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Difference between revisions of "What is the definition of Pure Mathematics?"

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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).
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This article is not finished. Everyone is welcomed to edit, BUT ONLY IN GOOD WAYS! The AoPS Secret Governemtn has a backup of this page.
  
 
= What is the definition of Pure Mathematics? =
 
= What is the definition of Pure Mathematics? =
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= Arithmetic =
 
= Arithmetic =
−
=== Definition ==
+
== Definition ==
−
''The branch of mathematics dealing with the properties and manipulation of constants''(a number that can not changed. This is the oppisite of a varible which is most of the tie repersented by a letter. This letter is most of the time <math>x</math>,<math>y</math> or <math>z</math> however it can be really any non-used symbol.).
+
''The branch of mathematics dealing with the properties and manipulation of constants''(a number that can not changed. This is the opposite of a variable which is most of the tie represented by a letter. This letter is most of the time <math>x</math>,<math>y</math> or <math>z</math> however it can be really any non-used symbol.).
  
 
=== Operations ===
 
=== Operations ===
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<math>f(3)=6</math>), negative numbers and zero were defined. A negitive number is a number below zero.
 
<math>f(3)=6</math>), negative numbers and zero were defined. A negitive number is a number below zero.
 
   
 
   
−
Division is the inverse of multiplication. To make this a well-defined function everywhere, fractions were defined. 3 divided by 2 is <math>\frac{3}{2}</math>.
+
The division is the inverse of multiplication. To make this a well-defined function everywhere, fractions were defined. 3 divided by 2 is <math>\frac{3}{2}</math>.
  
−
<math>n^{\text{th}}</math> roots and logarithms are the inverses of exponentiation. To make these well-defined functions everywhere, irrational numbers were defined. An irrational number is a number that can not be be expresd as <math>\frac{a}{b}.
+
<math>n^{\text{th}}</math> roots and logarithms are the inverses of exponentiation. To make these well-defined functions everywhere, irrational numbers were defined. An irrational number is a number that can not be expressed as <math>\frac{a}{b}</math>.
  
 
==== Negative numbers ====
 
==== Negative numbers ====
  
−
</math>a<math> and </math>b<math> are positive. (Above zero)
+
<math>a</math> and <math>b</math> are positive. (Above zero)
  
−
1. </math>(-a)(-b)=ab<math> Note:  
+
1. <math>(-a)(-b)=ab</math> Note:  
  
−
2. </math>(a)(-b)=-ab<math>
+
2. <math>(a)(-b)=-ab</math>
  
 
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 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 </math>(a)(-b)<math> is repeated subtraction. Therefore it is negative.
+
Proof for 2: Since <math>ab</math> is repeated addition then <math>(a)(-b)</math> is repeated subtraction. Therefore it is negative.
  
 
==== Fractions ====
 
==== Fractions ====
−
A fraction a number that can be expressed as two numbers divided. For example, five divided by four is </math>\frac{5}{4}<math>.  
+
A fraction a number that can be expressed as two numbers divided. For example, five divided by four is <math>\frac{5}{4}</math>.  
 
===== Simplifing fractions =====
 
===== Simplifing fractions =====
−
Find common factors in each half of the fration and then divide top and bottam of the fraction by that factor.
+
Find common factors in each half of the fraction and then divide top and bottom of the fraction by that factor.
  
−
</math>\frac{10}{5}=\frac{1}{2}<math>
+
<math>\frac{10}{5}=\frac{1}{2}</math>
  
−
</math>\frac{15}{3}=\frac{5}{1}<math>
+
<math>\frac{15}{3}=\frac{5}{1}</math>
  
−
</math>\frac{28}{2}=\frac{14}{1}<math>
+
<math>\frac{28}{2}=\frac{14}{1}</math>
 
===== Adding fractions =====
 
===== Adding fractions =====
−
Here is how you add fractions. Is they have the same bottom half then </math>\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}<math>. However, if the two or three or n fractions do not have the same bottom half you make them. </math>\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}<math>. Subtracting fractions is the same exept everything has a minus symbol. Remember to fully simplify.
+
Here is how you add fractions. Is they have the same bottom half then <math>\frac{a}{b}+\frac{c}{b}=\frac{a+c}{b}</math>. However, if the two or three or n fractions do not have the same bottom half you make them. <math>\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}</math>. Subtracting fractions is the same except everything has a minus symbol. Remember to fully simplify.
  
−
===== Multilpying fractions =====
+
===== Multiplying fractions =====
−
</math>\frac{a}{b}\frac{c}{d}=\frac{ac}{bd}<math>.  
+
<math>\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}</math>.  
  
−
</math>\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ac}{bd}<math>.
+
<math>\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ac}{bd}</math>.
  
 
Remember to fully simplify.
 
Remember to fully simplify.
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Most square roots are irrational. An irrational number is a number that can not be expresed a a fraction.
 
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 =====
 
===== Proof that any nonperfect square positive integer is irrational =====
−
Let us assume that </math>\sqrt{n}<math> is rational where </math>n<math> is a nonperfect square positive integer. Then it can be written as </math>\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2<math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore </math>\sqrt{n}<math> is irrational.
+
Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
  
 
===== pi and e =====
 
===== pi and e =====
−
Some irrational numbers are limits. That means they are the sum of smaller and smaller fraction going infinitely long. Pi or </math>\pi<math> (go to the geometry part of this article) is the limit </math>\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...=\frac{\pi}{4}<math>. e is the limit </math>\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}...<math>  More at the counting part of the article.
+
Some irrational numbers are limits. That means they are the sum of smaller and smaller fraction going infinitely long. Pi or <math>\pi</math> (go to the geometry part of this article) is the limit <math>\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}...=\frac{\pi}{4}</math>. e is the limit <math>\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}...</math>  More at the counting part of the article.
  
