[急]other數學問題

Q4a)Suppose that √3 is a rational number.
So, we have √3 = p/q , where p and q are relatively prime integers.
Hence, we have 3 = p²/q²
3q² = p²
Then, p² is divisible by 3 and hence p is divisible by 3.
Therefore, p=3r for some integer r.
Hence, we have 3q² = (3r)²
q² = 3r²
So, q² is divisible by 3 and hence q is divisible by 3.
It leads to a contradiction since p and q are relatively prime.
Thus, we have √3 is an irrational number.b)Suppose that the difference between a rational m and an irrational n is a rational k.
m - n = k
n = m - k leads to a contradiction since the difference between two rational numbers= m - k is rational. c)By a) , √3 is an irrational number.
By b) , √3 - 1 is an irrational number.
But √3 - (√3 - 1) = 1 is a rational number.Q5a)(1)i) a1 ≥ 1 since the boxer has to fight at least one match on day 1.
ii) a31 ≤ 46 since no more than a total of 46 matches within a period of 31 days.
iii) a1 < a2 < ⋯ < a30 < a31 since aj - ai ≥ j - i ≥ 1 for at least one match a day.(2)Consider two number list:
List 1 : a1 < a2 < a3 < ... < a30 < a31
List 2 : a1 + 15 < a2 + 15 < a3 + 15 < ... < a30 + 15 < a31 + 15There are total 62 numbers in two lists above.
The smallest number = a1 ≥ 1 ,
the greatest number = a31 + 15 ≤ 46 + 15 = 61.
By Pigeon Hole Principle, List 1 exist a number aj = ai + 15 at List 2.
Then aj - ai = 15 , so there is a period of consecutive days during day i+1 to
day j which the boxer has to fight exactly 15 matches.b)Consider there are 0 to 10 total 11 remainders of the 12 numbers when divided by 11.
By Pigeon Hole Principle, it must exist two of them have the same remainders
when divided by 11, then their difference is divisible by 11.