
Patrick B. answered 08/09/20
Math and computer tutor/teacher
(a)
x>3 and y>4 are given...
Multiplication property of inequalities says:
xy > 12 > 11.
(b)
for all x in D,
the range is f(x) = x
so the range is D , which is subset of R
(c)
Suppose x is odd AND y is odd
then x = 2m+1 and y=2n+1 for integers m and n
xy = (2m+1)(2n+1) = 4mn + 2m + 2n + 1
= 2(2mn+m+n) +1
by closure property of multiplicaiton and addition
over the integers, K=2mn+m+n is also an integer.
So then xy = 2K+1 for integer K
which is ODD
That is a contradiction of the GIVEN
that xy is even.
(2)
Suppose x is ODD GIVEN that xy is even and y is odd.
Then
x = 2n+1 and y = 2m+1 for integer m and n
xy = (2n+1)(2m+1)
= 4mn + 2n + 2m + 1
= 2(2mn+n+m) + 1
Again, by closure property of
multiplication and addition over
the integers, K=2mn+n+m is also an integer.
So then xy = 2K+1 is odd which is
a CONTRADICTION
(2B) Union of countable sets is countable
(2c) x+y = 3
0 = x^2y
0 = x^2 ( 3-x)
x = 0 or x=3
(0,3) and (3,0)