goldbach's conjecture is that every nonzero even number greater than 2 can be written as the sum of 2 primes. no one has ever proved that this is always true, but no one has found a conterexample, either.
1. show that goldbach's conjecture is true for the first 5 even numbers greater than 2.
2. give a way that someone could disprove goldbach's conjecture.
2007-05-12 05:20:22 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
1) [included a few more than 5 examples ... notice that some can be done in more than one way...]
4 = 2 + 2
6 = 3 + 3
8 = 3 + 5
10 = 3 + 7 ................... = 5 + 5
12 = 5 + 7
14 = 3 + 11...................... = 7 + 7
16 = 3 + 13 .........................= 5 +11
18 = 5 + 13 .............................= 7 + 11
20 = 3 + 17 .............................= 7 + 13
et al ...
2) .. to DIS prove the conjecture find an even number that is NOT the sum of any two prime number smaller than itself.
good luck :)
ps: "1" is not a prime number
2007-05-12 05:22:49 · answer #1 · answered by atheistforthebirthofjesus 6 · 0⤊ 0⤋
4 = 2+2
6 = 3+3
8 = 3+5
10 = 3+7 = 5+5
12 = 5+7.
To disprove Goldbach's conjecture one would
have to find an even number that cannot
be written as the sum of 2 primes.
Good luck finding one, since the conjecture
has been verified for all even numbers
less than 10^ 18.
What would really help in proving the
conjecture is to find a number
beyond which the conjecture is true.
This has been done for the "weak Goldbach
conjecture", which states
Every odd number greater than 5 can
be written as the sum of 3 primes.
This is known to be true for n > e ^ 3100.
There was a recent article in MATHEMATICS
OF COMPUTATION on these conjectures.
2007-05-12 06:06:05 · answer #2 · answered by steiner1745 7 · 0⤊ 0⤋
(1)
4 = 1 + 3
6 = 1 + 5
8 = 3 + 5
10 = 3 + 7
12 = 5 + 7
(2)
To disprove Goldbach's conjecture, you would just have to find one even number greater than 2 that could not be written as the sum of two primes.
2007-05-12 05:27:16 · answer #3 · answered by rrabbit 4 · 0⤊ 1⤋
1. 1+3=4
1+5=6
3+5=8
5+5= 10
5+7=12
2. cant help ya here...
2007-05-12 05:25:33 · answer #4 · answered by txmama423 3 · 0⤊ 1⤋
You've got good answers. I just want to point out that non zero and greater than 2 is a redundance.
2007-05-12 05:45:25 · answer #5 · answered by Steiner 7 · 0⤊ 0⤋
If we admit that the park has a rectangular shape, then its area is 17 * 11. As the area of the park is a product of more that two factors (11,17,1 and itself), it's a composite number.
2016-05-21 03:12:08 · answer #6 · answered by ? 3 · 0⤊ 0⤋