And Arithmetic

[2 + 2] = 4 5 + 5 = 10 [Shahriar] + cupcakes = heaven I learned arithmetic in [elementary school]. ❋ Areeyan (2007)

Man 1: "[Who's that]?" Man 2: "That's [Susie]." Man 1: "Man, she's [fine]...I'm gonna need that arithmetic." ❋ Casmith (2011)

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. ❋ Ben Faulding (2006)

That [test] was [harder] than [Chinese arithmetic]. ❋ Dridian (2015)

When [Katrina] was [sucking my dick] and it got [harder] than fucking chinese arithmetic. ❋ Carl Pickens (2007)

Me: [1+1]=[69] ...mental arithmetic sir. Teacher: F......[Fuck off] ❋ Bigfatbozwithbigfatballbag (2019)

Liberals and [climate] [activists] who believe in [carbon arithmetic] should stop reproducing. I think having more children is a wonderful thing. ❋ Sexydimma (2022)

I should have consulted with the arithmetic dog before [ordering] [the metal] for my project, he always remembers [to carry] the one. ❋ Meat_man (2010)

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. ❋ Some Punk Kid (2005)