolve(49,a,32,c),unit_del(b,45)].
105 c1 = 1 | c1 @ x = 2 @ x.  [resolve(51,b,42,b)].
646 c1 = 1 | member(c1 @ b,f).  [resolve(105,b,29,b(flip))].
768 c1 = 1 | maps_under(c1,b,f).  [resolve(646,b,32,c),unit_del(c,46)].
821 c1 = 1 | f4(b) != b.  [resolve(768,b,36,a)].
822 c1 = 1 | member(c1 @ f4(b),f).  [resolve(768,b,34,a)].
839 c1 = 1 | member(2 @ f4(b),f).  [para(51(b,1),822(b,1,1)),merge(b)].
853 c1 = 1 | 2 @ f4(b) = 2 @ b.  [resolve(839,b,27,a),flip(b),flip(c),unit_del(b,48)].
3898 c1 = 1 | f4(b) = b.  [resolve(853,b,17,a),flip(b)].
4129 c1 = 1.  [resolve(3898,b,821,b),merge(b)].
4130 -maps_under(1,x,f) | f4(x) != x.  [back_rewrite(36),rewrite([4129(1)])].
4131 -maps_under(1,x,f) | member(1 @ f4(x),f).  [back_rewrite(34),rewrite([4129(1),4129(4)])].
4132 member(1 @ f4(a),f).  [resolve(4131,a,96,a)].
4151 1 @ f4(a) = 1 @ a.  [resolve(4132,a,27,a),flip(a),flip(b),unit_del(b(flip),48)].
4152 f4(a) != a.  [resolve(4130,a,96,a)].
4153 x @ f4(a) != y @ a.  [ur(17,b,4152,a),flip(a)].
4154$F.  [resolve(4153,a,4151,a)].

If ,however ,as you pointed out the above is in cryptic ,you have to translate it into an ordinary formal proof the one written in logical books.

Then we can argue for the validity of your proof.

Formal proofs are quite easy to check

I don't understand what you're talking about τρικλινο.

The last line is garbled because it's basically a line that says "the proof is finished".

The way that these formal proofs proceed is to assume all the given information,
as well as the negation of the goal, and proceed to a contradiction. That's why
the last line is always \$F. (Also, remember to escape your \$ with "\\$" so
that phpBB doesn't flip out)

The cryptic-looking []'ed expressions are the justifications, which are cryptic
looking because they are programmatic output.

Also, my possible mistake, even Prover9 dislikes line-by-line proofs so
much that it occasionally does a slightly "higher-order" step (as some
of the steps listed below are not "pure"). There is an
associated program that can rewrite the proofs with a fuller justification
(using more steps). Would you like me to provide that instead?

[code]
Clause Justifications
After the initial stage of the output, each clause in the file has an integer identifier (ID) and a justification that may refer to IDs of other clauses. A justification is a list consisting of one primary step and some number of secondary steps. Most primary steps are inference rules applied to given clauses, and most secondary steps consist of simplification, rewriting, or orienting equalities.

Many of the types of step refer to positions of literals or terms in the parent clauses. Literals are identified by the characters 'a' (first literal), 'b' (second literal), etc. Terms are identified by the literal identifier followed by a sequence of integers giving the position of the term within the literal. For example, the position 'c,1,3,2' means third literal, first argument, third argument, second argument. Negation signs on literals are not included in the sequence.

Primary Steps.

