Skip to main content

Ooo Vectors

I had to learn basic things about vectors. So thought why not write a blog post :P? 

Also special thanks to papa 💫

 So here we go..

PHYSICAL QUANTITY:-

A quantity that can be measured and expressed in units. We have two types of Physical Quantity
  • Scalar:- A physical quantity that has only magnitude. Example: Mass, speed.
  • Vectors:- A physical quantity that has both magnitude and direction. Example: Displacement, velocity. 

Now, what are vectors?

VECTORS:-
  • A physical quantity that has both magnitude and direction.
  • In geometry, a vector is a directed line segment, whose length is the MAGNITUDE of the vector and the arrow indicating the DIRECTION.
  • The MAGNITUDE of a vector $(\vec{A})$ is the absolute value of a vector and is indicated as $\boxed{|\vec{A}|}.$
REPRESENTATION OF VECTORS:-
  • A vector pointing from $A$ to $B$ is written as $\vec{AB}.$
  • We usually define a single point $O$ as the ORIGIN. Then we associate every point $P$ with the vector $\vec{OP}$ or $\vec{P}$
MAGNITUDE OF VECTORS:-
  • It's just the distance between the initial point $A$ and endpoint $B.$
  • We consider the cartesian plane, let $A$ has coordinate $(x_0,y_0)$ and $B$ has coordinate $(x_1,y_1).$ 
  • Then  $\vec{AB}= ~~\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}.$
  • Letting $A$ as the origin and end point $B=~~(x,y),$ we get $\vec{AB}=~~\sqrt{x^2+y^2}$ 
EQUALITY OF VECTORS:-
  • Two vectors are called equal if their magnitudes and direction are the same. 
  • So a vector from $(0,0)$ to $(0,1)$ is same as $(0,2)$ to $(0,3).$
ADDITION OF VECTORS:- 
  • We use the Parallelogram Law to add vectors.
  • The vector pointing to $(x,y)$ from $(0,0)$ is denoted as $\langle x,y \rangle.$
  • If we have two vectors $\langle x_1,y_1 \rangle,~~\langle x_2,y_2 \rangle ,$ then adding we get $\langle x_1+x_2,~~y_1+y_2 \rangle.$
  • We also have another way. Let $\vec{AB}=\vec{a},~~ \vec{AC}= \vec{b}.$ Let $D$ be a point such that $ABDC$ is a Parallelogram. We say  $\vec{AB}+\vec{AC}=\vec{AD}.$
  • Now, what about the magnitude of $\vec{AD}?$
  • Let's drop the perpendicular from $D$ to $AB$ and call the feet $E.$ 
  • Suppose $\angle CAB=\theta$ and $\angle DAB=\alpha .$
  • So we get $$AD^2=~~(AB+BE)^2+(DE)^2$$ $$= (|\vec{a}|+\vec{b}\cos \theta)^2+~~(|\vec{b}|\sin \theta)^2$$ $$= |\vec{a}|^2+|\vec{a}||\vec{b}|\cos \theta+|\vec{b}|^2.$$
  • Hence, the magnitude of $$\boxed{ \vec{a}+ \vec{b}=~~\sqrt{ |\vec{a}|^2+|\vec{a}||\vec{b}|\cos \theta+|\vec{b}|^2}}$$

FUN FACT:- The sum of the coordinate of the points of a parallelogram are always equal.

Note:- We also have triangle law of addition, but that gives the same result ( here we basically join one head of the vector and one vector's tail and the resultant vector form the third side of the triangles, whose two sides are the two given vectors. )

More like, we have $\vec{BD}=\vec{b}$ and we are applying the triangle law. I think parallelogram is more famous (?)

MULTIPLICATION OF A VECTOR BY A SCALAR:- 
  • Suppose $\vec{a}$ is a vector and $k$ be any real number ( A scalar quantity).
  • We define $\vec{b}=k\vec{a}.$ And the magnitude would be $k|\vec{a}|.$
  • The vector $\vec{b}$ has same direction if $k$ is positive, opposite direction if $k$ is negetive.
SUBTRACTION OF VECTORS:- 
  • Let $\vec{a}$ and $\vec{b}$ be two vectors. we define $\vec{a}-\vec{b}$ as the sum of vector $\vec{a}$ and $\vec{-b}$
  • Note that $\vec{-b}$ is just the vector $\vec{b}$ but with invited direction.

