2+2 because i really know im just trying something real quick

4

Explanation:

hmm 2 + 2 can be written as 1 + 1 + 1 + 1. This is equivalent to 1(4). Since 1 * 4 = 4 our answer is 4.

That's a pretty straightforward way of solving. Now to something more theoretical:

Here is something I found on Qoura. Remember! I didn't come up with this, but it's a really cool explanation:

The set of integers is an (infinite) group with respect to addition. Since 2 is an integer, the sum of 2 and 2 must also be an integer. Suppose, for the sake of contradiction, that 2 + 2 < 4. We have 2 > 0. Adding two to both sides, we get 2+2 > 0+2. Since 0 is the identity element for addition, we have 2 + 2 > 2. Hence 2 < 2 + 2 < 4 so 2 + 2 must equal 3. Since 3 is prime, by Fermat’s Little Theorem we have the following for a ∈ Z +: a ^3−1 ≡ 1 (mod 3) a^3−1 ≡ 1 (mod 2 + 2) But a^3−1 = a^3/a = a × a × a/ a = a × a and, for a = 2, 2 × 2 ≡ 0 (mod 2 + 2) since, by the definition of multiplication, 2 × 2 = 2 + 2. So, we have 2^3−1 ≡ 0 (mod 2 + 2) 2^2+2−1 ≡ 0 (mod 2 + 2) This is a contradiction to Fermat’s Little Theorem, so 2 + 2 must not be prime. But 3 is prime. Hence 2+2 must not equal 3, and therefore 2+2 ≥ 4. It remains to show that 2 + 2 is not greater than 4.

To prove this we need the following lemma:

Lemma. ∀a ∈ Z, if a > 4, ∃b ∈ Z such that b > 0 and a − 2 = 2 + b. If a were a solution to the equation 2 + 2 = a, then we would have a − 2 = 2 + 0. The lemma states that this cannot hold for any a > 4, and so a = 4, as desired.

The proof is by induction over a. Our base case is a = 5. Let 5−2 = 2+b. Five is the 5th Fibonacci number, and 2 is the 3rd Fibonacci number. Therefore, by the definition of Fibonacci numbers, 5 − 2 must be the 4th Fibonacci number. Letting fi denote the ith Fibonacci number, then we have fi − fi−1 > 0 for i≠ 2, because f2 − f1 = f0 and f0 = 0, but f1 = 1, and the Fibonacci sequence is non decreasing. Hence (5 − 2) − 2 > 0. Now, suppose that ∃b ∈ Z such that b > 0 and (k − 1) − 2 = 2 + b. We need to prove that, for some b’ > 0, k − 2 = 2 + b’ . Our inductive hypothesis is equivalent to: k − 1 − 2 = 2 + b

k − 1 − 2 + 1 = 2 + b + 1

k − 2 = 2 + (b + 1)

Since 1 > 0 and b > 0, we have (b + 1) > 0. Thus, letting b’ = b + 1, we have a nonnegative solution to k − 2 = 2 + b’ , as desired. By Lemma , it is not the case that 2 + 2 > 4. Hence 2 + 2 ≤ 4. We also have 2 + 2 ≥ 4.

Therefore, 2 + 2 = 4