 
=== Exponent rules ===
 
=== Exponent rules ===
 
Listed below are some important properties of exponents:
 
Listed below are some important properties of exponents:
−
# </math> b^x\cdot b^y = b^{x+y}<math>
+
# <math> b^x\cdot b^y = b^{x+y}</math>
−
# </math> b^{-x}=\frac 1{b^x} <math>
+
# <math> b^{-x}=\frac 1{b^x} </math>
−
# </math> \frac{b^x}{b^y}=b^{x-y} <math>
+
# <math> \frac{b^x}{b^y}=b^{x-y} </math>
−
# </math> (b^x)^y = b^{xy} <math>
+
# <math> (b^x)^y = b^{xy} </math>
−
# </math> (ab)^x = a^x b^x <math>
+
# <math> (ab)^x = a^x b^x </math>
−
# </math> b^0 = 1 <math> (if </math>b \neq 0<math>.  </math>0^0<math> is undefined.)
+
# <math> b^0 = 1 </math> (if <math>b \neq 0</math>.  <math>0^0</math> is undefined.)
  
 
Here are explanations of the properties listed above:
 
Here are explanations of the properties listed above:
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# This is described in the previous section.
 
# This is described in the previous section.
 
# This results from using the previous two properties.
 
# This results from using the previous two properties.
−
# We are multiplying </math>b^x<math> by itself '''y''' times, which is the same as multiplying '''b''' by itself '''xy''' times.
+
# We are multiplying <math>b^x</math> by itself '''y''' times, which is the same as multiplying '''b''' by itself '''xy''' times.
 
# After multiplying '''ab''' by itself '''x''' times, we can collect '''a''' and '''b''' terms, thus establishing the property.
 
# After multiplying '''ab''' by itself '''x''' times, we can collect '''a''' and '''b''' terms, thus establishing the property.
−
#Hoping that property #1 will be true when </math>y=0<math>, we see that </math>b^x\cdot b^0<math> should (hopefully) be equal to </math>b^x<math>.  Thus, we ''define'' </math>b^0<math> to be equal to </math>1<math> in order to make this be true.
+
#Hoping that property #1 will be true when <math>y=0</math>, we see that <math>b^x\cdot b^0</math> should (hopefully) be equal to <math>b^x</math>.  Thus, we ''define'' <math>b^0</math> to be equal to <math>1</math> in order to make this be true.
 +
 
 +
hello
  
 
= Algebra =
 
= Algebra =
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== One-variable linear equations ==
 
== One-variable linear equations ==
 
=== Definition ===
 
=== Definition ===
−
'A One-variable linear equation is an equation that comes in the form </math>ax+b=c<math>. </math>a<math>, </math>b<math>, and </math>c<math> are constants and </math>x<math> is the varible in which to solve for
+
'A One-variable linear equation is an equation that comes in the form <math>ax+b=c</math>. <math>a</math>, <math>b</math>, and <math>c</math> are constants and <math>x</math> is the varible in which to solve for
 
=== Problems ===
 
=== Problems ===
 
==== Problem 1 ====
 
==== Problem 1 ====
−
Solve for </math>x<math>: </math>x+3=7<math>
+
Solve for <math>x</math>: <math>x+3=7</math>
 
==== Solution 1 ====
 
==== Solution 1 ====
−
Subtract 3 from both sides you get </math>x=4<math>
+
Subtract 3 from both sides you get <math>x=4</math>
 
==== Problem 2 ====
 
==== Problem 2 ====
−
Solve for </math>x<math>: </math>3x+2=98<math>
+
Solve for <math>x</math>: <math>3x+2=98</math>
 
==== Solution 2 ====
 
==== Solution 2 ====
−
Subtract two: </math>3x=96<math>. Divide by three: </math>x=32<math>
+
Subtract two: <math>3x=96</math>. Divide by three: <math>x=32</math>
 
==== Problem 3 ====
 
==== Problem 3 ====
−
Solve for </math>x<math>: </math>\sqrt{x+3}=5<math>
+
Solve for <math>x</math>: <math>\sqrt{x+3}=5</math>
 
==== Solution 3 ====
 
==== Solution 3 ====
−
When dealing with square roots you just square roots you square them because that makes no more square roots. So </math>\sqrt{x+3}=5<math> squared is </math>x+3=25<math> which tells us that </math>x<math> is </math>22<math>.
+
When dealing with square roots you just square roots you square them because that makes no more square roots. So <math>\sqrt{x+3}=5</math> squared is <math>x+3=25</math> which tells us that <math>x</math> is <math>22</math>.
 
==== Problem 4 ====
 
==== Problem 4 ====
−
Solve for </math>x<math>: </math>\frac{4}{x}=8<math>
+
Solve for <math>x</math>: <math>\frac{4}{x}=8</math>
 
==== Solution 4 ====
 
==== Solution 4 ====
−
When dealing with fractions you multiply by the denaminator: </math>4=8x<math> which tells us that </math>x=\frac{1}{2}<math>.
+
When dealing with fractions you multiply by the denaminator: <math>4=8x</math> which tells us that <math>x=\frac{1}{2}</math>.
  
 
== Quadratics ==
 
== Quadratics ==
 
=== Defination ===
 
=== Defination ===
−
''A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is </math>ax^2 + bx + c = 0<math> with </math>a<math>, </math>b<math>, and </math>c<math> being constants, or numerical coefficients, and </math>x<math> is an unknown variable.''
+
''A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared. The standard form is <math>ax^2 + bx + c = 0</math> with <math>a</math>, <math>b</math>, and <math>c</math> being constants, or numerical coefficients, and <math>x</math> is an unknown variable.''
  
 
=== The answer is always... ===
 
=== The answer is always... ===
−
</math>ax^2+bx+c=0<math>
+
<math>ax^2+bx+c=0</math>
  
−
</math>x^2+\frac{b}{a}x+\frac{c}{a}=0<math>
+
<math>x^2+\frac{b}{a}x+\frac{c}{a}=0</math>
  
−
</math>x^2+\frac{b}{a}=0-\frac{c}{a}<math>
+
<math>x^2+\frac{b}{a}=0-\frac{c}{a}</math>
  
−
</math>x^2+\frac{b}{a}+(\frac{b}{2a})^2=0-\frac{c}{a}+(\frac{b}{2a})^2<math>
+
<math>x^2+\frac{b}{a}+(\frac{b}{2a})^2=0-\frac{c}{a}+(\frac{b}{2a})^2</math>
  
  
−
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.
+
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 (<math>x</math>) and the base of the last term (<math>\frac{b}{2a}</math>). Add these two and raise everything to the second.
  