    * assumption -- input formula.
    * clausify -- from CNF translation of a non-clausal assumption.
    * goal -- input formula.
    * deny -- from CNF translation of the negation of a goal.
    * resolve(59,b,47,c) -- resolve the second literal of clause 59 with the third literal of clause 47.
    * hyper(59, b,47,a, c,38,a) -- hyperresolution; interpret the list as a clause ID followed by a sequence of triples, <literal,clause-ID,literal> the inference is presented as a sequence of binary resolution steps. In the example shown, start with clause 59; then resolve literal b with clause 47 on literal a; with the result of the first step, resolve literal c with clause 38 on literal a. The special case "xx" means resolution with x=x.
    * ur(39, a,48,a, b,88,a, c,87,a, d,86,a) -- unit-resulting resolution; the list is interpreted as in hyperresolution.
    * para(47(a,1),28(a,1,2,2,1)) -- paramodulate from the clause 47 into clause 28 at the positions shown.
    * copy(59) -- copy clause 59.
    * back_rewite(59) -- copy clause 59.
    * back_unit_del(59) -- copy clause 59.
    * new_symbol(59) -- introduce a new constant (see parameter new_constants).
    * factor(59,b,c) -- factor clause 59 by unifying the second and third literals.
    * xx_res(59,b) -- resolve the second literal of clause 59 with x=x.
    * propositional -- not used in standard proofs.
    * instantiate -- not used in standard proofs.
    * ivy -- not used in standard proofs.

Secondary Steps (each assumes a working clause, which is either the result of a primary step or a previous secondary step).

    * rewrite([38(5,R),47(5),59(6,R)]) -- rewriting (demodulation) with equations 38, 47, then 59; the arguments (5), (5), and (6) identify the positions of the rewritten subterms (in an obscure way), and the argument R indicates that the demodulator is used backward (right-to-left).
    * flip(c) -- the third literal is an equality that has been flipped by the term ordering. This does not necessarily mean that the equality is orientable by the primary term ordering, e.g., KBO.
    * merge(d) -- the fourth literal has been removed because it was identical to a preceding literal.
    * unit_del(b,38) -- the second literal has been removed because it was an instance of the negation clause 38 (which is a unit clause).
    * xx(b) -- the second literal has been removed because it was an instance of x!=x.
[/code]

My most sincere apologies; here is the proof again, renumbered so that the lines are consecutive, and
will [i]all[/i] steps properly expanded.

[code]
Proof of first part:

1 (all x all y all z (member(x,y * z) <-> (exists a exists b (x = a @ b & member(a,y) & member(b,z))))) # label(definition_of_cartesian_product) # label(non_clause).  [assumption].
2 (all x all y (subset(x,y) <-> (all z (member(z,x) -> member(z,y))))) # label(definition_of_subset) # label(non_clause).  [assumption].
3 (all x (member(x,A) <-> x = 1 | x = 2)) # label(construction_of_A) # label(non_clause).  [assumption].
4 (all x (member(x,B) <-> x = a | x = b)) # label(construction_of_B) # label(non_clause).  [assumption].
5 (all x (member(x,f) <-> x = 1 @ a | x = 2 @ b)) # label(construction_of_f) # label(non_clause).  [assumption].
6 subset(f,A * B) # label(non_clause) # label(goal).  [goal].
7 member(x,y * z) | u @ w != x | -member(u,y) | -member(w,z) # label(definition_of_cartesian_product).  [clausify(1)].
8 subset(x,y) | member(f3(x,y),x) # label(definition_of_subset).  [clausify(2)].
9 subset(x,y) | -member(f3(x,y),y) # label(definition_of_subset).  [clausify(2)].
10 member(x,A) | 1 != x # label(construction_of_A).  [clausify(3)].
11 member(x,A) | 2 != x # label(construction_of_A).  [clausify(3)].
12 member(x,B) | a != x # label(construction_of_B).  [clausify(4)].
13 member(x,B) | b != x # label(construction_of_B).  [clausify(4)].
14 -member(x,f) | 1 @ a = x | 2 @ b = x # label(construction_of_f).  [clausify(5)].
15 -subset(f,A * B).  [deny(6)].
16 member(1,A).  [xx_res(10,b)].
17 member(2,A).  [xx_res(11,b)].
18 member(a,B).  [xx_res(12,b)].
19 member(b,B).  [xx_res(13,b)].
20 member(f3(f,A * B),f).  [resolve(15,a,8,a)].
21 -member(f3(f,A * B),A * B).  [resolve(9,a,15,a)].
22A x @ y != f3(f,A * B) | -member(x,A) | -member(y,B).  [resolve(7,a,21,a)].
22B 1 @ x != f3(f,A * B) | -member(x,B).  [resolve(22A,b,16,a)].
22C 1 @ a != f3(f,A * B).  [resolve(22B,b,18,a)].
22 f3(f,A * B) != 1 @ a.  [copy(22C),flip(a)].
23A x @ y != f3(f,A * B) | -member(x,A) | -member(y,B).  [resolve(7,a,21,a)].
23B 2 @ x != f3(f,A * B) | -member(x,B).  [resolve(23A,b,17,a)].
23C 2 @ b != f3(f,A * B).  [resolve(23B,b,19,a)].
23 f3(f,A * B) != 2 @ b.  [copy(23C),flip(a)].
24A -member(f3(f,A * B),f) | 2 @ b = f3(f,A * B).  [resolve(14,b,22,a(flip))].
24B -member(f3(f,A * B),f).  [resolve(24A,b,23,a(flip))].
24 $F.  [resolve(20,a,24B,a)].

