Difference between revisions of "Talk:Gmaas"

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HOW DARE THEY! THIS IS HORRIBLE!!!!!! THIS IS NOT GOOD AT ALL!!!! - asdf334 ;( currently mourning this loss
HOW DARE THEY! THIS IS HORRIBLE!!!!!! THIS IS NOT GOOD AT ALL!!!! - asdf334 ;( currently mourning this loss
= What is the definition of Pure Mathematics? =
= What is the definition of Pure Mathematics? =

Revision as of 17:00, 25 August 2019

Follow the following steps to summon Gmaas:

1. Draw a circle and circumscribe it with a regular hexagon and an equilateral triangle.

2. Write the numerical value of $17^{36}$ along the edge of the circle.

3. Write the numerical value of $33^{29}$ along the edge of the hexagon.

4. Write the numerical value of $\cot(0)$ 5 centimeters above the hexagon.

5. Write the numerical value of $\tan(\frac{\pi}{2} rad)$ 5 centimeters below the hexagon.

6. Write the numerical value of $\ln(0)$ inside the circle.

7. Write the numerical value of the melting point of water inside the circle. Include units and write 15 significant digits.

8. Write the numerical value of the gravitational constant inside the circle. Include units and write 25 significant digits.

9. Carry the paper with the circle and hexagon in your hand.

10. Recite the value of $\pi^{e^2}$ and include $\pi^{e^2}$ (rounded to the nearest septillionth) digits.

11. Travel at the speed of light with that paper and Gmaas will be summoned.

EDIT: The author congratulates the previous editor for his research on the cutting edge of Gmaasology. He/she has been awarded the Nobel Prize in Gmaasology for his discovery. However, the aforementioned summoning has several problems: it does not specify the dimensions of the hexagon or the circle; the fact that it has to be drawn on a paper; the size of the paper; the pressure of the water being melted; whether $\frac{\pi}{2}$ radians, degrees, or gradians; the exact location if the value of $17^{36}$ and $33^{29}$ in the diagram and the base those numbers should be written in; the units of the melting point of water or the units of the gravitational constant; or the location where Gmaas will be summoned. He might be summoned on the other side of the universe.

The person who wrote the steps to summon Gmaas says:

When I followed this procedure to summon Gmaas, I used a paper that is 15 inches long and 11 inches wide. It is currently unknown whether paper of other dimensions can be used to summon Gmaas. For step 7, the pressure of the water being melted is at 611.66 Pascals, which matches the triple point of water. Also, the units of the melting points of water and the gravitational constant should be written in terms of SI base units. The circle used should have a radius of 5 inches. If you follow this procedure, Gmaas will be summoned for about 3.53 nanoseconds at a random position in the planet you are on. The exact time Gmaas will be summoned depends on how accurately you follow the procedure. For example, if the melting point of water and the gravitational constant are accurate to 200 significant digits, then Gmaas will be summoned for 77.8 microseconds. If they are accurate to 400 significant digits, Gmaas will be summoned for 1.7 seconds. The function that relates the number of significant digits and the time Gmaas will be summoned is still unknown.

EDIT: Based on the two data points, the function that relates the number of significant digits and the time Gmaas will be summoned is $t=.0081111s+.5444$.



This article is not finished. Colball is the only person allowed to edit. The plan for me is to make this page longer than Gmaas's. (Sorry, Gmaas).


NOOOOO. THEY DELETED THE PAGE!!! https://artofproblemsolving.com/wiki/index.php?title=Gmaas#Known_Facts_About_gmaas

HOW DARE THEY! THIS IS HORRIBLE!!!!!! THIS IS NOT GOOD AT ALL!!!! - asdf334 ;( currently mourning this loss


What is the definition of Pure Mathematics?

Oh, easy, you say it is just the study of numbers.

That may be true for some areas of math. However, what about geometry, trigonometry, and calculus? And what is the definition of numbers? Now you go to the dictionary and say The relationship between measurements and quantities using numbers and symbols. This is, however, not fully true because this definition also uses applied mathematics. We want pure mathematics.

