1. (a) for all x in the set of integers, (x² x) implies,
x² - x 0 or, x(x -1) 0,
or, (x 0 & x 1) or (x 0 & x 1)
or, (x 1) or (x 1) and for x = 0, we have trivially, 0² 0
So, we have covered all of the set of integers by the above statement.
So, the statement is true universally in the mentioned domain.
(b) by the statement, for all x, (x > 0 v x < 0) implies x is a non-zero integer.
But, the domain is the set of all integers, which include 0 as well.
So, the element 0 serves as a counter-example.
(c) for all x (x = 1) does not hold in the set of integers for any element other than 1.
So, for example, the integer 2 serves as a counter-example.
2. Let, a and b be two arbitrary odd numbers.
Then, a = 2s + 1 and b = 2t + 1 for integers s & t.
So, a² - b² = (2s+1)² - (2t+1)² = 4s² + 4s + 1 - 4t² - 4t - 1 = 4(s²+t²+s+t)
So, a² - b² is a multiple of 4, hence, always divisible by 4.
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