RESOLUTION OF VECTORS:-

  • In the above diagram, we have $\vec{a}=\vec{OA}=\vec{OB}+\vec{OC}.$
  • But note that $OB=|\vec{a}|\cos \alpha$ and $OC=|\vec{a}|\cos \beta.$
Now, we define $\hat{i},\hat{j},\hat {k}$ as the unit vector along $OX,OY,OX$ (It's like a unit which is introduced to simplify algebric manupilations, we will understand that later)
  • So $\vec{a}=|\vec{a}|\cos\alpha\hat{i}+~~|\vec{a}|\cos \beta\hat {j}.$
This we talked in $X-Y$ plane, right? Now, what about $X-Y-Z$ plane? Well.. it's the same!
  • We have $$\boxed{\vec{a}=|\vec{a}|\cos\alpha\hat{i}+ ~~| \vec{a}|\cos \beta\hat {j}+~~|\vec{a}|\cos\gamma\hat{k} }$$ where $\vec{a}$ makes angle of $\gamma$ on $Z$ axis. 
By this we can see that any vector in $\Bbb {R_3}$ can be expressed as a linear combination of $3$ unit vectors i.e $\hat{i},\hat{j},\hat {k}.$

P.S. I actually love the next two parts and the above part :P . It's math u see :O

DOT PRODUCT or SCALAR PRODUCT:-
  • We define, $\vec{a}\cdot \vec{b}=|\vec{a}|\cdot |\vec{b}|\cos\theta$ where $\theta $ is the angle between vectors $\vec{a}$ and $ \vec{b}.$
Here are few examples which will help you :

$$ \hat{i}\cdot \hat{i}=1$$
$$\hat{j}\cdot \hat{j}=1 $$
$$ \hat{k}\cdot \hat{k}=1$$
$$\hat{i}\cdot \hat{j}=0$$
$$ \hat{j}\cdot \hat{k}=0$$
$$ \hat{i}\cdot \hat{k}=0$$

Well.. this makes sense since $\sin 90=0$ and $\sin 0=1.$

Also, a fun thing to note is that the dot product is commutative and distributive too. Why am I saying this? We will understand when I say about cross-product...

Anyways here's a fun example to solve!

Example:- Find $$(\hat{i}+\hat{2j})\cdot(3\hat{i}-5{j})$$

Solution:- Note that $$(\hat{i}+\hat{2j})\cdot(3\hat{i}-5{j})$$ $$=3\hat{i}\cdot\hat{i}-5\hat{j}{i}+2\hat{j}\cdot 3\hat{i}-10\hat{j}\cdot\hat{j}= 3-0+0-10=-7.$$

CROSS PRODUCT or VECTOR PRODUCT:-
  • The cross product or the vector product of the two vectors is $$\boxed{|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin\theta}$$ where $\theta$ is the angle between vector $\vec{a}$ and $\vec{b}$
  • The direction of vector $\vec{a}\times \vec{b}$ is perpendicular to both $\vec{a}$ and $\vec{b}.$

Here are few examples which will help you :

$$ \hat{i}\times \hat{i}=0$$
$$\hat{j}\times \hat{j}=0 $$
$$ \hat{k}\times \hat{k}=0$$
$$\hat{i}\times \hat{j}=\hat{k}$$
$$ \hat{j}\times\hat{k}=\hat{i}$$
$$ \hat{k}\times\hat{i}=\hat{j}$$
$$\hat{j}\times \hat{i}=-\hat{k}$$
$$ \hat{k}\times\hat{j}=-\hat{i}$$
$$ \hat{i}\times \hat{k}=-\hat{j}$$

Well, by now you must have understood that Cross product isn't commutative. But it does follow distributive law.

Anyways here's another fun example to solve!

Example:- Find $(2\hat{i}+3\hat{j})\times(2\hat{i}+3\hat{j})$

Solution:- Won't write :P , but from this we can conclude that for any vector $\vec{u}$ we have $\vec{u}\times \vec{u}=0.$

AREA, DETERMINANTS and CROSS PRODDUCTS:-
  • Note that $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin\theta$
  • But we have $$Area=\frac 12 ab\sin\theta=\frac{|\vec{a}||\vec{b}|\sin\theta}{2}=\frac{\vec{a}\times \vec{b}|}{2} $$.
Note that $|\vec{a}\times \vec{b}|$ is scalar



Now, let's do a problem to understand this "Area" topic better.