−
</math>(x+\frac{b}{2a})^2=0-\frac{c}{a}+\frac{b^2}{2a^2}=\frac{b^2-4ac}{4a^2}<math>
+
<math>(x+\frac{b}{2a})^2=0-\frac{c}{a}+\frac{b^2}{2a^2}=\frac{b^2-4ac}{4a^2}</math>
  
−
</math>x+\frac{b}{2a}=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}<math>
+
<math>x+\frac{b}{2a}=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}</math>
  
−
</math>x=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}+\frac{b}{2a}<math>
+
<math>x=\frac{\sqrt{\pm b^2-4ac}}{\pm 2a}+\frac{b}{2a}</math>
  
−
When simplified the ''Quadratic Formula'' is  </math>{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}<math>
+
When simplified the ''Quadratic Formula'' is  <math>{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}</math>
  
  
 
===== Problem =====
 
===== Problem =====
−
Find all </math>x<math> if </math>x^2+x+1=0<math>
+
Find all <math>x</math> if <math>x^2+x+1=0</math>
 
===== Solution =====
 
===== Solution =====
−
Using our formula we see that </math>x=\frac{-1 \pm \sqrt{1^2-1(1)(1)}}{2(1)}=-\frac{1}{2}<math>
+
Using our formula we see that <math>x=\frac{-1 \pm \sqrt{1^2-1(1)(1)}}{2(1)}=-\frac{1}{2}</math>
  
 
== i ==
 
== i ==
−
</math>i=\sqrt{-1}<math>.
+
<math>i=\sqrt{-1}</math>.
  
−
</math>xi=\sqrt{-x^2}<math>.
+
<math>xi=\sqrt{-x^2}</math>.
  
−
Numbers like this are called ''imaginary numbers''. ''Impossible'', you say. But no. Solve </math>x^2+1=0<math>. You get </math>i<math>. </math>oi<math> is </math>o<math>. So zero is both real and imaginary. (real means ''not imaginary'')
+
Numbers like this are called ''imaginary numbers''. ''Impossible'', you say. But no. Solve <math>x^2+1=0</math>. You get <math>i</math>. <math>oi</math> is <math>o</math>. So zero is both real and imaginary. (real means ''not imaginary'')
  
 
=== Powers of i ===
 
=== Powers of i ===
  
−
</math>i^0=1<math>
+
<math>i^0=1</math>
−
</math>i^1=i<math>
+
<math>i^1=i</math>
−
</math>i^2=-1<math>
+
<math>i^2=-1</math>
−
</math>i^3=-i<math>
+
<math>i^3=-i</math>
−
</math>i^4=1<math>
+
<math>i^4=1</math>
−
</math>i^5=i<math>
+
<math>i^5=i</math>
−
</math>i^6=-1<math>
+
<math>i^6=-1</math>
−
</math>i^7=-i<math>
+
<math>i^7=-i</math>
  
 
The pattern repeats.  
 
The pattern repeats.  
  
 
=== Complex numbers ===
 
=== Complex numbers ===
−
A complex number is </math>ai+b<math>, where a and b are real. All numbers are complex becuase a and/or/never b can be zero.
+
A complex number is <math>ai+b</math>, where a and b are real. All numbers are complex because a and/or/never b can be zero.
  
−
</math>ai+b+ci+d=<math> Complex
+
<math>ai+b+ci+d=</math> Complex
  
−
</math>(ai+b)(ci+d)=<math> Complex
+
<math>(ai+b)(ci+d)=</math> Complex
  
−
</math>\frac{ai+b}{ci+d}=<math> Complex
+
<math>\frac{ai+b}{ci+d}=</math> Complex
  
−
</math>(ai+b)-(ci+d)=<math> Complex
+
<math>(ai+b)-(ci+d)=</math> Complex
  
 
== Systems of equations ==
 
== Systems of equations ==
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{| class="wikitable" style="margin: 1em auto 1em auto"
 
{| class="wikitable" style="margin: 1em auto 1em auto"
−
| </math>2a - 3b<math> || </math>= 4<math>
+
| <math>2a - 3b</math> || <math>= 4</math>
 
|-
 
|-
−
| </math>3a - 2b<math> || </math>= 3<math>
+
| <math>3a - 2b</math> || <math>= 3</math>
 
|}
 
|}
  
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===== Problem =====
 
===== Problem =====
−
Find the ordered pair </math>(x,y)<math> for which
+
Find the ordered pair <math>(x,y)</math> for which
  
 
{| class="wikitable" style="margin: 1em auto 1em auto"
 
{| class="wikitable" style="margin: 1em auto 1em auto"
−
| </math>x - 12y<math> || </math>= 2<math>
+
| <math>x - 12y</math> || <math>= 2</math>
 
|-
 
|-
−
| </math>3x + 6y<math> || </math>= 6<math>
+
| <math>3x + 6y</math> || <math>= 6</math>
 
|}
 
|}
  
 
===== Solution =====
 
===== Solution =====
−
We can eliminate </math>y<math> by adding twice the second equation to the first:
+
We can eliminate <math>y</math> by adding twice the second equation to the first:
  
 
{| class="wikitable" style="margin: 1em auto 1em auto"
 
{| class="wikitable" style="margin: 1em auto 1em auto"
−
| || </math>x - 12y= 2<math>
+
| || <math>x - 12y= 2</math>
 
|-
 
|-
−
| </math>+2(<math> || </math>3x + 6y = 6)<math>
+
| <math>+2(</math> || <math>3x + 6y = 6)</math>
 
|-
 
|-
−
| || </math>\overline{7x + 0=14}<math>
+
| || <math>\overline{7x + 0=14}</math>
 
|}
 
|}
  
−
Thus </math>x=2<math>.  We can then plug in for </math>x<math> in either of the equations:  <center></math> (2)-12y = 2 \Rightarrow y = 0<math>.</center>
+
Thus <math>x=2</math>.  We can then plug in for <math>x</math> in either of the equations:  <center><math> (2)-12y = 2 \Rightarrow y = 0</math>.</center>
  
−
Thus, the solution to the system is </math>(2,0)<math>.
+
Thus, the solution to the system is <math>(2,0)</math>.
  