Also, most of these definitions miss one area of math. Chaos Theory. What is Chaos Theory? Chaos Theory is a recently discovered area of math where nothing can be predicted but nothing is random. We are only at the beginning of learning it. For example, can a butterfly that flaps its wings in Brazil trigger a tornado in Texas?

Some definitions hit almost all the areas of math, but some are too broad and logic, for example, often fits into the definition.

We can, however, define some areas of math but not the whole thing. For example, the definition of geometry is Geometry is concerned with the properties and relations of points, lines, surfaces, solids, and higher dimensional analogs. Or the definition of probability is the extent to which an event is likely to occur.



The branch of mathematics dealing with the properties and manipulation of constants.


Arithmetic starts with one thing which without it no arithmetic can survive: Counting Positive Integers. 1,2,3,4,5...

Addition and addition repeated

Addition is combining these integers. The symbol for this is +. $a+b=b+a$

Multiplication is repeating addition. The symbol for this is $\cdot$. $ab=ba$

Exponentiation is repeated Multiplication. The symbol for a to the power of b is $a^b$. $a^b \neq b^a$.


Subtraction is the inverse of addition. To make this a well-defined function everywhere, negative numbers and zero were defined.

Division is the inverse of multiplication. To make this a well-defined function everywhere, non-integral fractions were defined.

$n^{\text{th}}$ roots and logarithms are the inverses of exponentiation. To make these well-defined functions everywhere, irrational numbers were defined.

Negative numbers

$a$ and $b$ are positive.

1. $(-a)(-b)=ab$

2. $(a)(-b)=-ab$

Proof for 1: This is, in fact, the reason why the negative numbers were introduced: so that each positive number would have an additive inverse. ... The fact that the product of two negatives is positive is therefore related to the fact that the inverse of the inverse of a positive number is that positive number back again.

Proof for 2: Since ab is repeated addition then $(a)(-b)$ is repeated subtraction. Therefore it is negative.


A fraction a number that can be expressed as two numbers divided. For example, five divided by four is $\frac{5}{4}$.

Simplifing fractions

Find common factors in each half of the fration and then divide top and bottam of the fraction by that factor.




Adding fractions

Here is how you add fractions. Is they have the same bottom half then $\frac{a}{b}+\frac{c}{b}=\frac{a+b}{c}$. However, if the two or three or n fractions do not have the same bottom half you make them. $\frac{a}{b}+\frac{c}{d}=\frac{bd}{ad+bc}$. Subtracting fractions is the same exept everything has a minus symbol. Remember to fully simplify.

Multilpying fractions



Remember to fully simplify.

Irrational numbers

Most square roots are irrational. An irrational number is a number that can not be expresed a a fraction.

Proof that any nonperfect square positive integer is irrational

Let us assume that $\sqrt{n}$ is rational where $n$ is a nonperfect square positive integer. Then it can be written as $\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2$. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore $\sqrt{n}$ is irrational.

pi and e

Some irrational numbers are limits. That means they are the sum of smaller and smaller fraction going infinitely long. Pi or $\pi$ (go to the geometry part of this article) is the limit $\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...=\frac{\pi}{4}$. e is the limit $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}...$ More at the counting part of the article.

Exponent rules

Listed below are some important properties of exponents:

  1. $b^x\cdot b^y = b^{x+y}$
  2. $b^{-x}=\frac 1{b^x}$
  3. $\frac{b^x}{b^y}=b^{x-y}$
  4. $(b^x)^y = b^{xy}$
  5. $(ab)^x = a^x b^x$
  6. $b^0 = 1$ (if $b \neq 0$. $0^0$ is undefined.)

Here are explanations of the properties listed above:

  1. On both sides, we are multiplying b together x+y times. Thus, they are equivalent.
  2. This is described in the previous section.
  3. This results from using the previous two properties.
  4. We are multiplying $b^x$ by itself y times, which is the same as multiplying b by itself xy times.
  5. After multiplying ab by itself x times, we can collect a and b terms, thus establishing the property.
  6. Hoping that property #1 will be true when $y=0$, we see that $b^x\cdot b^0$ should (hopefully) be equal to $b^x$. Thus, we define $b^0$ to be equal to $1$ in order to make this be true.



The part of mathematics in which letters and other general symbols are used to represent numbers and quantities in formulae and equations.