Examples:- Find the area of triangle with vertices $(-1,-3),~~(4,1),~~(-2,3).$

Solution:- We know that transition doesn't affect the area so..
$$(-1,-3)\rightarrow (0,0)$$
$$(4,1)\rightarrow (5,4)$$
$$(-2,3)\rightarrow (-1,6)$$

Now, we locate these points in the coordinate plane.



Clearly we get $[ABC]=[DBFE]-[ABD]-[ACE]-[BCF]=36-3-10-3=17.$

Now, plotting these points take a bit of time right? So we present two more solutions.

Second solution:-  Remember cross-products? We are going to use that!

We have $\langle 5,4\rangle = 5\hat{i}+4\hat{j}.$ Similarly, we have $\langle 1,6\rangle =-\hat{i}+6\hat{j}.$

So taking cross product, we get $|(5\hat{i}+4\hat{j})\times (\hat{i}+6\hat{j})|=|34\hat{k}|=34.$

So area is $\frac{34}{2}=17.$

Third Solution:- Now we will use matrix :)


Another problem!

Find the area of the triangle (-1,-2,-3),(1,2,3),(3,2,1)$

Solution:- Again transitioning doesn't change the area, so..

$$(-1,-2,-3)\rightarrow (0,0,0)$$
$$(1,2,3)\rightarrow (2,4,6)$$
$$(3,2,1)\rightarrow (4,4,4)$$

Now, we use heron's formula.

Calculating the distances between each point, we get the distance $\sqrt{56},\sqrt{8},\sqrt{48}.$

So we get $s= \sqrt{14}+\sqrt{2}+2\sqrt{3}.$

And so we get $$A=\sqrt{\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}\right)\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}-\sqrt{56}\right)\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}-\sqrt{8}\right)\left(\sqrt{14}+\sqrt{2}+2\sqrt{3}-\sqrt{48}\right)}$$ $$=4\sqrt{6}.$$

Second Solution:- 

We use matrix which is way faster.

Now taking magnitude, using the fact that cross factor is two times the area i.e

Area=$\frac12(\sqrt{64+16^2+64})= 4\sqrt{6}.$


Anyways hope this post helps someone :) If you liked this post and want more posts like these then do subscribe to Sunaina Thinks Absurd.

Sunaina💜
Labels:

Comments

  1. Abdul AhlawatApril 30, 2021 at 8:12 PM

    Don’t clickbait like that smh

    ReplyDelete
  2. UnknownMay 1, 2021 at 7:15 AM

    Simply brilliant... The geometrical interpretation has made it more meaningful...and hence lively😊... CONGRATULATIONS & Best Wishes to a budding Mathematician... Sunaina🎊

    ReplyDelete
  3. UnknownMay 6, 2021 at 7:52 PM

    Hey, best wishes for INMO camp. But in which class you are in now?

    ReplyDelete
  4. Sunaina Pati May 7, 2021 at 10:30 AM

    Thanks, we just havig MOOC (also if you want to see the lectures, you can see HBCSE's channel in Youtube). Also I am in Grade 11( just entered ).

    ReplyDelete
    Replies
    1. Mukul BorahMay 8, 2021 at 5:31 PM

      Great. I hope will get select for IMO. Your video lectures?

      Delete
  5. UnknownAugust 2, 2021 at 1:43 AM

    You are at grade 11?I saw your account on stackexchange.You are learning that math alone or there is olympiad instructor that is helping you?Thank you.

    ReplyDelete
    Replies
    1. Sunaina Pati August 2, 2021 at 8:07 AM

      Well, I was doing OTIS last year, where I took instructor( I won't reveal his name for privacy reasons), but if you follow my blog, you will know who he is!

      Delete
  6. Bandwala pagol chhagolJanuary 3, 2022 at 12:31 PM

    Did you by nychance learn vector bashing :P ?

    ReplyDelete
    Replies
    1. Sunaina Pati January 3, 2022 at 1:43 PM

      :O Not really, there was one G2 which was solvable using Vectors.. but that wasn't a vector bashing :P But sounds fun to learn!