============================== end of proof ==========================

Proof of second part:

1 (all x all y all z all u (x @ y = z @ u <-> x = z & y = u)) # label(ordered_pair_equality) # label(non_clause).  [assumption].
2 (all x (member(x,A) <-> x = 1 | x = 2)) # label(construction_of_A) # label(non_clause).  [assumption].
3 (all x (member(x,B) <-> x = a | x = b)) # label(construction_of_B) # label(non_clause).  [assumption].
4 (all x (member(x,f) <-> x = 1 @ a | x = 2 @ b)) # label(construction_of_f) # label(non_clause).  [assumption].
5 (all x all y all g (maps_under(x,y,g) <-> member(y,B) & member(x @ y,g))) # label(non_clause).  [assumption].
6 (all x (member(x,A) -> (exists y (maps_under(x,y,f) & (all z (member(x @ z,f) -> y = z)))))) # label(non_clause) # label(goal).  [goal].
7 x @ y != z @ u | x = z # label(ordered_pair_equality).  [clausify(1)].
8 x @ y != z @ u | u = y # label(ordered_pair_equality).  [clausify(1)].
9 x @ y = z @ u | x != z | u != y # label(ordered_pair_equality).  [clausify(1)].
10 -member(x,A) | 1 = x | 2 = x # label(construction_of_A).  [clausify(2)].
11 member(x,B) | a != x # label(construction_of_B).  [clausify(3)].
12 member(x,B) | b != x # label(construction_of_B).  [clausify(3)].
13 1 != 2 # label(1_and_2_distinct).  [assumption].
14 2 != 1.  [copy(13),flip(a)].
15 -member(x,f) | 1 @ a = x | 2 @ b = x # label(construction_of_f).  [clausify(4)].
16 member(x,f) | 1 @ a != x # label(construction_of_f).  [clausify(4)].
17 member(x,f) | 2 @ b != x # label(construction_of_f).  [clausify(4)].
18 maps_under(x,y,z) | -member(y,B) | -member(x @ y,z).  [clausify(5)].
19 member(c1,A).  [deny(6)].
20 -maps_under(c1,x,f) | member(c1 @ f4(x),f).  [deny(6)].
21 -maps_under(c1,x,f) | x != f4(x).  [deny(6)].
22 -maps_under(c1,x,f) | f4(x) != x.  [copy(21),flip(b)].
23 x @ y = z @ y | x != z.  [xx_res(9,c)].
24 member(a,B).  [xx_res(11,b)].
25 member(b,B).  [xx_res(12,b)].
26 2 @ x != 1 @ y.  [resolve(7,b,14,a)].
27 member(1 @ a,f).  [xx_res(16,b)].
28A 1 = c1 | 2 = c1.  [resolve(19,a,10,a)].
28B c1 = 1 | 2 = c1.  [copy(28A),flip(a)].
28 c1 = 1 | c1 = 2.  [copy(28B),flip(b)].
29A maps_under(1,a,f) | -member(a,B).  [resolve(27,a,18,c)].
