Hey Zillcellix u ban evading?
- Thread starter FireZ
- Start date
Did you just call Zillcellix out to be a vader?
FireZ pulling a Qverload?
yes I did gonna do a check and if it checks out.......
Oh dear.
jesus christ FireZ thinks he is gonna get mod rank if he reports ban evaders.. fucking staff hunter faggot
Qverload pls ban FireZ for staff hunting
yeah 2 similar ip ranges and your poss arent totally different from deweys.
- Joined
- Apr 25, 2012
- Sythe Gold
- 1
- Messages
- 357
Dude.. What ip range does that fall to? My permanent address is in around auburn, Washington lol. That's my SVU and PayPal details
I feel like leanbean now trying to explain to the community lol.
Also the posts? What do you mean the way we speak?
Certain retarded ban evaders use the exact same pattern when selling items.
Hey guys its X
I'm selling y
please contact me on (same skype every fuckin time)
although I'm new on the forums, im not stupid!
Welcome to 2011.
No shit sherlock that was 2012
Mods (accusers) contradicting the case lol
Does his IP match my svu real identity?
Prove that he's me bitches.
Theorem 1. Two plus two equals four.
Proof of Theorem 1. 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 1. ∀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. So, it only remains to prove our lemma. 2 + 2 = 4
Proof of Lemma 1. 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 6= 2, because f2 − f1 = f0 and f0 = 0, but f1 = 1, and the Fibonacci
sequence is nondecreasing. 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 > 0, k − 2 = 2 + b
0
. 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
0 = b + 1, we
have a nonnegative solution to k − 2 = 2 + b
0
, as desired.
By Lemma 1, it is not the case that 2 + 2 > 4. Hence 2 + 2 ≤ 4. We also
have 2 + 2 ≥ 4. Therefore,
2 + 2 = 4
Proof of Theorem 1. 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 1. ∀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. So, it only remains to prove our lemma. 2 + 2 = 4
Proof of Lemma 1. 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 6= 2, because f2 − f1 = f0 and f0 = 0, but f1 = 1, and the Fibonacci
sequence is nondecreasing. 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 > 0, k − 2 = 2 + b
0
. 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
0 = b + 1, we
have a nonnegative solution to k − 2 = 2 + b
0
, as desired.
By Lemma 1, it is not the case that 2 + 2 > 4. Hence 2 + 2 ≤ 4. We also
have 2 + 2 ≥ 4. Therefore,
2 + 2 = 4