r****f
发帖数: 672 | 1 前面的I63有一个逻辑表述上面的错误，即他说所有素数的乘积加一p1p2...pr+1是素数。
正确表述应该是：
Suppose that p1，p2，.. pr are all of the primes.
Let P = p1p2...pr+1 and let p be a prime dividing P;
then p can not be any of p1, p2, ..., pr, otherwise p would divide the
difference P-p1p2...pr=1, which is impossible.
So this prime p is still another prime, and p1, p2, ..., pr would not be all
of the primes. | f*******i
发帖数: 1049 | 2 Of course this is the original proof. But under the (false) assumption, you
actually did prove P=p1p2...+1 is a prime?
Why? b/c if P is composite, "let p be a prime dividing P;
then p can not be any of p1, p2, ..., pr, otherwise p would divide the
difference P-p1p2...pr=1, which is impossible". And then again p must be one
of pj, since they are the only primes , contradiction hence P is not
composite=> P
is prime.
I know this is redundant and tautological, but you cannot say "product of
all prime p1p2...pr+1 is prime." is a common logical mistake | r****f
发帖数: 672 | 3 你仔细体会一下我贴的描述跟I63的区别
you
one
【在 f*******i 的大作中提到】
: Of course this is the original proof. But under the (false) assumption, you
: actually did prove P=p1p2...+1 is a prime?
: Why? b/c if P is composite, "let p be a prime dividing P;
: then p can not be any of p1, p2, ..., pr, otherwise p would divide the
: difference P-p1p2...pr=1, which is impossible". And then again p must be one
: of pj, since they are the only primes , contradiction hence P is not
: composite=> P
: is prime.
: I know this is redundant and tautological, but you cannot say "product of
: all prime p1p2...pr+1 is prime." is a common logical mistake
| w********a
发帖数: 215 | | l*3
发帖数: 2279 | 5 证明看不懂, 不要瞎掰掰.
前提假设已经说了素数只有p_1,p_2,...,p_k这些, 并且易知其中每一个都不整除N, 且
N大于1, 那么根据素数的定义
"a是素数 <=> a是大于1的自然数, 且a不被任何小于a的素数整除"
就可以推出N是素数.
这里的 "N是素数", 当然是指 "如果假设成立, 那么N是素数"
你不要死盯着你认可的方法, 这天底下正确的方法多得很, 你要说别人错了, 你得说明
确.
我什么时候脱离前提假定, 直接说过 "给k个素数, 他们的连乘积再加一也是素数" ?
数。
all
【在 r****f 的大作中提到】
: 前面的I63有一个逻辑表述上面的错误，即他说所有素数的乘积加一p1p2...pr+1是素数。
: 正确表述应该是：
: Suppose that p1，p2，.. pr are all of the primes.
: Let P = p1p2...pr+1 and let p be a prime dividing P;
: then p can not be any of p1, p2, ..., pr, otherwise p would divide the
: difference P-p1p2...pr=1, which is impossible.
: So this prime p is still another prime, and p1, p2, ..., pr would not be all
: of the primes.
| l*3
发帖数: 2279 | 6 你仔细体会一下以下两句话的区别:
"如果素数只有p_1,p_2,...,p_k这k个, 那么N=p_1*p_2*...*p_k+1是素数"
"给k个素数p_1,p_2,...,p_k, 则N=p_1*p_2*...*p_k+1是素数"
你反驳的到底是哪一句?
数。
all
【在 r****f 的大作中提到】
: 前面的I63有一个逻辑表述上面的错误，即他说所有素数的乘积加一p1p2...pr+1是素数。
: 正确表述应该是：
: Suppose that p1，p2，.. pr are all of the primes.
: Let P = p1p2...pr+1 and let p be a prime dividing P;
: then p can not be any of p1, p2, ..., pr, otherwise p would divide the
: difference P-p1p2...pr=1, which is impossible.
: So this prime p is still another prime, and p1, p2, ..., pr would not be all
: of the primes.
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