一道国外的大1数学题,帮帮忙谢谢
2个回答
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(a) the product (called P) has 2 as one of it's factors, so P is also even according to definition.
(b)
Not sure what's asked. Assume it asked to prove 3 is always a factor of n(2n+1)(4n+1):
(i) n = 3k (n, k are integers)
It's clear, 3 is a factor of n(2n + 1)(4n + 1)
(ii) n = 3k + 1
2n + 1 = 6k + 3 = 3(2k + 1), 3 is a factor
(iii) n = 3k + 2
4n + 1 = 12k + 9 = 3(4k + 3), 3 is a factor
(3)
Suppose √3 is a rational number, then it can be expressed in the form of a/b, here a and b are both integers and the only common factor is 1.
√3 = a/b
3 = a²/b²
a² = 3b² (1)
3 is a factor of 3b², so a² must has 3 as a foctor, let a = 3c
(3c)² = 3b²
b² = 3c²
Follow the same reasoning, b has 3 as a factor; so a and b have 1 and 3 as common factors. This contradicts the initial assumption, so √3 is not a rational number.
(b)
Not sure what's asked. Assume it asked to prove 3 is always a factor of n(2n+1)(4n+1):
(i) n = 3k (n, k are integers)
It's clear, 3 is a factor of n(2n + 1)(4n + 1)
(ii) n = 3k + 1
2n + 1 = 6k + 3 = 3(2k + 1), 3 is a factor
(iii) n = 3k + 2
4n + 1 = 12k + 9 = 3(4k + 3), 3 is a factor
(3)
Suppose √3 is a rational number, then it can be expressed in the form of a/b, here a and b are both integers and the only common factor is 1.
√3 = a/b
3 = a²/b²
a² = 3b² (1)
3 is a factor of 3b², so a² must has 3 as a foctor, let a = 3c
(3c)² = 3b²
b² = 3c²
Follow the same reasoning, b has 3 as a factor; so a and b have 1 and 3 as common factors. This contradicts the initial assumption, so √3 is not a rational number.
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