 
==== Substitution ====
 
==== Substitution ====
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===== Problem =====
 
===== Problem =====
−
Find the ordered pair </math>(x,y)<math> for which
+
Find the ordered pair <math>(x,y)</math> for which
  
 
{| class="wikitable" style="margin: 1em auto 1em auto"
 
{| class="wikitable" style="margin: 1em auto 1em auto"
−
| </math>x - 12y<math> || </math>= 2<math>
+
| <math>x - 12y</math> || <math>= 2</math>
 
|-
 
|-
−
| </math>3x + 6y<math> || </math>= 6<math>
+
| <math>3x + 6y</math> || <math>= 6</math>
 
|}
 
|}
  
 
===== Solution =====
 
===== Solution =====
−
The first equation can be solved for </math>x<math>:  <center></math> x = 12y + 2. <math></center>
+
The first equation can be solved for <math>x</math>:  <center><math> x = 12y + 2. </math></center>
  
−
Plugging this into the second equation yields <center></math>3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0. <math></center>
+
Plugging this into the second equation yields <center><math>3(12y + 2) + 6y = 6 \Leftrightarrow 42 y = 0. </math></center>
  
−
Thus </math>y=0<math>.  Plugging this into either of the equations and solving for </math>x<math> yields </math>x=2<math>.
+
Thus <math>y=0</math>.  Plugging this into either of the equations and solving for <math>x</math> yields <math>x=2</math>.
  
 
== Graphing Equations ==
 
== Graphing Equations ==
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=== Problem ===
 
=== Problem ===
−
Graph the line </math>2x + 3y = 24<math>.
+
Graph the line <math>2x + 3y = 24</math>.
  
 
=== Solution ===
 
=== Solution ===
−
To graph a line, it is necesasry to find two points </math>(x,y)<math> that satisfy </math>2x + 3y = 24<math>.  Letting </math>x=0<math> gives </math>3y = 24\Leftrightarrow y = 8<math>.  So </math>(0,8)<math> is one point on the graph.
+
To graph a line, it is necesasry to find two points <math>(x,y)</math> that satisfy <math>2x + 3y = 24</math>.  Letting <math>x=0</math> gives <math>3y = 24\Leftrightarrow y = 8</math>.  So <math>(0,8)</math> is one point on the graph.
  
−
Find another point by letting </math>y=0<math>.  Plugging this in and solving gives </math>x=12<math>.  So </math>(12,0)<math> is our other point.
+
Find another point by letting <math>y=0</math>.  Plugging this in and solving gives <math>x=12</math>.  So <math>(12,0)</math> is our other point.
  
 
Now plot these in the coordinate plane and draw a line through them:
 
Now plot these in the coordinate plane and draw a line through them:
Line 264: Line 266:
 
''The branch of mathematics that deals with the properties and relationships of numbers, especially the positive integers''.
 
''The branch of mathematics that deals with the properties and relationships of numbers, especially the positive integers''.
 
== Vocab ==
 
== Vocab ==
−
Factor: A factor (f) of n is an integer in which </math>fx=n<math> where </math>x<math> is an integer.
+
Factor: A factor (f) of n is an integer in which <math>fx=n</math> where <math>x</math> is an integer.
  
 
Multiple: The inverse of factors.
 
Multiple: The inverse of factors.
  
 
== Primes ==
 
== Primes ==
−
A '''prime number''' (or simply '''prime''') is a positive integer </math>p>1<math> whose only positive divisors are 1 and itself.   
+
A '''prime number''' (or simply '''prime''') is a positive integer <math>p>1</math> whose only positive divisors are 1 and itself.   
−
Note that </math>1<math> is usually defined as being neither prime nor composite because it is its only factor among the natural number numbers.  
+
Note that <math>1</math> 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 </math>p_1, p_2, \ldots, p_n<math>, then the number </math>N = p_1p_2\cdots p_n + 1<math> is not divisible by any of them, but </math>N<math> must [[#Importance of Primes|have]] a prime factor, which leads to a direct contradiction.  
+
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 <math>p_1, p_2, \ldots, p_n</math>, then the number <math>N = p_1p_2\cdots p_n + 1</math> is not divisible by any of them, but <math>N</math> must [[#Importance of Primes|have]] a prime factor, which leads to a direct contradiction.  
  
 
=== Techniques to Check for Prime Numbers ===
 
=== Techniques to Check for Prime Numbers ===
  
 
==== Divisibility ====
 
==== Divisibility ====
−
A prime number is only divisible by one or itself, so a number </math>n<math> is prime if and only if </math>n<math> is not divisible by any integer greater than </math>1<math> and less than </math>n<math>.  One only needs to check integers up to </math>\sqrt{n}<math> because dividing larger numbers would result in a quotient smaller than </math>\sqrt{n}<math>.
+
A prime number is only divisible by one or itself, so a number <math>n</math> is prime if and only if <math>n</math> is not divisible by any integer greater than <math>1</math> and less than <math>n</math>.  One only needs to check integers up to <math>\sqrt{n}</math> because dividing larger numbers would result in a quotient smaller than <math>\sqrt{n}</math>.
  
 
==== Modular Arithmetic ====
 
==== Modular Arithmetic ====
 
Modular arithmetic can help determine if a number is not prime.
 
Modular arithmetic can help determine if a number is not prime.
  
−
* If a number not equal to </math>2,3<math> is congruent to </math>0,2,3,4,6 \pmod{6}<math>, then the number is not prime.
+
* If a number not equal to <math>2,3</math> is congruent to <math>0,2,3,4,6 \pmod{6}</math>, then the number is not prime.
−
* If a number not equal to </math>2,5<math> ends with an even digit or </math>5<math>, then the number is not prime.
+
* If a number not equal to <math>2,5</math> ends with an even digit or <math>5</math>, then the number is not prime.
  
  
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==== Mersenne Primes ====
 
==== Mersenne Primes ====
  
−
A Mersenne prime is a prime of the form </math>2^n-1<math>. 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.
+
A Mersenne prime is a prime of the form <math>2^n-1</math>. 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 ====
 
==== Twin Primes ====
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=== Advanced Definition ===
 
=== 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 </math>p=ab<math> with </math>a,b\in R<math>, then exactly one of ''a'' and ''b'' is a unit.
+
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 <math>p=ab</math> with <math>a,b\in R</math>, then exactly one of ''a'' and ''b'' is a unit.
  
 
== LCM and GCD ==
 
== LCM and GCD ==
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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.
 
