Answer:
See explanation below
Step-by-step explanation:
(a) For all positive integers, n is even if and only if 3n^2 + 8 is even.
If n is even, then it can be written as n = 2k for some k,
⇔ 3n²+8 = 3(2k)²+8 ⇔3(4k²) + 8 ⇔12k²+ 8 ⇔ 2 (6k+4) which is even since it has the form 2r.
(b) If a and b are rational numbers, then a^2+ b^2> 2ab
We know that all squares are positive. Even the square of a negative number is positive ((-2)²= 4).
Therefore, we can say that (a-b)²≥0
⇒a²-2ab+b² ≥ 0
⇒a² + b² ≥ 2ab
(c) If n is an even integer, then n + 1 is odd.
If n is an even integer, then it can be written as n = 2k
Then, n + 1 would be (2k) +1 which clearly is odd.
d) Every odd number is the difference of two perfect squares.
Let n = 2k +1 for some k. Now take k² and (k+1)².
The difference of these two numbers is
(k+1)²- k² = k² +2k +1 - k² = 2k + 1, which is odd.
Therefore, every odd number is the difference of two perfect squares.