29 maps_under(1,a,f).  [resolve(24,a,29A,b)].
30 c1 = 1 | c1 @ x = 2 @ x.  [resolve(28,b,23,b)].
31 c1 = 1 | member(c1 @ b,f).  [resolve(30,b,17,b(flip))].
32A c1 = 1 | maps_under(c1,b,f) | -member(b,B).  [resolve(31,b,18,c)].
32 c1 = 1 | maps_under(c1,b,f).  [resolve(25,a,32A,c)].
33 c1 = 1 | f4(b) != b.  [resolve(32,b,22,a)].
34 c1 = 1 | member(c1 @ f4(b),f).  [resolve(32,b,20,a)].
35A c1 = 1 | c1 = 1 | member(2 @ f4(b),f).  [para(28(b,1),34(b,1,1))].
35 c1 = 1 | member(2 @ f4(b),f).  [copy(35A),merge(b)].
36A c1 = 1 | 1 @ a = 2 @ f4(b) | 2 @ b = 2 @ f4(b).  [resolve(35,b,15,a)].
36B c1 = 1 | 2 @ f4(b) = 1 @ a | 2 @ b = 2 @ f4(b).  [copy(36A),flip(b)].
36C c1 = 1 | 2 @ f4(b) = 1 @ a | 2 @ f4(b) = 2 @ b.  [copy(36B),flip(c)].
36 c1 = 1 | 2 @ f4(b) = 2 @ b.  [resolve(26,a,36C,b)].
37A c1 = 1 | b = f4(b).  [resolve(36,b,8,a)].
37 c1 = 1 | f4(b) = b.  [copy(37A),flip(b)].
38A c1 = 1 | c1 = 1.  [resolve(37,b,33,b)].
38 c1 = 1.  [copy(38A),merge(b)].
39 -maps_under(1,x,f) | f4(x) != x.  [para(38(a,1),22(a,1))].
40A -maps_under(1,x,f) | member(c1 @ f4(x),f).  [para(38(a,1),20(a,1))].
40 -maps_under(1,x,f) | member(1 @ f4(x),f).  [para(38(a,1),40A(b,1,1))].
41 member(1 @ f4(a),f).  [resolve(40,a,29,a)].
42A 1 @ a = 1 @ f4(a) | 2 @ b = 1 @ f4(a).  [resolve(41,a,15,a)].
42B 1 @ f4(a) = 1 @ a | 2 @ b = 1 @ f4(a).  [copy(42A),flip(a)].
42C 1 @ f4(a) = 1 @ a | 1 @ f4(a) = 2 @ b.  [copy(42B),flip(b)].
42 1 @ f4(a) = 1 @ a.  [resolve(26,a,42C,b(flip))].
43 f4(a) != a.  [resolve(39,a,29,a)].
44A x @ a != y @ f4(a).  [resolve(8,b,43,a)].
44 x @ f4(a) != y @ a.  [copy(44A),flip(a)].
45 $F.  [resolve(44,a,42,a)].
============================= end of proof ===============
[/code]

Just for teh lulz:

Maybe you should first define, what you mean by $ \{1,2\}$ and $ \{a,b\}$.

For example, $ x\equal{}\{1,2\} \Leftrightarrow \forall y . y\in \{1,2\} \rightarrow (y\equal{}\{\{\}\} \vee y\equal{}\{\{\},\{\{\}\}\})$ or $ x\equal{}\{1,2\} \Leftrightarrow x \in 3 \wedge 0 \neq x$. Depending on this, it can be easier or harder to prove that $ \{1,2\}$ is a set anyway (it could be a proper class).

[quote="Schoppenhauer"]Just for teh lulz:

Maybe you should first define, what you mean by $ \{1,2\}$ and $ \{a,b\}$.