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 ====
 
==== 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 </math>2 \cdot 2 \cdot 2 \cdot 3<math>. The prime factorization of 18 is </math>2 \cdot 3 \cdot 3<math>. In GCD you take the prime factorization of each number and then find which primes match up.
+
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 <math>2 \cdot 2 \cdot 2 \cdot 3</math>. The prime factorization of 18 is <math>2 \cdot 3 \cdot 3</math>. In GCD you take the prime factorization of each number and then find which primes match up.
  
 
=== LCM ===
 
=== LCM ===
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What is the LCM of 8 and 5?
 
What is the LCM of 8 and 5?
 
====== Solution ======
 
====== Solution ======
 +
1.
 
The way to solve LCM's with no common factors is to multiply them. The answer is just 40.
 
The way to solve LCM's with no common factors is to multiply them. The answer is just 40.
 +
 +
2.
 +
Note that <math>8=2^3,</math> <math>5=5^1.</math> Taking the maximum of all such prime factors yields <math>40=2^3\cdot 5.</math>
 +
 +
This is a pathetic solution, but this is the more general solution.
 +
 
===== Problem 2 =====
 
===== Problem 2 =====
 
====== Problem ======
 
====== Problem ======
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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.
 
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 ====
 
==== 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, </math>{23_{8}}<math> means 23 in base 8 (which is equal to 19 in base 10).
+
A base can be any whole number bigger than 1. The base of a number may be written next to the number: for instance, <math>{23_{8}}</math> 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:
 
Here are some examples of how numbers are written in varying bases, compared to decimal:
  
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== How to ==
 
== How to ==
 
The way to mesure the extent to which an event is likely to occur is to count the number of equally likely occur is to count the wanted options over the number of equally likelly possible options.
 
The way to mesure the extent to which an event is likely to occur is to count the number of equally likely occur is to count the wanted options over the number of equally likelly possible options.
−
For example, the probibility of rolling a 6 on a 6 sided dice is </math>\frac{1}{6}<math> because there are six equally likely possibilitys and only one of them is a success.
+
For example, the probibility of rolling a 6 on a 6 sided dice is <math>\frac{1}{6}</math> because there are six equally likely possibilitys and only one of them is a success.
  
  
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Pascal's Identity states that
 
Pascal's Identity states that
  
−
</math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}<math>
+
<math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>
  
−
for any positive integers </math>k<math> and </math>n<math>.  Here, </math>\binom{n}{k}<math> is the binomial coefficient </math>\binom{n}{k} = nCk = C_k^n<math>.
+
for any positive integers <math>k</math> and <math>n</math>.  Here, <math>\binom{n}{k}</math> is the binomial coefficient <math>\binom{n}{k} = nCk = C_k^n</math>.
  
−
Remember that </math>\binom{n}{r}=\frac{n!}{k!(n-k)!}<math>. </math>\binom{n}{r}<math> means the number of ways to pick r thing from n things where order does not matter. </math>n!=1 \cdot 2 \cdot 3 \cdot...\cdot n<math>.
+
Remember that <math>\binom{n}{r}=\frac{n!}{k!(n-k)!}</math>. <math>\binom{n}{r}</math> means the number of ways to pick r thing from n things where order does not matter. <math>n!=1 \cdot 2 \cdot 3 \cdot...\cdot n</math>.
  
 
=== Proving it ===
 
=== Proving it ===
−
If </math>k > n<math> then </math>\binom{n}{k} = 0 = \binom{n - 1}{k - 1} + \binom{n - 1}{k}<math> and so the result is pretty clear.
+
If <math>k > n</math> then <math>\binom{n}{k} = 0 = \binom{n - 1}{k - 1} + \binom{n - 1}{k}</math> and so the result is pretty clear.
−
So assume </math>k \leq n<math>.  Then
+
So assume <math>k \leq n</math>.  Then
  
 
<cmath>\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)!}\\
 
<cmath>\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)!}\\
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Pascal's Triangle is really combinations. It looks something like this if it is depicted as combinations:
 
Pascal's Triangle is really combinations. It looks something like this if it is depicted as combinations:
  
−
                               </math>\binom{0}{0}<math>                         
+
                               <math>\binom{0}{0}</math>                         
−
                 </math>\binom{1}{0}<math>                  </math>\binom{1}{1}<math>
+
                 <math>\binom{1}{0}</math>                  <math>\binom{1}{1}</math>
−
     </math>\binom{2}{0}<math>                    </math>\binom{2}{1}<math>                </math>\binom{2}{2}<math>
+
     <math>\binom{2}{0}</math>                    <math>\binom{2}{1}</math>                <math>\binom{2}{2}</math>
  
 
And on and on...
 
And on and on...
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==== Proof ====
 
==== Proof ====
  
−
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:  </math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}<math>. Its Pascals Identity! Therefore each row looks something like this:
+
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:  <math>{n \choose k}={n-1\choose k-1}+{n-1\choose k}</math>. Its Pascals Identity! Therefore each row looks something like this:
  
−
</math>\binom{n}{0}  \binom{n}{1}  \binom{n}{2} ... \binom{n}{n}<math>
+
<math>\binom{n}{0}  \binom{n}{1}  \binom{n}{2} ... \binom{n}{n}</math>
  
 
== Patterns and Properties ==
 
== Patterns and Properties ==
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=== Binomial Theorem ===
 
=== Binomial Theorem ===
 
Let's multiply out some binomials. Try it yourself and it will not be fun:
 
Let's multiply out some binomials. Try it yourself and it will not be fun:
−
</math>(x+y)^0=1<math>
+
<math>(x+y)^0=1</math>
  
−
</math>(x+y)^1=1x+1y<math>
+
<math>(x+y)^1=1x+1y</math>
  
−
</math>(x+y)^2=1x^2+2xy+1y^2<math>
+
<math>(x+y)^2=1x^2+2xy+1y^2</math>
  
−
</math>(x+y)^2=1x^3+3x^2y+3y^2x+1^3<math>
+
<math>(x+y)^2=1x^3+3x^2y+3y^2x+1^3</math>
  
 
If you take away the x's and y's you get:
 
If you take away the x's and y's you get:
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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:
 
