#Bezout's Identity

[adeadpumpkin]

Thanks for the tag ehehehe >:]

"Théorème de Bézout" (I think it translate it translate to "Bezout's identity" I think ?)

Sorry for the math jumpscare I needed it for a class,,,

Tagging eheheh ( feel free to not do it if you don't want too <3) : @bytesofaffection @laul-self-fif @celestialdandy (and anyone who see it really)

last google search, go

um. Tag four people.

what do they make sewer tunnels out of

@ncc1701ohno @affixjoy @the-magpieprince @twinkboimler

[isomorphismes]

What is an algebraic variety?

The shape of the answers to equations.

Start with an equation like

Please take notice of three things:

the "+1" is used to shift the "zero line" or "answer line" up and down. So in fact "zero" is arbitrary.

This specific equation's left-hand side (which does not correspond to the picture; the picture is just supposed to be "an arbitrary continuous function") factors into

which factors into

I didn't specify what "x" is--it could belong to any coherent corpus of generalised "numbers" where "multiplicative identity" ("one") is meaningful

could be matrices, functional space, something I haven't thought of yet

tropical geometry. redefine "times" as ^ and "plus" as v

set theory union intersection

it is just a shorthand for

calculus 101. polynomials.

so not the price train of a narrowly-held company (all jumps)

fundamental theorem of algebra: Copf; is closed

as a mapping to zeroes

letting the LHS handle all the info (instead of x^2=16) is better organisation

affine -- geometric -- "forget the origin"

affine variety

kernel -- what's thrown in the trash. linear algebra.

intersections. system of equations. pretend a physical system. like going to the moon. 

intersection. it has to solve all of them.

bezout's theorem.

the body

[jarviscodinghub]

ASSIGNMENT-2 CS 2214: DISCRETE STRUCTURES FOR COMPUTING solved

ASSIGNMENT-2 CS 2214: DISCRETE STRUCTURES FOR COMPUTING solved

1. (25 points) Let p be a prime number and a an integer not divisible by p. Prove that there exists an integer b such that ab ≡ 1 (mod p). Conclude that every non-zero element in Zp has a multiplicative inverse. Hint: Use Bezout’s identity. 2. (25 points) Let x be an integer such that x leaves a remainder 3 when divided by 5, leaves a remainder 7 when divided by 12 and it leaves a remainder 8…

[mathematicianadda]

Choosing a Discrete Mathematics Topic to Explain

Instead of a final exam, my teacher decided to assign us homework where we choose a theorem or a concept that we've covered in class and explain it. We then prepare an exercise about our topic and solve the exercise.

I LOVE this homework, as it gives me so much freedom. Also, since it replaces the final exam, I think I am allowed to go more in-depth on a topic, rather than the usual 1-2 page assignments.

However, I don't know what to pick because I want my topic and exercise to be something interesting and I am a little picky. Do you guys have any suggestions for a topic and an interesting exercise? I am not asking you to solve the exercise for me. I would like to work on it on my own. My only request is that it is not something that is done to death i.e. that is given as an exercise in every Discrete Mathematics class. By the way, this is my first year as a maths undergraduate, so please go easy on me.

What we've covered in class: Number Theory (Modular Arithmetic, Fermat's Little Theorem, Bezout's Identity, Chinese Remainder Theorem, Fundamental Theorem of Arithmetic, etc.), Induction and Recursion, Counting (Pigeonhole Principle, The Binomial Theorem, etc.), Discrete Probability (Bayes' Theorem, Expected Value and Variance), Advanced Counting Techniques (Solving Linear Recurrence Relations, Generating Functions, Inclusion-Exclusion Principle), Graphs (Euler and Hamilton Paths, Planar Graphs, Graph Coloring), Trees.

submitted by /u/dnzszr [link] [comments] from math https://ift.tt/2Yk8c3F https://ift.tt/eA8V8J from Blogger https://ift.tt/3fMcW87

[coraniaid]

OK, now that the poll is over I can talk about how I solved it and the potentially useful trick I missed:

Two of the given numbers are easy to rule out: 231 and 209 cannot be prime. A number is divisible by 3 if and only if the sum of its decimal digits is divisible by 3 (which rules out 231 since 2+3+1=6) and divisible by 11 if and only if the alternating sum of its decimal digits is divisible by 11 (which rules out 209, since 2-0+9=11; in fact 231 is also divisible by 11 since 2-3+1=0).