For example, $ x \equal{} \{1,2\} \Leftrightarrow \forall y . y\in \{1,2\} \rightarrow (y \equal{} \{\{\}\} \vee y \equal{} \{\{\},\{\{\}\}\})$ or $ x \equal{} \{1,2\} \Leftrightarrow x \in 3 \wedge 0 \neq x$. Depending on this, it can be easier or harder to prove that $ \{1,2\}$ is a set anyway (it could be a proper class).[/quote]

True facts.

[quote="τρικλινο"]UAUA!!!    AKUNA MA TATA!!!![/quote]

Thankfully, mathematics is [i]not[/i] a problem free philosophy.

[quote="Schoppenhauer"][quote="τρικλινο"]Definition: {x,y}=w$ \Longleftrightarrow\forall z(z\in w\Longleftrightarrow z \equal{} x\vee z \equal{} y)$ & w is a set

Theorem : $ z\in \{x,y\}\Longleftrightarrow z \equal{} x\vee z \equal{} y$

proof:

in the above definition put w={x,y}[/quote]You'll get a cake for that.[/quote]

Not me ,Ratrick Suppes.

The above was taken from his book: AXIOMATIC SET THEORY ,PAGE 31

[quote="τρικλινο"][quote="Schoppenhauer"]Just for teh lulz:

Maybe you should first define, what you mean by $ \{1,2\}$ and $ \{a,b\}$.

For example, $ x \equal{} \{1,2\} \Leftrightarrow \forall y . y\in \{1,2\} \rightarrow (y \equal{} \{\{\}\} \vee y \equal{} \{\{\},\{\{\}\}\})$ or $ x \equal{} \{1,2\} \Leftrightarrow x \in 3 \wedge 0 \neq x$. Depending on this, it can be easier or harder to prove that $ \{1,2\}$ is a set anyway (it could be a proper class).[/quote]

An equivalence [b]P[/b] INTRODUCING  a new n-place operation symbol [b]O[/b] IS a proper definition if and only if [b]P[/b] is of the form:

......................................[b]O[/b]($ v_{1},.............v_{n}$) = [b]w[/b]$ \Longleftrightarrow$[b]Q[/b]..........................................................

AND THE FOLLOWING RESTRICTIONS ARE SATISFIED:1) $ v_{1},........v_{n}$ are distinct variables.

2)[b]Q[/b] HAS NO FREE VARIABLES other than $ v_{1},.............v_{n}, w$

3)[b]Q[/b] is a formula in which the only non-logical constants are the primitive or previously defined symbols of set theory,and

4)The formula ($ \exists ! w$)[b]Q[/b] is derivable from the axioms and preceding definitions.

[b]Does your definition satisfy all that??[/b][/quote]

Why do you feel the need to put bold everywhere? [b]Do you feel people are too stupid to
read your entire post?[/b]

(I'm sorry if the above comes off harsh, but I'm honestly tired of seeing all-caps
and bold shouting -- call caps is always unnecessary when bold and italics are
available, and shouldn't be used to shout, but sparingly for emphasis or for
titles of books, etc.)

[quote="τρικλινο"]I use bolds and caps only for emphasis and sometimes caps unintentionally, because simply the keyboard was stuck there[/quote]

If it's unintentional, OK. But please keep in mind that it's a bit offensive when it looks like shouting; other things would
include multiple question marks/exclamations/etc.

For example:
UNDERSTAND????

would probably be offensive, as opposed to

Understand?

or

Does that make sense?


Vertical alignment of Latex is messed up in several threads. For instance, see

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=272494
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=272398
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=271829
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=269313

I'll probably post this somewhere else in this forum also, but I have also having difficulty with LaTeX. Specificall 2 digit exponets. They appear as the first digit as normal exponet size, but the 2 digit is normal number size.

See [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=276790]here.[/url]

And a mod should probably split this and bluecarneal's post off into either the test form or the $ \text{\LaTeX}$ forum.

Learn to use {}.

x^{12} becomes $ x^{12}.$

e^{-x^{2k}} becomes $ e^{-x^{2k}}.$

The LaTeX forum of which Kent speaks is http://www.artofproblemsolving.com/Forum/index.php?f=123 (123, very easy to remember); he also might have recommended the search feature http://www.artofproblemsolving.com/Forum/search.php (linked to from the top of the forum).