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 </math>(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}<math>. Repeatedly using the distributive property, we see that for a term </math>a^m b^{n-m}<math>, we must choose </math>m<math> of the </math>n<math> terms to contribute an </math>a<math> to the term, and then each of the other </math>n-m<math> terms of the product must contribute a </math>b<math>. Thus, the coefficient of </math>a^m b^{n-m}<math> is the number of ways to choose </math>m<math> objects from a set of size </math>n<math>, or </math>\binom{n}{m}<math>. Extending this to all possible values of </math>m<math> from </math>0<math> to </math>n<math>, we see that </math>(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}<math>, as claimed.
+
We can write <math>(a+b)^n=\underbrace{ (a+b)\cdot(a+b)\cdot(a+b)\cdot\cdots\cdot(a+b) }_{n}</math>. Repeatedly using the distributive property, we see that for a term <math>a^m b^{n-m}</math>, we must choose <math>m</math> of the <math>n</math> terms to contribute an <math>a</math> to the term, and then each of the other <math>n-m</math> terms of the product must contribute a <math>b</math>. Thus, the coefficient of <math>a^m b^{n-m}</math> is the number of ways to choose <math>m</math> objects from a set of size <math>n</math>, or <math>\binom{n}{m}</math>. Extending this to all possible values of <math>m</math> from <math>0</math> to <math>n</math>, we see that <math>(a+b)^n = \sum_{m=0}^{n}{\binom{n}{m}}\cdot a^m\cdot b^{n-m}</math>, as claimed.
  
−
Similarly, the coefficients of </math>(x+y)^n<math> will be the entries of the </math>n^\text{th}<math> row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS]
+
Similarly, the coefficients of <math>(x+y)^n</math> will be the entries of the <math>n^\text{th}</math> row of Pascal's Triangle. This is explained further in the Counting and Probability textbook [AoPS]
 
===== In real life =====
 
===== In real life =====
 
It is really only used for multipling out binomials. More usage at  
 
It is really only used for multipling out binomials. More usage at  
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===== Theorem =====
 
===== Theorem =====
 
====== Theorem ======
 
====== Theorem ======
−
It states that </math>\binom{n}{0}+\binom{n}{1}+...+\binom{n}{n}=2^n<math>
+
It states that <math>\binom{n}{0}+\binom{n}{1}+...+\binom{n}{n}=2^n</math>
  
 
====== Why do we need it? ======
 
====== Why do we need it? ======
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===== Proofs =====
 
===== Proofs =====
 
====== Subset proof ======
 
====== 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 </math>(2)(2)(2)(2)<math>...n times. </math>2^n<math>.
+
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 <math>(2)(2)(2)(2)</math>...n times. <math>2^n</math>.
  
 
====== Alternate proof ======
 
====== Alternate proof ======
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If you look at the numbers in the third diagonal you see that they are triangle numbers.
 
If you look at the numbers in the third diagonal you see that they are triangle numbers.
 
===== Proof =====
 
===== Proof =====
−
Now we can make an equation: </math>\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}<math>
+
Now we can make an equation: <math>\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}</math>
  
 
==== Hockey stick ====
 
==== Hockey stick ====
−
For </math>n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}$.
+
For <math>n,r\in\mathbb{N}, n>r,\sum^n_{i=r}{i\choose r}={n+1\choose r+1}</math>.
  
  
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https://www.google.com/search?q=classification+of+quadrilaterals&rlz=1C1JZAP_enUS854US854&tbm=isch&source=iu&ictx=1&fir=cTz-Mz0bfMvP3M%253A%252Cr9XExtRx3Y9j7M%252C_&vet=1&usg=AI4_-kR3H3HLnjvGfeFONyn0jA2dKvnQmQ&sa=X&ved=2ahUKEwi_h7_h7f_iAhXUvp4KHbYkCrAQ9QEwAHoECAEQAw#imgrc=vAqIbgZXY5hQqM:&vet=1
 
https://www.google.com/search?q=classification+of+quadrilaterals&rlz=1C1JZAP_enUS854US854&tbm=isch&source=iu&ictx=1&fir=cTz-Mz0bfMvP3M%253A%252Cr9XExtRx3Y9j7M%252C_&vet=1&usg=AI4_-kR3H3HLnjvGfeFONyn0jA2dKvnQmQ&sa=X&ved=2ahUKEwi_h7_h7f_iAhXUvp4KHbYkCrAQ9QEwAHoECAEQAw#imgrc=vAqIbgZXY5hQqM:&vet=1
 
= The Proof Part of this article =
 
= The Proof Part of this article =
−
This part shows what Proves ussually look like
+
This part shows what Proofs usually look like
 
== Proof that any nonperfect square positive integer is irrational ==
 
== Proof that any nonperfect square positive integer is irrational ==
 
Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
 
Let us assume that <math>\sqrt{n}</math> is rational where <math>n</math> is a nonperfect square positive integer. Then it can be written as <math>\frac{p}{q}=\sqrt{n} \Rightarrow \frac{p^2}{q^2}=n \Rightarrow (q^2)n=p^2</math>. But no perfect square times a nonperfect square positive integer is a perfect square. Therefore <math>\sqrt{n}</math> is irrational.
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Proof that <math>1+2+3...+n+1+2+3...+(n-1)=n^2</math>
 
Proof that <math>1+2+3...+n+1+2+3...+(n-1)=n^2</math>
 
=== Proof 1 ===
 
=== Proof 1 ===
−
<math>1+2+3...+n+1+2+3...+(n-1)=n^2 \Rightarrow \frac{n(n+1)}{2}+\frac{n-1(n)}{2}=n^2 \Rightarrow \frac{n(n+1)+n-1(n)}{2}=n^2 \Rightarrow \frac{n^2+n+n^2-n}{2}=n^2 \Rightarrow \frac{2n^2}{2}=n^2 \Rightarrow n^2=n^2</math>.
+
<math>1+2+3...+n+1+2+3...+(n-1)=n^2 \Rightarrow \frac{n(n+1)}{2}+\frac{n-1(n)}{2}=n^2</math>
 +
<math>\Rightarrow \frac{n(n+1)+n-1(n)}{2}=n^2 \Rightarrow \frac{n^2+n+n^2-n}{2}=n^2 \Rightarrow \frac{2n^2}{2}=n^2 \Rightarrow n^2=n^2</math>.
 
=== Proof 2 ===
 
=== Proof 2 ===
 
The <math>1+2+\cdots+n</math> part refers to an <math>n</math> by <math>n</math> square cut by its diagonal, and includes all the squares on the diagonal. The  <math>1+2+\cdots+ n-1</math> part refers to an <math>n</math> by <math>n</math> square cut by its diagonal, but doesn't include the squares on the diagonal. Putting these together gives us a <math>n</math> by <math>n</math> square.
 