The next divisibility test I know is this: a number of the form 10t+d is divisible by 7 if and only if t-2d is divisible by seven (for example: 133 is divisible by 7 because 13-6=7). Unfortunately neither 227 nor 323 are divisible by 7 -- this test gives 22-14=8 which obviously isn't divisible by 7, and 32-6=26 which is also not divisible by 7 -- so this doesn't actually help us (but note in passing that 231 is also divisible by 7, since 23-2=21).

However, the test for divisibility by 7 is actually one of an infinite family of tests that works for all primes other than 2 or 5. There's a basic result in number theory called Bezout's Identity which says that if a and b are two non-zero numbers then there will always exist integers x and y such that ax + by = gcd(a,b), where gcd(a,b) denotes the greatest common divisior of a and b. In particular, if p is a prime other than 2 or 5, then gcd(10,p) = 1 and Bezout's identity tells us that we can always find x and y such that 10x+py = 1. A bit of simple algebraic manipulation of 10t+d using this identity then gives us that 10t+d is a multiple of p if and only if t+x*d is a multiple of p.

For example, for p=7 we have x=-2 and y=3 and this gives us the test for divisibility by 7 we just used above. For p=13 we have x=4 and y=-3, which means that 10t+d is divisible by 13 if and only if t+4d is divisible by 13. For p=17 we have x=-5 and y=3, which means that 10t+d is divisible by 17 if and only if t-5d is divisible by 17. And so on and so forth for higher primes as needed.

Now, using the generalized version of this test we can check that neither 227 nor 323 is divisible by 13 either. For 227 the test gives 22+28=50 which obviously isn't divisible by 13, and for 323 the test gives 32+12=44 which obviously isn't divisible by 13 either.

The lazy approach at this point -- which is what I did -- is to keep on testing, and if you do this you'll find that 227 isn't divisible by 17 (22-35=-13 is not divisible by 17) but that 323 is divisible by 17 (32-15=17 exactly). So the prime we're lookig for must be 227, which is the answer.

A (slightly) smarter approach though, is to realize we don't have to test for divisibility by 17 at all. As soon as we've confirmed that 227 isn't divisible by 13 we know it must be prime, because 227 is smaller than 256=16^2 and so any prime factors it has other than itself must be smaller than 16. It doesn't have any prime factors smaller than 16, so we're done (as long as we assume the poll question is fair and only one of the four numbers is prime).

That all being said, I was kind of surprised at first to see that, although the poll narrowly picked the correct answer, the two options I'd dismissed as obviously not prime right at the start (231 and 209) both received far more votes than 323, even though it took me much more thinking to rule out 323 as an option.

But then, well after the fact, I realized there's a trick here that I completely missed and other people obviously spotted: 323 is one less than a perfect square (specificially 323=324-1=18^2-1). But that means 323 is the difference of two squares and we can factorize it (if any number n = a^2 - b^2 we can factor it as n = (a+b)*(a-b), so 323 = 18^2 - 1^2 = 19*17).

I'm not sure that that's really easier to see than the divibility by 3 trick (I do not have 18^2 memorized, though it's simple to calculate as 4*81), but it's certainly a lot more sensible than what I actually did.

Anyway, fun poll, not a walrus in sight, would vote in again.

pls reblog for sample size etc

follow for more occasional useless polls :)

[learning2program]

Interesting fact about the GCD

Today, we will be utilizing the Euclidean algorithm module we wrote earlier to study an interesting fact about GCD.

I learned a cool new fact today that I thought I’d share. It uses the Euclidean algorithm module that I wrote up earlier like I said, so if you haven’t see that yet you should read it here.