One-variable linear equations


'A One-variable linear equation is an equation that comes in the form $ax+b=c$. $a$, $b$, and $c$ are constants and $x$ is the varible'.

The answer is always...




When there are fractions in the equation, you multiply both sides by the LMC of the fractions and then you solve. More at the number theory part of this article.



A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is $ax^2 + bx + c = 0$ with $a$, $b$, and $c$ being constants, or numerical coefficients, and $x$ is an unknown variable.

The answer is always...





Since the left side of the equation right above is a perfect square, you can factor the left side by using the coefficient of the first term (x) and the base of the last term(b/2a). Add these two and raise everything to the second.


$x+\frac{b}{2a}=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}$

$x=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}+\frac{b}{2a}$

When simplified the Quadratic Formula is ${x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$




Numbers like this are called imaginary numbers. Impossible, you say. But no. Solve $x^2+1=0$. You get $i$. $oi$ is $o$. So zero is both real and imaginary. (real means not imaginary)

Powers of i

$i^0=1$ $i^1=i$ $i^2=-1$ $i^3=-i$ $i^4=1$ $i^5=i$ $i^6=-1$ $i^7=-i$

The pattern repeats.

Complex numbers

A complex number is $ai+b$, where a and b are real. All numbers are complex becuase a and/or/never b can be zero.

$ai+b+ci+d=$ Complex

$(ai+b)(ci+d)=$ Complex

$\frac{ai+b}{ci+d}=$ Complex

$(ai+b)-(ci+d)=$ Complex

Systems of equations

A system of equations is a set of equations which share the same variables. An example of a system of equations is

$2a - 3b$ $= 4$
$3a - 2b$ $= 3$

Solving Linear Systems

A system of linear equations is where all of the variables are to the power 1. There are three elementary ways to solve a system of linear equations.

Gaussian Elimination

Gaussian elimination involves eliminating variables from the system by adding constant multiples of two or more of the equations together. Let's look at an example:


Find the ordered pair $(x,y)$ for which

$x - 12y$ $= 2$
$3x + 6y$ $= 6$

We can eliminate $y$ by adding twice the second equation to the first:

$x - 12y= 2$
$+2($ $3x + 6y = 6)$
$\overline{7x + 0=14}$
Thus $x=2$. We can then plug in for $x$ in either of the equations:
$(2)-12y = 2 \Rightarrow y = 0$.

Thus, the solution to the system is $(2,0)$.


The second method, substitution, requires solving for a variable and then plugging that variable into another equation therefore reducing the number of variables. We'll show how to solve the same problem from the elimination section using substitution.


Find the ordered pair $(x,y)$ for which

$x - 12y$ $= 2$
$3x + 6y$ $= 6$
The first equation can be solved for $x$:
$x = 12y + 2.$
Plugging this into the second equation yields
$3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0.$

Thus $y=0$. Plugging this into either of the equations and solving for $x$ yields $x=2$.


There are many more types of algebra: inequalities, polynomials, graphing equations, arithmetic, and geometric sequence.

Algebra is a broad and diverse area of math in which this is just a short introduction.

Number Theory


The branch of mathematics that deals with the properties and relationships of numbers, especially the positive integers.


A prime number (or simply prime) is a positive integer $p>1$ whose only positive divisors are 1 and itself. Note that $1$ is usually defined as being neither prime nor composite because it is its only factor among the natural number numbers.

There are an infinite number of prime numbers. A standard proof attributed to Euclid notes that if there are a finite set of prime numbers $p_1, p_2, \ldots, p_n$, then the number $N = p_1p_2\cdots p_n + 1$ is not divisible by any of them, but $N$ must have a prime factor, which leads to a direct contradiction.

Techniques to Check for Prime Numbers


A prime number is only divisible by one or itself, so a number $n$ is prime if and only if $n$ is not divisible by any integer greater than $1$ and less than $n$. One only needs to check integers up to $\sqrt{n}$ because dividing larger numbers would result in a quotient smaller than $\sqrt{n}$.

Modular Arithmetic

Modular arithmetic can help determine if a number is not prime.