The <math>1+2+\cdots+n</math> part refers to an <math>n</math> by <math>n</math> square cut by its diagonal, and includes all the squares on the diagonal. The  <math>1+2+\cdots+ n-1</math> part refers to an <math>n</math> by <math>n</math> square cut by its diagonal, but doesn't include the squares on the diagonal. Putting these together gives us a <math>n</math> by <math>n</math> square.
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And that is <math>n^2</math>.
 
And that is <math>n^2</math>.
 
[[Category:Math]]
 
[[Category:Math]]
 +
[[Category:Definition]]
 +
[[Category:Mathematics]]

Latest revision as of 08:20, 28 September 2024

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Contents

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.

Arithmetic

Definition

The branch of mathematics dealing with the properties and manipulation of constants(a number that can not changed. This is the opposite of a variable which is most of the tie represented by a letter. This letter is most of the time $x$,$y$ or $z$ however it can be really any non-used symbol.).

Operations

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

Addition and addition repeated

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

Multiplication is repeating addition. The symbol for this is $\cdot$. $a \cdot b=b \cdot a$. Note: When using letters, you can just say $ab=ba$.

Exponentiation is repeated Multiplication. The symbol for a to the exponent of b is $a^b$. $a^b \neq b^a$. The $\neq$ symbol means Not Equal.

Inverse

Subtraction is the inverse (remember that inverse means oppisite) of addition. To make this a well-defined function (the symbol for a funtion of a certian number($x$) is $f(x)=$blah blah blah where a function is a relationship or expression involving one or more variables. For example, if $f(x)=x+3$ then $f(3)=6$), negative numbers and zero were defined. A negitive number is a number below zero.

The division is the inverse of multiplication. To make this a well-defined function everywhere, fractions were defined. 3 divided by 2 is $\frac{3}{2}$.

$n^{\text{th}}$ roots and logarithms are the inverses of exponentiation. To make these well-defined functions everywhere, irrational numbers were defined. An irrational number is a number that can not be expressed as $\frac{a}{b}$.

Negative numbers

$a$ and $b$ are positive. (Above zero)

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

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.

Fractions

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 fraction and then divide top and bottom of the fraction by that factor.

$\frac{10}{5}=\frac{1}{2}$

$\frac{15}{3}=\frac{5}{1}$

$\frac{28}{2}=\frac{14}{1}$

Adding fractions

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

Multiplying fractions

$\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$.

$\frac{\frac{a}{b}}{\frac{c}{d}}=\frac{ac}{bd}$.

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.

hello

Algebra

Definition

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

Definition

'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 in which to solve for

Problems

Problem 1

Solve for $x$: $x+3=7$

Solution 1

Subtract 3 from both sides you get $x=4$

Problem 2

Solve for $x$: $3x+2=98$

Solution 2

Subtract two: $3x=96$. Divide by three: $x=32$

Problem 3

Solve for $x$: $\sqrt{x+3}=5$

Solution 3

When dealing with square roots you just square roots you square them because that makes no more square roots. So $\sqrt{x+3}=5$ squared is $x+3=25$ which tells us that $x$ is $22$.

Problem 4

Solve for $x$: $\frac{4}{x}=8$

Solution 4

When dealing with fractions you multiply by the denaminator: $4=8x$ which tells us that $x=\frac{1}{2}$.

Quadratics

Defination

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...

$ax^2+bx+c=0$

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

$x^2+\frac{b}{a}=0-\frac{c}{a}$

$x^2+\frac{b}{a}+(\frac{b}{2a})^2=0-\frac{c}{a}+(\frac{b}{2a})^2$


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 ($\frac{b}{2a}$). Add these two and raise everything to the second.

$(x+\frac{b}{2a})^2=0-\frac{c}{a}+\frac{b^2}{2a^2}=\frac{b^2-4ac}{4a^2}$

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

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

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


Problem

Find all $x$ if $x^2+x+1=0$

Solution

Using our formula we see that $x=\frac{-1 \pm \sqrt{1^2-1(1)(1)}}{2(1)}=-\frac{1}{2}$

i

$i=\sqrt{-1}$.

$xi=\sqrt{-x^2}$.

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 because 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:

Problem

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

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

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 $(2,0)$.

Substitution

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.

Problem

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

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

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 $x=2$.

Graphing Equations

Graphing Lines

Given two distinct points on a line, one can construct the whole line. So one way to graph a line given its equation is to just find two points on it and to draw a straight line through them.

Problem

Graph the line $2x + 3y = 24$.

Solution

To graph a line, it is necesasry to find two points $(x,y)$ that satisfy $2x + 3y = 24$. Letting $x=0$ gives $3y = 24\Leftrightarrow y = 8$. So $(0,8)$ is one point on the graph.

Find another point by letting $y=0$. Plugging this in and solving gives $x=12$. So $(12,0)$ is our other point.

Now plot these in the coordinate plane and draw a line through them:

Twopoints2.PNG

The arrowheads on the ends of the line segment indicate that the line goes on infinitely in both directions. Note: This is just a short intro to graphing. More at https://artofproblemsolving.com/wiki/index.php/Graph_of_a_function.

Algebra

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

Definition

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

Vocab

Factor: A factor (f) of n is an integer in which $fx=n$ where $x$ is an integer.

Multiple: The inverse of factors.

Primes

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

Divisibility

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.

LCM and GCD

GCD

Definition

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.

LCM

Definition

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.

Problems

Problem 1
Problem

What is the LCM of 8 and 5?

Solution

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

2. Note that $8=2^3,$ $5=5^1.$ Taking the maximum of all such prime factors yields $40=2^3\cdot 5.$

This is a pathetic solution, but this is the more general solution.

Problem 2
Problem

What is the LCM of 15 and 10.

Solution

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
Problem

Find the LCM of 2, 3 and 5.

Solution

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

Bases

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.

Probibility

Definition

The extent to which an event is likely to occur.

How to

The way to mesure the extent to which an event is likely to occur is to count the number of equally likely occur is to count the wanted options over the number of equally likelly possible options. For example, the probibility of rolling a 6 on a 6 sided dice is $\frac{1}{6}$ because there are six equally likely possibilitys and only one of them is a success.