Today we will be studying the fact that

\[\sum_{\substack{1 \leq a \leq n\ gcd(a, n) =1}} a = \frac{n \phi(n)}{2}. \]

I’ll first cover some preliminaries though. First, one should know what I mean when I say

\[ \phi(n). \]

This is known as the Euler totient function, and is an interesting function in number theory. It counts the number of numbers coprime to your input which are less than your value. It has a variety of applications and a long history behind it which is outside the scope of this post, but you can see more here.

Next, we’ll need a theorem.

Theorem: If gcd(a, n) = 1, then gcd(n-a, n) = 1.

Proof:

By Bezout's Identity, we have that gcd(a, n) = 1 implies

\[ \alpha a + \beta n = 1.\]

Add and subtract alpha n to the left hand side to get

\[ \alpha a + \beta n - \alpha n + \alpha n = 1.\] By simple algebra, we get

\[ \alpha(a - n) + n(\beta + \alpha) = 1.\]

Let

\[ \gamma = \beta + \alpha\]

for notational simplicity. Then we have that

\[ \alpha(a-n) + n\gamma = 1.\]

By the converse of Bezout's Identity, this then gives us

\[\gcd(a-n, n) = 1.\]

However, note that

\[\gcd(-(a-n),n) = \gcd(a-n, n)\]

(this follows since if d | m, n then d | -m, n; so we have that (-m,n) and (m,n) have the same set of common divisors, which implies the same gcd), which then implies gcd(n-a, n) = 1. Q.E.D.

Using these two, the construction of the earlier claim is rather clear. We can use an argument similar to that of Gauss when counting the integers. Let

\[x_1, \ldots, x_{\phi(n)} \]

be the integers which are coprime to n in ascending order. Then we can arrange the integers to then be

\[ x_1, \ldots, x_{\phi(n)} \]

\[ x_{\phi(n)}, \ldots, x_1. \]

Add the integers vertically to then get $\phi(n)$ instances of n. This is clear from the theorem earlier. We however have added the sequence to itself, and so dividing it by two then gives us the sum of the coprime integers. Therefore, we have

\[\sum_{1 \leq a \leq n, gcd(a, n) =1} a = \frac{n \phi(n)}{2}. \]

One can then check this for the first however many integers using python. For simplicity, I will be only checking the first 30 integers; however, you can modify the file to be as large as you want (it will probably start slowing down a lot as you set the integer amount to be higher and higher, however). There is really only one function that we will be doing to use this, though we will import the gcd function from our euclideanalg module.

check(p): The function starts by constructing a list of all of the positive integers less than p which are coprime to p. Here, we use the gcd function to check if gcd(i, p) == 1 at each iteration of i. Next, we sum up each of the numbers in this new list that we’ve constructed and save it in a variable denoted sum. We take the length of the list (which is equivalent to the Euler totient of p) and then calculate the value

\[\frac{n \phi(n)}{2}. \]

If the sum of the coprimes and this value are equivalent, the function returns true; otherwise, the function returns false (however, by the prior claims, we have shown that this should never return false). We now have a tangible way to view whether or not this claim is true! The code for the function follows

def check(p): print('checking ' + str(p)) primes = [] for i in range(1, p): if gcd(p, i) == 1: primes.append(i) sum = 0 print('coprimes: ' + str(primes)) for i in primes: sum += i print('sum of coprimes: ' + str(sum)) #now to actually check tot = len(primes) print('euler totient function: ' + str(tot)) total = (p * tot)/2 if total == sum: return True else: return False

As always, you can find this on the github page here.

[relatively-dense]

Isn't the second one kind of cheating? If you know Z is a PID you have almost certainly used Bezout's identity already...

top 10 math proofs that make you go “what the fuck???”

