Set 3.2 #7
Let x=52.46721672167216......
100000x=5246721.67216721......
10x= 
524.67216721......
100000x-10x=5246197
Therefore, x=5246197 / 99990
Set 3.2 #15
True.
Proof: Let r and s be two arbitrary rational numbers , such that r = a/b, and s =
c/d
(Neither b nor d is zero ). Then
r – s
= a/b – c/d
= ad/bd – bc/bd
= (ad – bc) / bd
where bd is not zero, neither. Since (ad-bc) and bd are both integers, r - s is
still a
quotient of integers. Therefore r-s is rational.
Set 3.2 #21
True.
α is any odd integer. Let α=2k+1, where k is any integer.
α ^2 + α
= (2k+1)^2 +2k+1
= 4k^2+4k+1+2k+1
= 4k^2+6k+2
= 2 (2k^2 + 3k +1)
Since (2k^2 + 3k +1) can be any integer, and α ^2 + α is 2 times this integer.
By
definition, α ^2 + α is an even number.
Set 3.2 #29
Since x^2+bx+c = (x-r)(x-s), we have r+s = -b and rs=c, by
algebraic equivalence . r and
s are two solutions . Without losing generality, suppose that r is a rational
solution.
Besides, it is true that s =c/r, where both c and r are rational numbers. By
definition, the
quotient of two rational numbers is also rational.
Set 3.3 #25
False.
For example, a=10, b=5, and c=6.
a | bc, because 10 | 30. But 10 ┼ 5, and 10 ┼ 6.
Set 3.3 #26
True.
For all integers a and b, if a | b, then 
k,
such that ak=b. For a^2 and b^2, there exists k,
such that a^2k^2 = b^2. Therefore, a^2 | b^2.
Set 3.4 #24
Proof: Any two consecutive integers are of one even and
one odd. Without losing
generality, we let the smaller one a=2k, and the bigger one b=2k+1. The case
that a=2k-
1, and b=2k has the same proving steps .
The product of a and b is c =ab=2k*(2k+1)=2*(2k^2+k), where
(2k^2+k) is an integer.
By definition, the number c is an even number.
Set 3.5 #4
The floor of -32/5 is -7, and the ceiling of -32/5 is -6.
Set 3.6 #9
As sume that for all real numbers x and y, if x is
irrational and y is rational, then x-y is
rational. Let y = a/b, and x-y=c/d, where a, b, c, and d are integers (b, d are
non-zero).
Then x can be written as c/d+a/b, which has been proved in problem 3.2 #15 that
is a
rational number. This contradicts to the fact that x is irrational, therefore,
the assumption
does not hold. Hence, x-y is irrational.
Set 3.6 #18
Prove by contraposition.
Suppose both the two numbers of a and b is greater than or equal to 25, namely,
a>=25,
and b>=25. Then a+b >=25+25 =50. Therefore, if the sum of two real number is
less
than 50, then at least one of the number is less than 25.