  • If a number not equal to $2,3$ is congruent to $0,2,3,4,6 \pmod{6}$, then the number is not prime.
  • If a number not equal to $2,5$ ends with an even digit or $5$, then the number is not prime.

Importance of Primes

According to the Fundamental Theorem of Arithmetic, there is exactly one unique way to factor a positive integer into a product of primes. This unique prime factorization plays an important role in solving many kinds of number theory problems.

Mersenne Primes

A Mersenne prime is a prime of the form $2^n-1$. For such a number to be prime, n must itself be prime. Compared to other numbers of comparable sizes, Mersenne numbers are easy to check for primality because of the https://en.wikipedia.org/wiki/Lucas%E2%80%93Lehmer_primality_test Lucas-Lehmer test, severely reducing the amount of computation needed.

Twin Primes

Two primes that differ by exactly 2 are known as twin primes. The following are the first few pairs of twin primes:
3, 5
5, 7
11, 13
17, 19
29, 31
41, 43

It is not known whether or not there are infinitely many pairs of twin primes. This is known as the Twin Prime Conjecture, which is a specific instance of the Hardy-Littlewood conjecture.

Gaussian Primes

A Gaussian prime is a prime that extends the idea of the traditional prime to the Gaussian integer. One can define this term for any ring, especially number rings.

Advanced Definition

When the need arises to include negative divisors, a prime is defined as an integer p whose only divisors are 1, -1, p, and -p. More generally, if R is an integral domain, then a nonzero element p of R is a prime if whenever we write $p=ab$ with $a,b\in R$, then exactly one of a and b is a unit.




The GCD or Greatest Common Divisor of multiple numbers is the largest number that is a factor of all those multiple numbers. The answer can be and usually is 1.

Prime Factorization

The way to solve most GCD's is with using Prime Factorization. Remember that the way to find the prime factorization of a number is to find which primes products hit that certain number. The prime factorization of 24 is $2 \cdot 2 \cdot 2 \cdot 3$. The prime factorization of 18 is $2 \cdot 3 \cdot 3$. In GCD you take the prime factorization of each number and then find which primes match up.



The LCM or Least Common Multiple of two or three or n numbers is the smallest integer in which both numbers are factors of it. For example, the LCM of 10 and 5 is 10.


Problem 1

What is the LCM of 8 and 5?


The way to solve LCM's with no common factors is to multiply them. The answer is just 40.

Problem 2

What is the LCM of 15 and 10.


The Way to solve this is see that the answer would be the first multiple of 5 that is divisble by 15 and 10. That is just 30.

Problem 3

Find the LCM of 2, 3 and 5.


It is the same with three numbers. Multiply them to get 30. Same as last time.


In mathematics, a base or radix is the number of different digits or a combination of digits and letters that a system of counting uses to represent numbers. For example, the most common base used today is the decimal system. Because "dec" means 10, it uses the 10 digits from 0 to 9. Most people think that we most often use base 10 because we have 10 fingers.

Some Numbers in other bases

A base can be any whole number bigger than 1. The base of a number may be written next to the number: for instance, ${23_{8}}$ means 23 in base 8 (which is equal to 19 in base 10). Here are some examples of how numbers are written in varying bases, compared to decimal:

In Arabic numbers (decimal, or base 10), there are 10 digits: 0,1,2,3,4,5,6,7,8,9. You need one digit each to count up to 9, but two digits for ten, and three digits for a hundred, which is ten times ten. In Binary, base 2, you need two digits for two, as you only have two digits, 0 and 1. Base 5 has five digits, and the number five becomes 10. For base 16, you will need sixteen digits, and there are only ten numerals. So we use the letters A,B,C,D,E,F. These represent the decimal numbers 10, 11, 12, 13, 14 and 15. Look at the table below and find the pattern for these bases.

Number Theory

This is the end of the number theory part of this article. Number Theory, however, has much more to than what I have just shown you.

Pascal's Triangle and Combinatorics

Pascal's Identity


Pascal's Identity states that

${n \choose k}={n-1\choose k-1}+{n-1\choose k}$

for any positive integers $k$ and $n$. Here, $\binom{n}{k}$ is the binomial coefficient $\binom{n}{k} = nCk = C_k^n$.