[jarviscodinghub]

ASSIGNMENT-2 CS 2214: DISCRETE STRUCTURES FOR COMPUTING solved

ASSIGNMENT-2 CS 2214: DISCRETE STRUCTURES FOR COMPUTING solved

1. (25 points) Let p be a prime number and a an integer not divisible by p. Prove that there exists an integer b such that ab ≡ 1 (mod p). Conclude that every non-zero element in Zp has a multiplicative inverse. Hint: Use Bezout’s identity. 2. (25 points) Let x be an integer such that x leaves a remainder 3 when divided by 5, leaves a remainder 7 when divided by 12 and it leaves a remainder 8…

[jarviscodinghub]

ASSIGNMENT-2 CS 2214: DISCRETE STRUCTURES FOR COMPUTING solution

ASSIGNMENT-2 CS 2214: DISCRETE STRUCTURES FOR COMPUTING solution

1. (25 points) Let p be a prime number and a an integer not divisible by p. Prove that there exists an integer b such that ab ≡ 1 (mod p). Conclude that every non-zero element in Zp has a multiplicative inverse. Hint: Use Bezout’s identity. 2. (25 points) Let x be an integer such that x leaves a remainder 3 when divided by 5, leaves a remainder 7 when divided by 12 and it leaves a remainder 8…

[lilith-hazel-mathematics]

This is actually a form of Partial Fraction Decomposition! You can obtain a proof for PFD over the integers in the exact same way as you'd do over the polynomials. Basically it relies on Bezout's lemma, which is ultimately derived from Euclidean division. According to Bezout's lemma, given any coprime a,b, there exists x,y such that ax+by=1. As a simple corollary, for every c there exists x,y such that ax+by=c. As a fun fact, Bezout actually proved this for polynomials first, even though it's most well known about the integers.

For polynomials, PFD guarantees that every rational function can be expressed as a polynomial plus a sum of fractions, where the denominators are powers of irreducibles, each irreducible appears at most once, and the numerators have lesser degree than their denominator. The analogous statement for integers is almost identical: every rational number can be written as an integer plus a sum of fractions, where the denominators are powers of irreducibles (AKA primes), each irreducible appears at most once, and the numerators are lesser than their denominators. You can prove the integer version almost identically to how we'd prove of the polynomial version.

To prove PFD over the polynomials, it suffices to show that every rational function A(x)/B(x) is expressible as a sum of fractions, where the denominators are powers of irreducibles. We prove this inductively on the degree of the denominator B. If B is already a power of an irreducible, we are done. Otherwise, we can rewrite B=C*P^n, where P is an irreducible polynomial and C is coprime to P. Now we can apply Bezout's lemma to find X,Y such that XP^n + YC=A. We see that A/B = (XP^n + YC)/(CP^n) = X/C + Y/P^n. It's clear that C has lesser degree than B, so apply inductive premise to write X/C in the desired form. Since Y/P^n is itself a fraction with an irreducible power as its denominator, the induction is concluded. If needed (it's never needed) we can merge all the fractions having compatible denominators to ensure that each irreducible appears only once. As the last step, apply Euclidean Division to reduce each fraction so that the numerator has smaller degree than the denominator; this introduces a polynomial to the summation.

The proof of PFD over the integers basically writes itself by analogy. Suppose a,b are integers, then we can write a/b as a sum of fractions, where the denominators are powers of primes. Prove this by strong induction on b. In case b is a power of a prime, we're done. Otherwise write b=c*p^n where p is prime and c is coprime to p. Now apply Bezout's lemma to find x,y such that a=xp^n+yc. We see that a/b = x/c + y/p^n, and by applying inductive premise to x/c, we prove our claim and complete the induction. If needed (it's never needed) we can merge all the fractions having compatible denominators to ensure that each prime appears only once. As the last step, apply Euclidean division to reduce each fraction so that the numerator is positive and lesser than the denominator; this introduces an integer to the summation, and each fraction will fall within the [0,1) interval.

math challenge.

Prove that any rational number can be written uniquely as the sum of an integer, a fraction between 0 and 1 whose denominator is a power of 2, a fraction between 0 and 1 whose denominator is a power of 3, a fraction between 0 and 1 whose denominator is a power of 5, and so on through all prime powers.

(note: when we say between 0 and 1, 0 is included and 1 isn't)

If you already know this please tell me what it is called

[suinicide]

Let a and b be integers with greatest common divisor d. Then there exist integers x and y such that ax + by = d. Moreover, the integers of the form az + bt are exactly the multiples of d. - bezout's identity

everyone has their own "lorem ipsum" proof, that they fill in when they need a proof to throw into a thing. mine is the like, bézout's identity. whats yours