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

elevator arithmetic

The mental process of estimating the respective weights of the other passengers in an elevator. Then comparing it to the maximum weight posted next to the fire inspection sign to see if its safe.

After that big fat guy and the kid in the wheel chair got on, I redid the Elevator Arithmetic and decided I was going to take the stairs.

by Ben Faulding April 4, 2006

chinese arithmetic

A term describing anything that is very hard to do.
This term has been around for years.

That test was harder than Chinese arithmetic.

by Dridian January 3, 2016

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

Asthmatic-Hitler

When a girl is giving you fellatio, and you put her in a leg lock so she begins to choke on your penis. While stuck, you take two fingers and smear poop on her upper lip, giving her a Hitler moustache. When you release her she will be wheezing on the ground with said moustache, and will have received the Asthmatic-Hitler.

Steve: Yo I'm pretty sure I'm going to jail
David: For what?
Steve: I gave that girl the asthmatic-hitler and she died

David: Sucks man..

by Normal dude 44 March 7, 2011

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

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