✔ 最佳答案
有理數 (rational number) is a real number which can be expressed as a ratio of two integers .
So, rational number include
(i) natural number: 1,2,3,.....
It is because 1 can be represented by 1/1, 2 can be represented 2/1
which are ratio of two integers
(ii) 0: It is because 0 can be represented by 0/a where a is a number not equal to 0
(iii) negative number: -1,-2,-3
It is because -1 can be represented by -1/1, 2 can be represented -2/1
which are ratio of two integers
(iv) fraction by definition (fraction is a ratio of two integers)
For example 1/2, 4/5, 100/99 are all rational number
無理數 (irrational number)
If a real number which can not be expressed as a ratio of two integers , then it is a
irrational number
Classic examples include √2, log3 , π
Obviously a number can either rational or irrational, but not both. This means that rational number and irrational number divide the real number into two sets.
Appendix proof that √2 is irrational
Proof
Assume that √2 is a rational number. This would mean that there exist integers a and b such that a / b = √2.
Then √2 can be written as an irreducible fraction (the fraction is shortened as much as possible) a / b such that a and b are coprime integers and (a / b)2 = 2.
It follows that a2 / b2 = 2 and a2 = 2 b2.
Therefore a2 is even because it is equal to 2 b2 which is also even.
It follows that a must be even (odd square numbers have odd square roots and even square numbers have even square roots).
Because a is even, there exists an integer k that fulfills: a = 2k.
We insert the last equation of (3) in (6): (2k)2 = 2b2 is equivalent to 4k2 = 2b2 is equivalent to 2k2 = b2.
Because 2k2 is even it follows that b2 is also even which means that b is even because only even numbers have even squares.
By (5) and (8) a and b are both even, which contradicts that a / b is irreducible as stated in (2).
Since we have found a contradiction, the assumption (1) that √2 is a rational number must be false; that is to say, √2 is irrational