Arithnostic

A person who believes you cannot prove that math really exists

"You cant prove that math really exists, I'm an Arithnostic"
"What? D-!? That's religious persecution!"
Daren Streblow

by oldschoolgamer98 October 26, 2008

arithmedick

Used to count how many penises one has encountered in their lifetime.

Betty used arithmedick to count the number of cocks she penetrated during Spring Break.

by GlazeHer April 26, 2014

arithmetic

Arithmetic is a form of Elementary school math.

2 + 2 = 4
5 + 5 = 10
Shahriar + cupcakes = heaven

I learned arithmetic in elementary school.

by Areeyan February 7, 2007

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

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

arithmatic

You're a dumbass who can't spell arithmetic.

You meant to do a search for arithmetic, not "arithmatic", you moron.

by Cixelsid September 19, 2009