This is also happening in other subfora. For example, see these threads

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=57381
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=273724
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=51&t=261698
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=57&t=184458

EDIT: The problem seems solved now.

This is about vertical alignments, so matrices and other tall bits of $ \text{\LaTeX}$ show it best.

Some better examples:
[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=131959]The problem.[/url]
Normal display in post #[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1454013#1454013]2[/url] and the rest of the first page. Bad alignment in post #[url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1476316#1476316]26[/url] and the rest of the second page.

Hmm, I'm going to guess we may have had a small bug with the LaTeX stuff for a few days. Regenerating the images in that post 26 that jmerry mentioned and the topic luis mentioned made everything nice and clean.

A new one, seen [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1513372#1513372]here[/url]: $ (x\minus{}2)(x^{2}\plus{}2x\plus{}4)$. I think the vertical alignment is off as in the others from this thread, but it's the non-italic $ x$ that really caught my eye.


A right triangle is formed in the first quadrant by the x- and y-axes and a line through the point (1,2) [the line passes through the points (0,y) and (x,0)]

Write the length of the hypotenuse as a function of x.

And if you're looking for the minimum... it's the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?&t=24520]pole-corner problem[/url].

Right now I'm just trying to figure out how I can write the length of the hypotenuse in terms of x - I don't care about fully solving it at the moment

That comes down to getting $ y$ in terms of $ x$. Draw the picture, including lines parallel to the axis through $ (1,2)$, and look for similar right triangles.

I got $ y\equal{}\frac{2x}{x\minus{}1}$

Is that right?

[quote="Quevvy"]I got $ y \equal{} \frac {2x}{x \minus{} 1}$

Is that right?[/quote]I don't think so. For one thing, $ (1,2)$ isn't a point on that equation, while it should be; for another thing, the equation you have is not even that of a line.

[quote="jmerry"]No, it's right. This is an equation relating the two points $ (x,0)$ and $ (0,y)$, not the equation of a line.[/quote]Ah. You meant $ x$ and $ y$ as the length and width. Yes, that is correct.  :blush:


[quote="philB"]A point inside a square is positioned so that the distances to the for vertices are 27, 21, 6 and x units. If x is a whole number, what is the value of x?[/quote]
[hide="British Flag Theorem"]$27^{2}+6^{2}=21^{2}+x^{2}$
$729+36=441+x^{2}$
$324=x^{2}$
$x=18$[/hide]

[quote="philB"]A point inside a square is positioned so that the distances to the for vertices are 27, 21, 6 and x units. If x is a whole number, what is the value of x?[/quote]
[hide]Draw altitudes from that point to all of the sides.[/hide]
or, going a bit more obscure,
[hide]
Using some obscure theorem:
If there's a point inside a rectangle (technically it doesn't even have to be in the square, or even that plane really) called P, and the vertices are A, B, C, and D, then we know that
$PA^{2}+PC^{2}=BP^{2}+DP^{2}$
[/hide]
Using one of the 2 above methods, we see that our answer is
[hide]18[/hide]

Yes.

Extention: Derive the British Flag Theorem.

[hide="hint"]
As diophantient said, drop perpendiculars from all the sides. Assign some variables, then Pythagorean Galore![/hide]

one: what exactly is the british flag thm.  and extension is not extention

[hide]the perpendiculars are a, b, c, d going clockwise. a^2+d^2 equals the square of top left, a^2+b^2 equals square of top right, b^2+c^2 equals square of bottom right, and c^2+d^2 is bottom left. just add them up and you get they are equal [/hide]


Why are the electrons of metallic elements oxidized in the order p, s, d? Shouldn't the higher energy level electrons be lost first (d, p, s)?

The outermost electrons, regardless of their energy, are lost first.

Does that help?

Well, remember that the s orbital is very close in energy to the d orbital of the shell below it. In a cation, the d orbital is more stable, and so the s is removed first.