      Are you giving INMO? :O

      Delete

Post a Comment

Popular posts from this blog

Geometry ( Finally!!!)

 This is just such an unfair blog.  Like if one goes through this blog, one can notice how dominated  Algebra is!! Like 6 out of 9 blog post is Algebra dominated -_- Where as I am not a fan of Algebra, compared to other genres of Olympiad Math(as of now). And this was just injustice for Synthetic Geo. So this time , go geo!!!!!!!!!!!  These problems are randomly from A Beautiful Journey through Olympiad Geometry.  Also perhaps I will post geo after March, because I am studying combi.  Problem:  Let $ABC$ be an acute triangle where $\angle BAC = 60^{\circ}$. Prove that if the Euler’s line of $\triangle ABC$ intersects $AB$ and $AC$ at $D$ and $E$, respectively, then $\triangle ADE$ is equilateral. Solution:  Since $\angle A=60^{\circ}$ , we get $AH=2R\cos A=R=AO$. So $\angle EHA=\angle DOA.$ Also it's well known that $H$ and $O $ isogonal conjugates.$\angle OAD =\angle EAH.$ By $ASA$ congruence, we get $AE=AD.$ Hence $\triangle ADE$ is equilateral....
3 comments
Read more

Problems I did this week [Jan8-Jan14]

Yeyy!! I am being so consistent with my posts~~ Here are a few problems I did the past week and yeah INMO going to happen soon :) All the best to everyone who is writing!  I wont be trying any new problems and will simply revise stuffs :) Some problems here are hard. Try them yourself and yeah~~Solutions (with sources) are given at the end! Problems discussed in the blog post Problem1: Let $ABC$ be a triangle whose incircle $\omega$ touches sides $BC, CA, AB$ at $D,E,F$ respectively. Let $H$ be the orthocenter of $DEF$ and let altitude $DH$ intersect $\omega$ again at $P$ and $EF$ intersect $BC$ at $L$. Let the circumcircle of $BPC$ intersect $\omega$ again at $X$. Prove that points $L,D,H,X$ are concyclic. Problem 2: Let $ ABCD$ be a convex quadrangle, $ P$ the intersection of lines $ AB$ and $ CD$, $ Q$ the intersection of lines $ AD$ and $ BC$ and $ O$ the intersection of diagonals $ AC$ and $ BD$. Show that if $ \angle POQ= 90^\circ$ then $ PO$ is the bisector of $ \angle AOD$ ...
Post a Comment
Read more

Just spam combo problems cause why not

This post is mainly for Rohan Bhaiya. He gave me/EGMO contestants a lot and lots of problems. Here are solutions to a very few of them.  To Rohan Bhaiya: I just wrote the sketch/proofs here cause why not :P. I did a few more extra problems so yeah.  I sort of sorted the problems into different sub-areas, but it's just better to try all of them! I did try some more combo problems outside this but I tried them in my tablet and worked there itself. So latexing was tough. Algorithms  "Just find the algorithm" they said and they died.  References:  Algorithms Pset by Abhay Bestrapalli Algorithms by Cody Johnson Problem1: Suppose the positive integer $n$ is odd. First Al writes the numbers $1, 2,\dots, 2n$ on the blackboard. Then he picks any two numbers $a, b$ erases them, and writes, instead, $|a - b|$. Prove that an odd number will remain at the end.  Proof: Well, we go $\mod 2$. Note that $$|a-b|\equiv a+b\mod 2\implies \text{ the final number is }1+2+\dots ...
2 comments
Read more

IMO 2023 P2

IMO 2023 P2 Well, IMO 2023 Day 1 problems are out and I thought of trying the geometry problem which was P2.  Problem: Let $ABC$ be an acute-angled triangle with $AB < AC$. Let $\Omega$ be the circumcircle of $ABC$. Let $S$ be the midpoint of the arc $CB$ of $\Omega$ containing $A$. The perpendicular from $A$ to $BC$ meets $BS$ at $D$ and meets $\Omega$ again at $E \neq A$. The line through $D$ parallel to $BC$ meets line $BE$ at $L$. Denote the circumcircle of triangle $BDL$ by $\omega$. Let $\omega$ meet $\Omega$ again at $P \neq B$. Prove that the line tangent to $\omega$ at $P$ meets line $BS$ on the internal angle bisector of $\angle BAC$. Well, here's my proof, but I would rather call this my rough work tbh. There are comments in the end! Proof Define $A'$ as the antipode of $A$. And redefine $P=A'D\cap (ABC)$. Define $L=SP\cap (PDB)$.  Claim1: $L-B-E$ collinear Proof: Note that $$\angle SCA=\angle SCB-\angle ACB=90-A/2-C.$$ So $$\angle SPA=90-A/2-C\implies \ang...
1 comment
Read more