Remember that $\binom{n}{r}=\frac{n!}{k!(n-k)!}$. $\binom{n}{r}$ means the number of ways to pick r thing from n things where order does not matter. $n!=1 \cdot 2 \cdot 3 \cdot...\cdot n$.

Proving it

If $k > n$ then $\binom{n}{k} = 0 = \binom{n - 1}{k - 1} + \binom{n - 1}{k}$ and so the result is pretty clear. So assume $k \leq n$. Then

\begin{eqnarray*}\binom{n-1}{k-1}+\binom{n-1}{k}&=&\frac{(n-1)!}{(k-1)!(n-k)!}+\frac{(n-1)!}{k!(n-k-1)!}\\ &=&(n-1)!\left(\frac{k}{k!(n-k)!}+\frac{n-k}{k!(n-k)!}\right)\\ &=&(n-1)!\cdot \frac{n}{k!(n-k)!}\\ &=&\frac{n!}{k!(n-k)!}\\ &=&\binom{n}{k}. \qquad\qquad\square\end{eqnarray*} There we go. We proved it!

Why is it needed?

It's mostly just a cool thing to know. However, if you want to know how to use it in real life go to https://artofproblemsolving.com/videos/counting/chapter12/141. Or really any of the counting and probability videos.

Introduction to Pascal's Triangle

How to build it

Pascal's Triangle is a triangular array of numbers in which you start with two infinite diagonals of ones and each of the rest of the numbers is the sum of the two numbers above it. It looks something like this:

                                        1 1
                                       1 2 1
                                      1 3 3 1
                                     1 4 6 4 1

And on and on...



Pascal's Triangle is really combinations. It looks something like this if it is depicted as combinations:

                $\binom{1}{0}$                  $\binom{1}{1}$
   $\binom{2}{0}$                     $\binom{2}{1}$                $\binom{2}{2}$

And on and on...


If you look at the way we build the triangle, each number is the sum of the two numbers above it. Assuming that these combinations are true then each combination in the sum of the two combinations above it. In an equation, it would look something like this: ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$. Its Pascals Identity! Therefore each row looks something like this:

$\binom{n}{0}   \binom{n}{1}   \binom{n}{2} ... \binom{n}{n}$

Patterns and Properties

In addition to combinations, Pascal's Triangle has many more patterns and properties. See below. Be ready to be amazed.

Binomial Theorem

Let's multiply out some binomials. Try it yourself and it will not be fun: $(x+y)^0=1$




If you take away the x's and y's you get:

        1 1
       1 2 1
      1 3 3 1     

It's Pascal's Triangle!


Here are a number of different ways to prove the Binomial Theorem, for example by a straightforward application of mathematical induction. The Binomial Theorem also has a nice combinatorial proof:

We can write $(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}$. Repeatedly using the distributive property, we see that for a term $a^m b^{n-m}$, we must choose $m$ of the $n$ terms to contribute an $a$ to the term, and then each of the other $n-m$ terms of the product must contribute a $b$. Thus, the coefficient of $a^m b^{n-m}$ is the number of ways to choose $m$ objects from a set of size $n$, or $\binom{n}{m}$. Extending this to all possible values of $m$ from $0$ to $n$, we see that $(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}$, as claimed.

Similarly, the coefficients of $(x+y)^n$ will be the entries of the $n^\text{th}$ row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS]

In real life

It is really only used for multipling out binomials. More usage at https://artofproblemsolving.com/videos/counting/chapter14/126.

Powers of 2


It states that $\binom{n}{0}+\binom{n}{1}+...+\binom{n}{n}=2^n$

Why do we need it?

It is useful in many word problems (That means, yes, you can use it in real life) and it is just a cool thing to know. More at https://artofproblemsolving.com/videos/mathcounts/mc2010/419.

Subset proof

Say you have a word with n letters. How many subsets does it have in terms of n? Here is how you answer it: You ask the first letter Are you in or are you out? Same to the second letter. Same to the third. Same to the n. Each of the letters has two choices: In and Out. The would be $(2)(2)(2)(2)$...n times. $2^n$.

Alternate proof

If you look at the way we built the triangle you see that each number is row n-1 is added on twice in row n. This means that each row doubles. That means you get powers of two.

