The fundamental theorem of arithmetic states that {n: n is an element of N > 1} (the set of natural numbers, or positive integers, except the number 1) can be represented uniquely apart from rearrangement as the product of one or more prime numbers (a positive integer that's divisible only by 1 and itself). This theorem is also called the unique factorization theorem and is a corollary to Euclid's first theorem, or Euclid's principle, which states that if p is a prime number and p/ab is given (a does not equal 0; b does not equal 0), then p is divisible by a or p is divisible by b.
Proof: First prove that every integer n > 1 can be written as a product of primes by using inductive reasoning. Let n = 2. Since 2 is prime, n is a product of primes. Suppose n > 2, and the above proposition is true for N < n. If n is prime, then n is a product of primes. If n is composite, then n = ab, where a < n and b < n. Therefore, a and b are products of primes. Hence, n = ab is also a product of primes. Since that has been established, we can now prove that such a product is unique (except for order). Suppose n = p sub1 * p sub2 * ... * p subk = q sub1 * q sub2 * ... * q subr, where the p's and q's are primes. If so, then p sub1 is divisible by (q sub1 * ... * q subr) by Euclid's first theorem. What is the relationship between p sub1 and one of the q's? If the r in q subr equals 1, then p sub1 = q sub1 since the only divisors of q are + or - 1 and + or - q and p > 1, making p = q. What about the other factors in the divisor? If p does not divide q, then the greatest common denominator of p and q is 1 since the only divisors of p are + or - 1 and + or - p. Thus there are integers m and n so that 1 = am + bn. Multiplying by q subr yieds q subr = amq subr + bnq subr. Since we are saying that p is divisible by q, let's say the q sub1 * q subr = cp. Then q subr = amq subr + bnq subr = amq subr + bcm = m(aq subr + bc). Therefore, p is divisible by q sub1 of q sub2 * ... * q subr. If p sub1 is divisible by q sub1, then p sub1 = q sub 1. If this does not work the first time, then repeat the argument until you find an equality. Therefore, one of the p's must equal one of the q's. In any case, rearrange the q's so that p sub1 = q sub1, then p sub1 * p sub2 * ... * p subk= p sub1 * q sub2 * ... * q subr and p sub2 * ... * p subk = q sub2 * ... * q subr, and so on. By the same argument, we can rearrange the remaining q's so that p sub2 = q sub2. Thus n can be expressed uniquely as a product of primes regardless of order, making the fundamental theorem of arithmetic true.
by some punk kid August 15, 2005
When an inebriated man is still able to have a raging erection, we say that his fundamentals are looking strong.
When she saw his raging erection after he had drunk a bottle of Jack Daniels, she screamed in delight "The fundamentals are looking strong"
by BooBooKittyFuckReal August 10, 2011
A nice youtuber with over 5 subscribers thats my brother, I wanted to make tis so it would make his day so please dont reject it! His channel is Redstone And Skin Fundamentals please go show him some love!
by Sexy Youtuber October 17, 2020
a cult that holds very conservative views. Will literally shoot you if you use the NIV (KJV only), wear pants as a female, question leadership in the church, or are gay. They believe that women are under men and should never have any leadership roles. Basically they stay at home barefoot and pregnant. They also beat their children before they can even form words (spare the rod spoil the child). Many have schools so they can keep the children away from the world so they do not get corrupted. If you grew up in this denomination im sorry :(
by awaw54 December 4, 2021
The first fundamental theorem of calculus states that (just trying to remember here), with F(x) being the antiderivative of f(x), while u stand for the upper bound and l stands for the lower bound, the definite integral of a function f(x) is equal to F(u) - F(l).
"Solve the integral from 2 to 3 of x^2" Well, this should be easy. Just use the first fundamental theorem of calculus.
The antiderivative of x^2 = (x^3)/3. (3^3)/3 = 9, and (2^3)/3 = 8/3. 9 - 8/3 = 19/3, which is equal to 6.3333333... Time to search this up on good ol' Wolfram Alpha. *Checks* Alright!
The antiderivative of x^2 = (x^3)/3. (3^3)/3 = 9, and (2^3)/3 = 8/3. 9 - 8/3 = 19/3, which is equal to 6.3333333... Time to search this up on good ol' Wolfram Alpha. *Checks* Alright!
by CMCTT June 24, 2017
a school full of a bunch of white kids with money and teachers who think they're all that. The school is all walls and constantly smells like someone's bath and body works hand sanitizer.
by chickennoodles.12 September 6, 2021
A school in Pasadena, CA that has many hot hispanic girls. This school also has many emos and wannabe gangsters.
by Guccibeltnigga782 September 3, 2023