IMO Shortlist 2021 C1

 I am planning to do at least one ISL every day so that I do not lose my Olympiad touch (and also they are fun to think about!). Today, I tried the 2021 IMO shortlist C1.  (2021 ISL C1) Let $S$ be an infinite set of positive integers, such that there exist four pairwise distinct $a,b,c,d \in S$ with $\gcd(a,b) \neq \gcd(c,d)$. Prove that there exist three pairwise distinct $x,y,z \in S$ such that $\gcd(x,y)=\gcd(y,z) \neq \gcd(z,x)$. Suppose not. Then any $3$ elements $x,y,z\in S$ will be $(x,y)=(y,z)=(x,z)$ or $(x,y)\ne (y,z)\ne (x,z)$. There exists an infinite set $T$ such that $\forall x,y\in T,(x,y)=d,$ where $d$ is constant. Fix a random element $a$. Note that $(x,a)|a$. So $(x,a)\le a$.Since there are infinite elements and finite many possibilities for the gcd (atmost $a$). So $\exists$ set $T$ which is infinite such that $\forall b_1,b_2\in T$ $$(a,b_1)=(a,b_2)=d.$$ Note that if $(b_1,b_2)\ne d$ then we get a contradiction as we get a set satisfying the proble...
Post a Comment
Read more

Problems with meeting people!

Yeah, I did some problems and here are a few of them! I hope you guys try them! Putnam, 2018 B3 Find all positive integers $n < 10^{100}$ for which simultaneously $n$ divides $2^n$, $n-1$ divides $2^n - 1$, and $n-2$ divides $2^n - 2$. Proof We have $$n|2^n\implies n=2^a\implies 2^a-1|2^n-1\implies a|n\implies a=2^b$$ $$\implies 2^{2^b}-2|2^{2^a}-2\implies 2^b-1|2^a-1\implies b|a\implies b=2^c.$$ Then simply bounding. USAMO 1987 Determine all solutions in non-zero integers $a$ and $b$ of the equation $$(a^2+b)(a+b^2) = (a-b)^3.$$ Proof We get $$ 2b^2+(a^2-3a)b+(a+3a^2)=0\implies b = \frac{3a-a^2\pm\sqrt{a^4-6a^3-15a^2-8a}}{4}$$ $$\implies a^4-6a^3-15a^2-8a=a(a-8)(a+1)^2\text{ a perfect square}$$ $$\implies a(a-8)=k^2\implies a^2-8a-k^2=0\implies \implies a=\frac{8\pm\sqrt{64+4k^2}}{2}=4\pm\sqrt{16+k^2}. $$ $$ 16+k^2=m^2\implies (m-k)(m+k)=16.$$ Now just bash. USAMO 1988 Suppose that the set $\{1,2,\cdots, 1998\}$ has been partitioned into disjoint pairs $\{a_i,b_i\}$ ($1...
3 comments
Read more

Challenging myself? [Jan 15-Jan 27]

Ehh INMO was trash. I think I will get 17/0/0/0-1/3-5/10-14, which is def not good enough for qualifying from 12th grade. Well, I really feel sad but let's not talk about it and focus on EGMO rather.  INMO 2023 P1 Let $S$ be a finite set of positive integers. Assume that there are precisely 2023 ordered pairs $(x,y)$ in $S\times S$ so that the product $xy$ is a perfect square. Prove that one can find at least four distinct elements in $S$ so that none of their pairwise products is a perfect square. I will use Atul's sol, cause it's the exact same as mine.  Proof: Consider the graph $G$ induced by the elements of $S$ and edges being if the products are perfect squares. Note that if $xy = a^2$ and $xz = b^2$, then $yz = \left( \frac{ab}{x} \right)^2$, since its an integer and square of a rational number its a perfect square and so $yz$ is an edge too. So the graph is a bunch of disjoint cliques, say with sizes $c_1, c_2, \cdots, c_k$. Then $\sum_{i=1}^k c_i^2 = 2023$, which ...
Post a Comment
Read more