Triangle Numbers


If you look at the numbers in the third diagonal you see that they are triangle numbers.


Now we can make an equation: $\binom{n}{2}=1+2+3+...+(n-1) \Rightarrow \binom{n}{2}=\frac{n(n+1)}{2} \Rightarrow \frac{n!}{2!(n-2)!}=\frac{n(n+1)}{2} \Rightarrow \frac{n(n+1)}{2}=\frac{n(n+1)}{2}$

Hockey stick

For $n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}$.

[asy] int chew(int n,int r){  int res=1;  for(int i=0;i<r;++i){   res=quotient(res*(n-i),i+1);   }  return res;  } for(int n=0;n<9;++n){  for(int i=0;i<=n;++i){   if((i==2 && n<8)||(i==3 && n==8)){    if(n==8){label(string(chew(n,i)),(11+n/2-i,-n),p=red+2.5);}    else{label(string(chew(n,i)),(11+n/2-i,-n),p=blue+2);}    }   else{    label(string(chew(n,i)),(11+n/2-i,-n));    }   }  } [/asy]

This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself is highlighted, a hockey-stick shape is revealed.


Inductive Proof

This identity can be proven by induction on $n$.

Base Case Let $n=r$.

$\sum^n_{i=r}{i\choose r}=\sum^r_{i=r}{i\choose r}={r\choose r}=1={r+1\choose r+1}$.

Inductive Step Suppose, for some $k\in\mathbb{N}, k>r$, $\sum^k_{i=r}{i\choose r}={k+1\choose r+1}$. Then $\sum^{k+1}_{i=r}{i\choose r}=\left(\sum^k_{i=r}{i\choose r}\right)+{k+1\choose r}={k+1\choose r+1}+{k+1\choose r}={k+2\choose r+1}$.

Algebraic Proof

It can also be proven algebraically with Pascal's Identity, ${n \choose k}={n-1\choose k-1}+{n-1\choose k}$. Note that

${r \choose r}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+1 \choose r+1}+{r+1 \choose r}+{r+2 \choose r}+\cdots+{r+a \choose r}$ $={r+2 \choose r+1}+{r+2 \choose r}+\cdots+{r+a \choose r}=\cdots={r+a \choose r+1}+{r+a \choose r}={r+a+1 \choose r+1}$, which is equivalent to the desired result.

Combinatorial Proof 1

Imagine that we are distributing $n$ indistinguishable candies to $k$ distinguishable children. By a direct application of Balls and Urns, there are ${n+k-1\choose k-1}$ ways to do this. Alternatively, we can first give $0\le i\le n$ candies to the oldest child so that we are essentially giving $n-i$ candies to $k-1$ kids and again, with Balls and Urns, ${n+k-1\choose k-1}=\sum_{i=0}^n{n+k-2-i\choose k-2}$, which simplifies to the desired result.

Combinatorial Proof 2

We can form a committee of size $k+1$ from a group of $n+1$ people in ${{n+1}\choose{k+1}}$ ways. Now we hand out the numbers $1,2,3,\dots,n-k+1$ to $n-k+1$ of the $n+1$ people. We can divide this into $n-k+1$ disjoint cases. In general, in case $x$, $1\le x\le n-k+1$, person $x$ is on the committee and persons $1,2,3,\dots, x-1$ are not on the committee. This can be done in $\binom{n-x+1}{k}$ ways. Now we can sum the values of these $n-k+1$ disjoint cases, getting \[{{n+1}\choose {k+1}} ={{n}\choose{k}}+{{n-1}\choose{k}}+{{n-2}\choose{k}}+\hdots+{{k+1}\choose{k}}+{{k}\choose{k}}.\]



The branch of mathematics concerned with the properties and relations of points, lines, surfaces, solids, and higher dimensional analogs.


We are not going to go into zero-d because that means nothing (literally). It is just an infinitely small dot called a point. (Duh). One-D or One dimension either a line (infinitely long, straight and perfectly narrow in both directions) a, a ray (infinitely long line in one direction) or a line segment (any length, straight and perfectly narrow in both directions). This is still pretty boring. Ya, I thought so.




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