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25 April, 18:54

5. 30pts] Prove or disprove each of the following claims: (a) For all positive integers, n is even if and only if 3n^2 + 8 is even (b) If a and b are rational numbers, then a^2 + b^2> 2ab (c) If n iss an even integer, then n + 1 is odd. (d) Every odd number is the difference of two perfect squares.

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  1. 25 April, 21:37
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    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.
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