Some random problems

  I know, I know. Different font indeed. I have deleted a few of my MSE answers. I felt they weren't that good in quality. And a few questions are from my prev aops account which I have deactivated now. I also have posted 10 IOQM types of problems. These can be used while preparing for IOQM. Problem: Prove that $\dfrac{ab}{c^3}+\dfrac{bc}{a^3}+\dfrac{ca}{b^3}> \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$, where $a,b,c$  are different positive real numbers.  Proof: Note that by AM-GM $$\frac{ab}{c^3}+\frac{bc}{a^3}\ge \frac{2b}{ac}$$ and we also have $$\frac {b}{ac}+\frac{c}{ab}\ge \frac{2}{a}$$. Hence, $$\sum_{cyc}\frac{ab}{c^3}\ge\sum_{cyc}\frac{b}{ac}\ge\sum_{cyc}\frac{1}{a}$$ where everything we got is by applying AM-GM on $2$ terms and then dividing by $2$. USA TST 2007: Triangle $ABC$ which is inscribed in circle $\omega$. The tangent lines to $\omega$ at $B$ and $C$ meet at $T$. Point $S$ lies on ray $BC$ such that $AS$ is perpendicular to $AT$. Points $B_1$ and $C_1...
17 comments
Read more

Solving Random ISLs And Sharygin Solutions! And INMO happened!!

Some of the ISLs I did before INMO :P  [2005 G3]:  Let $ABCD$ be a parallelogram. A variable line $g$ through the vertex $A$ intersects the rays $BC$ and $DC$ at the points $X$ and $Y$, respectively. Let $K$ and $L$ be the $A$-excenters of the triangles $ABX$ and $ADY$. Show that the angle $\measuredangle KCL$ is independent of the line $g$ Solution: Note that $$\Delta LDK \sim \Delta XBK$$ and $$\Delta ADY\sim \Delta XCY.$$ So we have $$\frac{BK}{DY}=\frac{XK}{LY}$$ and $$\frac{DY}{CY}=\frac{AD}{XC}=\frac{AY}{XY}.$$ Hence $$\frac{BK}{CY}=\frac{AD}{XC}\times \frac{XK}{LY}\implies \frac{BK}{BC}=\frac{CY}{XC}\times \frac{XK}{LY}=\frac{AB}{BC}\times \frac{XK}{LY} $$ $$\frac{AB}{LY}\times \frac{XK}{BK}=\frac{AB}{LY}\times \frac{LY}{DY}=\frac{AB}{DL}$$ $$\implies \Delta CBK\sim \Delta LDK$$ And we are done. We get that $$\angle KCL=360-(\angle ACB+\angle DKC+\angle BCK)=\angle DAB/2 +180-\angle DAB=180-\angle DAB/2$$ Motivation: I took a hint on this. I had other angles but I did...
4 comments
Read more

Orders and Primitive roots

 Theory  We know what Fermat's little theorem states. If $p$ is a prime number, then for any integer $a$, the number $a^p − a$ is an integer multiple of $p$. In the notation of modular arithmetic, this is expressed as \[a^{p}\equiv a{\pmod {p}}.\] So, essentially, for every $(a,m)=1$, ${a}^{\phi (m)}\equiv 1 \pmod {m}$. But $\phi (m)$ isn't necessarily the smallest exponent. For example, we know $4^{12}\equiv 1\mod 13$ but so is $4^6$. So, we care about the "smallest" exponent $d$ such that $a^d\equiv 1\mod m$ given $(a,m)=1$.  Orders Given a prime $p$, the order of an integer $a$ modulo $p$, $p\nmid a$, is the smallest positive integer $d$, such that $a^d \equiv 1 \pmod p$. This is denoted $\text{ord}_p(a) = d$. If $p$ is a primes and $p\nmid a$, let $d$ be order of $a$ mod $p$. Then $a^n\equiv 1\pmod p\implies d|n$. Let $n=pd+r, r\ll d$. Which implies $a^r\equiv 1\pmod p.$ But $d$ is the smallest natural number. So $r=0$. So $d|n$. Show that $n$ divid...
7 comments
Read more