Adithya

Name of a weeb, discord user and he is a very smart person. He will give you answers but very indirectly to see how dumb you are.

1: hey what's the answer for b
2: alright so it cant be 1 2 or 4, what can it be?
1:uhh 1?
2:ah you are extremely dumb I see
1: HEY! stop acting like an Adithya

by Phineas#1366 November 10, 2020

arithmetard

A person who is incapable of even the most basic math. (also: arithmetarded)

Chad: How much do we leave for a tip?
Jeff: HUH?
Kirk: Don't ask him, he's an arithmetard.

by MikeBlitz March 12, 2009

Arithmesexual

Someone turned on by math, someone sexually attracted to math, usually an introvert.

"You see that arithmesexual over there?"
"Who her?"
"Yeah, apparently secant functions turn her on."
"What in fresh fuck?"

by Texas' Ace February 14, 2019

Adithya

Someone who is smart, handsome and tall. He Has a huge fanbase with mainly girls.

Damn Adithya

by randomboy321 November 22, 2021

aadithya

He is a red hot man, he is said to be a ladies man. His hotness was asked to be turned down by the Sun as it made it look less bright. He is appreciated for his creativity and many bully him as they are jealous of his hotness and cannot match up to him. He is said to be related to Henry Ford and his abs is said to have abs.

Oh, aadithya make me wanna go all the way around.

by A Fan of the Man May 31, 2021

Adithya

The best person in the world, everybody likes him and will kiss the ground he walks on. He can humiliate anyone in any sport, and makes Einstein look childish. In the near future he will assassinate Donald Trump and he will become the people's hero.

Person 1: Man, that guy is cool, what's his name.
Person 2: Slaps him in the face. YOU DON'T KNOW HIS NAME!? You noob it's Adithya.

by Luck Master 2017 January 17, 2017

Fundamental Theorem of Arithmetic

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 September 6, 2005