Pascal's Triangle

Pascal's Identity

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

And on and on...

Combinations

Combinations

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

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

And on and on...

Proof

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: $(x+y)^0=1$

$(x+y)^1=1x+1y$

$(x+y)^2=1x^2+2xy+1y^2$

$(x+y)^2=1x^3+3x^2y+3y^2x+1^3$

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

         1
        1 1
       1 2 1
      1 3 3 1

It's Pascal's Triangle!

Proof

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

Theorem
Theorem

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.

Proofs
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

Theorem

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

Proof

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.


Proof

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 \[{{n+1}\choose {k+1}} ={{n}\choose{k}}+{{n-1}\choose{k}}+{{n-2}\choose{k}}+\hdots+{{k+1}\choose{k}}+{{k}\choose{k}}.\]

Geometry

Studying the Geometry of Shapes

Zero-D

This is an infinitly small dot called a point. (Duh). However, mathematicians imagine them to be large enough to be drawn.

One-D

1D is pretty boring too. You just have rays, lines, and line segments.

Line

A infinitly long, infinitly narrow, perfectly striaght thing in with no starting or ending point. (I cant really use the word lines).

Ray

A infinitly long, infinitly narrow, perfectly striaght thing in with one starting point.

Line segment

A infinitly long, infinitly narrow, perfectly striaght thing in with two starting points.

Space

Plane

In mathematics, a plane is a flat, two-dimensional surface that extends infinitely far.

3D Space

Space is the infinitly three-dimensional extent in which objects and events have relative position and direction.

Angles

An angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle. Angles formed by two rays lie in a plane, but this plane does not have to be a Euclidean plane. Angles are also formed by the intersection of two planes in Euclidean and other spaces. These are called dihedral angles. Angles formed by the intersection of two curves in a plane are defined as the angle determined by the tangent rays at the point of intersection. Similar statements hold in space, for example, the spherical angle formed by two great circles on a sphere is the dihedral angle between the planes determined by the great circles. Most people mesure angles useing degrees. Degrees are out of 360. For example, 180° or 180 degrees is a straight line or a half turn.

An angle equal to 0° is called a zero angle.

An angle smaller than 90° but not zero is called acute.

An angle equal to 90° is called right.

An angle greater than 90° but less than 180° is called obtuse.

An angle equal to 180° is called striaght.

An angle greater than 180° but less than 360° is called obtuse.

An angle equal to 360° is called a full angle.

Angles $n$ and $m$ supplementary if $n+m=$180° Angles $n$ and $m$ Complementary if $n+m=$90°

Vertical angles: Each of the pairs of opposite angles made by two intersecting lines.

Area

Area

The area of a 2D object is the amount of square units in it.

Shapes

Polygons

A plane shape (two-dimensional) with straight sides.

Examples: triangles, rectangles and pentagons.

(Note: a circle is not a polygon because it has a curved side).

Triangles and their classification

Angle Classifacation

Acute Triangle All angles are less than 90°

Right Triangle Has a right angle (90°)

Obtuse Triangle Has an angle more than 90°

Side Classifacation

Equilateral Triangle Three equal sides

Isosceles Triangle Two equal sides Two equal angles

Scalene Triangle No equal sides No equal angles

Quadrilaterals and their classification

https://www.google.com/search?q=classification+of+quadrilaterals&rlz=1C1JZAP_enUS854US854&tbm=isch&source=iu&ictx=1&fir=cTz-Mz0bfMvP3M%253A%252Cr9XExtRx3Y9j7M%252C_&vet=1&usg=AI4_-kR3H3HLnjvGfeFONyn0jA2dKvnQmQ&sa=X&ved=2ahUKEwi_h7_h7f_iAhXUvp4KHbYkCrAQ9QEwAHoECAEQAw#imgrc=vAqIbgZXY5hQqM:&vet=1

The Proof Part of this article

This part shows what Proofs usually look like

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.

Proof of the Pythagorean Theorem

$ABCD$ and $EFGH$ are squares.

[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label("$A$", A, NNW); label("$B$", B, ENE); label("$C$", C, ESE); label("$D$", D, SSW); draw(E--F--G--H--cycle); label("$E$", E, N); label("$F$", F,SE); label("$G$", G, S); label("$H$", H, W); label("a", A--B,N); label("a", B--F,SE); label("a", C--G,S); label("a", H--D,W); label("b", E--B,N); label("b", F--C,SE); label("b", G--D,S); label("b", A--H,W); label("c", E--H,NW); label("c", E--F); label("c", F--G,SE); label("c", G--H,SW); [/asy]

$(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2$.

Proof that n choose 2 is 1+2+3...+(n-1)

$\binom{n}{2}=1+2+3...+n \Rightarrow \binom{n}{2}=\frac{n(n-1)}{2} \Rightarrow \frac{n(n-1)}{2}=\frac{n(n-1)}{2}$.Yay, I proved it!

Proof that $1+2+3...+n+1+2+3...+(n-1)=n^2$

Proof that $1+2+3...+n+1+2+3...+(n-1)=n^2$

Proof 1

$1+2+3...+n+1+2+3...+(n-1)=n^2 \Rightarrow \frac{n(n+1)}{2}+\frac{n-1(n)}{2}=n^2$ $\Rightarrow \frac{n(n+1)+n-1(n)}{2}=n^2 \Rightarrow \frac{n^2+n+n^2-n}{2}=n^2 \Rightarrow \frac{2n^2}{2}=n^2 \Rightarrow n^2=n^2$.

Proof 2

The $1+2+\cdots+n$ part refers to an $n$ by $n$ square cut by its diagonal, and includes all the squares on the diagonal. The $1+2+\cdots+ n-1$ part refers to an $n$ by $n$ square cut by its diagonal, but doesn't include the squares on the diagonal. Putting these together gives us a $n$ by $n$ square.

Proof 3

We proceed using induction. If $n = 1$, then we have $1+0=1^2$. Now assume that $n$ works. We prove that $n+1$ works. We add a $2n+1$ on both sides, such that the left side becomes $1+2+\cdots + (n+1)+1+2+\cdots + n = n^2 + 2n + 1 = (n+1)^2$ and we are done.

Proof 4

$1 2 3 4 5 ... n$

$0 1 2 3 4 ... (n-1)$

________________

$1 3 5 7 9 ... 2n-1$ And that is $n^2$.

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